Show that a ball dropped from a height h feet reaches the floor in 14h−−√ seconds. Then use this result to find the time, in seconds, the ball has been bouncing when it hits the floor for the first, second, third and fourth times:

Answer:

(a) 0.48006

(b) 0.3251

(c) 0.19489

(d) 97.5​% of people consumed less than 56 gallons of bottled​ water.

Step-by-step explanation:

We are given that the per capita consumption of bottled water is approximately normally distributed with a mean of 34.5 and a standard deviation of 11 gallons.

Let X = per capita consumption of bottled water

So, X ~ N([tex]\mu = 34.5,\sigma^{2} =11^{2}[/tex])

The z score probability distribution is given by;

            Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)

(a) Probability that someone consumed more than 35 gallons of bottled​ water = P(X > 35)

   P(X > 35) = P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{35-34.5}{11}[/tex] ) = P(Z > 0.05) = 1 - P(Z [tex]\leq[/tex] 0.05)

                                                   = 1 - 0.51994 = 0.48006

(b) Probability that someone consumed between 25 and 35 gallons of bottled​ water = P(25 < X < 35)

   P(25 < X < 35) = P(X < 35) - P(X [tex]\leq[/tex] 25)

   P(X < 35) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{35-34.5}{11}[/tex] ) = P(Z < 0.05) = 0.51994

   P(X [tex]\leq[/tex] 25) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{25-34.5}{11}[/tex] ) = P(Z [tex]\leq[/tex] -0.86) = 1 - P(Z < 0.86)

                                                   = 1 - 0.80511 = 0.19489

Therefore, P(25 < X < 35) = 0.51994 - 0.19489 = 0.3251

(c) Probability that someone consumed less than 25 gallons of bottled​ water = P(X < 25)

     P(X < 25) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{25-34.5}{11}[/tex] ) = P(Z < -0.86) = 1 - P(Z [tex]\leq[/tex] 0.86)

                                                   = 1 - 0.80511 = 0.19489

(d) We have to find that 97.5​% of people consumed less than how many gallons of bottled​ water, which means ;

    P(X < x) = 0.975

    P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{x-34.5}{11}[/tex] ) = 0.975

    P(Z > [tex]\frac{x-34.5}{11}[/tex] ) = 0.975

Now in the z table the critical value of X which have an area less than 0.975 is 1.96, i.e.;

           [tex]\frac{x-34.5}{11}[/tex] = 1.96

        [tex]x[/tex] - 34.5 = [tex]1.96 \times 11[/tex]

                  [tex]x[/tex]  = 34.5 + 21.56 = 56.06 ≈ 56 gallons of bottled water

So, 97.5​% of people consumed less than 56 gallons of bottled​ water.

a. The probability that someone consumed more than 35 gallons of bottled water is approximately 52.12%. b. The probability that someone consumed between 25 and 35 gallons of bottled water is approximately 9.58%. c. The probability that someone consumed less than 25 gallons of bottled water is approximately 42.51%. d. 97.5% of people consumed less than approximately 57.56 gallons of bottled water.

a. To find the probability that someone consumed more than 35 gallons of bottled water, we need to standardize the value of 35 using the formula z = (x - mean) / standard deviation, where x is the value we want to find the probability for. In this case, x = 35. Plugging in the values, we get z = (35 - 34.5) / 11 = 0.045. Using a standard normal distribution table, the probability (area under the curve) to the right of 0.045 is approximately 0.5212. Therefore, the probability that someone consumed more than 35 gallons of bottled water is 0.5212 or 52.12%.

b. To find the probability that someone consumed between 25 and 35 gallons of bottled water, we need to find the probability to the right of 25 (P(x > 25)) and subtract the probability to the right of 35 (P(x > 35)). Following the same process as in part (a), we find that P(x > 25) ≈ 0.5746 and P(x > 35) ≈ 0.4788. Therefore, the probability that someone consumed between 25 and 35 gallons of bottled water is approximately 0.5746 - 0.4788 = 0.0958 or 9.58%.

c. To find the probability that someone consumed less than 25 gallons of bottled water, we need to find the probability to the left of 25 (P(x < 25)). Using the standard normal distribution table, we find that P(x < 25) ≈ 0.4251. Therefore, the probability that someone consumed less than 25 gallons of bottled water is approximately 0.4251 or 42.51%.

d. To find the value at which 97.5% of people consumed less than, we need to find the z-score that corresponds to the 97.5th percentile of the standard normal distribution. Using the standard normal distribution table, we find that the z-score corresponding to the 97.5th percentile is approximately 1.96. Now we can use the formula z = (x - mean) / standard deviation to solve for x. Plugging in the values, we get 1.96 = (x - 34.5) / 11. Solving for x, we find x ≈ 1.96 * 11 + 34.5 ≈ 57.56. Therefore, 97.5% of people consumed less than approximately 57.56 gallons of bottled water.

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Answer:

c) NORM.INV(R, 1200, 350)

Step-by-step explanation:

Given that the daily demand for gasoline at a local gas station is normally distributed with a mean of 1200 gallons, and a standard deviation of 350 gallons.

X = demand for gasolene at a local gas station is N(1200, 350)

R is any random number between 0 and 1.

Daily demand for gasolene would be

X = Mean + std deviation * z value, where Z = normal inverse of a value between 0 and 1.

The norm inv (R, 1200, 350) for R between 0 and 1 gives all the values of X

Hence correct choice would be

Option c) NORM.INV(R, 1200, 350)

Answer:

Step-by-step explanation:

number of pennies = 5

number of dimes = 26

number of nickels = 18

number of quarters = 12

Total = 5 + 26 + 18 + 12 = 61

Probability to take out a quarter = 12 / 61 = 0.1967

Probability to take out a penny without replacement = 5 / 60 = 0.0833

Total probability to take out a quarter and then penny without replacement                                    

                               = 0.1967 x 0.0833 = 0.0164

The probability of pulling out a quarter and then a penny from the jar without replacement is 0.0131, calculated by multiplying the individual probabilities of each event.

To calculate the probability of pulling out a quarter and then a penny from the jar without replacement, we need to consider the total number of coins and the number of each kind of coin. Initially, the jar contains 5 pennies, 26 dimes, 18 nickels, and 12 quarters, for a total of 61 coins.

First, let's find the probability of picking a quarter. There are 12 quarters out of 61 coins, so the probability is 12/61. After picking a quarter, we do not replace it; thus, there are now 60 coins left in the jar, including only 4 pennies. Now, the probability of picking a penny is 4/60 or 1/15.

We find the combined probability of these two events occurring in sequence by multiplying their probabilities: (12/61) * (1/15).

Probability of picking a quarter then a penny without replacement = (12/61)  * (1/15)
= 12/(61 * 15)
= 12/915
= 0.0131 (when rounded to four decimal places)

Answer:

[tex]\text{Average diameter is 1.65 mm and we decide that it is not 1.65 mm.}[/tex]                

Step-by-step explanation:

We are given the following in the question:

The needle size should not be too big and too small.

The diameter of the needle should be 1.65 mm.

We design the null and the alternate hypothesis

[tex]H_{0}: \mu = 1.65\text{ mm}\\H_A: \mu \neq 1.65\text{ mm}[/tex]

Sample size, n = 35

Sample mean, [tex]\bar{x}[/tex] = 1.64 mm

Sample standard deviation, s = 0.07 mm

Type I error:

Thus, type I error in this study would mean we reject the null hypothesis that the average diameter is 1.65 mm but actually the average diameters of the needle is 1.65 mm.

Thus, average diameter is 1.65 mm and we decide that it is not 1.65 mm.

Answer:

Type I error will be Rejecting the null hypothesis that the average diameter of needles is 1.65 mm and assume that the average diameter of needles is different from 1.65 mm but the fact is that the null hypothesis was true that the average diameter of needles is 1.65 mm.

Step-by-step explanation:

We are given that a manufacturing process is producing hypodermic needles that will be used for blood donations. These needles need to have a diameter of 1.65 mm—too big and they would hurt the donor (even more than usual), too small and they would rupture the red blood cells, rendering the donated blood useless.

So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 1.65 mm

Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu\neq[/tex] 1.65 mm  

Also, the most recent random sample of 35 needles have an average diameter of 1.64 mm and a standard deviation of 0.07 mm.

Now, Type I error Type I error states that : Probability of rejecting null hypothesis given the fact that null hypothesis was true. It is the the probability of rejecting a true hypothesis.

So, in our question Type I error will be Rejecting the null hypothesis that the average diameter of needles is 1.65 mm and assume that the average diameter of needles is different from 1.65 mm but the fact is that the null hypothesis was true that the average diameter of needles is 1.65 mm.

(a) Rewrite y(t) = 4sin(2πt) + 15cos(2πt) using a sine term only: y(t) = √241sin(2πt - 1.249)

(b) Amplitude:  √241, Period: 0.5s, Frequency: 2Hz.

(c) Graph shows two sine waves, one with amplitude √241 and another with 15, out of phase, repeating every 0.5s.

(a) Expressing the function in terms of a sine term only:

To express the function y(t) = 4sin(2πt) + 15cos(2πt) in terms of a sine term only, we can use trigonometric identities to rewrite the cosine term in terms of sine:

cos(2πt) = sin(π/2 - 2πt)

Now, we can rewrite the function as follows:

y(t) = 4sin(2πt) + 15sin(π/2 - 2πt)

(b) Determining the amplitude, period, and frequency:

Amplitude (A): The amplitude of a sinusoidal function is the coefficient of the sine term. In this case, the amplitude is 4.

Period (T): The period of a sinusoidal function is the time it takes for one complete cycle. The period can be found using the formula T = 1/f, where f is the frequency. In this case, the frequency is 2, as we'll see in part (b). So, T = 1/2 = 0.5 seconds.

Frequency (f): The frequency of a sinusoidal function is the number of cycles per second (in hertz, Hz). In this case, the frequency is 2 Hz.

(c) Drawing the function in the time domain:

To draw the function y(t) = 4sin(2πt) + 15sin(π/2 - 2πt), you can follow these steps:

1. Create a set of axes with time (t) on the horizontal axis and y(t) on the vertical axis.

2. The amplitude of the first sine term is 4, so the first term oscillates between -4 and 4.

3. The second sine term has an amplitude of 15, but it is shifted in phase by π/2. This means it starts at its maximum value of 15 and goes down to -15.

4. The total function y(t) is the sum of these two sine terms. You'll see a combination of two sinusoidal waves, one with a smaller amplitude (4) and one with a larger amplitude (15), and they are out of phase with each other.

The period of the function is 0.5 seconds (as calculated in part (b)), so you'll see this pattern repeat every 0.5 seconds.

To learn more about Amplitude here

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Answer:

(a) 3.432

(b) 3.175

Step-by-step explanation:

The given data and corresponding weights is:

                                                  [tex]\begin{array}{cc}x_i&w_i\\3.2&6\\3.0&3\\2.5&2\\4.0&8\\\end{array}[/tex]

(a) The weighted mean is determined by the following expression:

[tex]M_W=\sum\frac{x_i*w_i}{w_i}\\M_W=\frac{3.2*6+3.0*3+2.5*2+4.0*8}{6+3+2+8}\\M_W=3.432[/tex]

(b) The simple mean of the given data is determined by adding up the four values dividing the result by 4:

[tex]M_S = \frac{3.2+3.0+2.5+4.0}{4}\\ M_S=3.175[/tex]

The value is lower than the weighted mean.

The weighted mean is 3.455, while the sample mean without weighting is 3.175.

To compute the weighted mean, we multiply each data value by its corresponding weight, then add up the results. We then divide this sum by the sum of the weights. In this case, the weighted mean can be calculated as follows:

Weighted Mean = (3.2 * 6 + 3.0 * 3 + 2.5 * 2 + 4.0 * 8) / (6 + 3 + 2 + 8) = 3.455

To compute the sample mean without weighting, we simply add up all the data values and divide the sum by the number of data values. In this case, the sample mean can be calculated as follows:

Sample Mean = (3.2 + 3.0 + 2.5 + 4.0) / 4 = 3.175

The difference between the two computations is that the weighted mean takes into account the importance of each data value by assigning weights to them, whereas the sample mean without weighting treats all data values equally.

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Answer:

(1) The probability of the event X = 2 is 0.0413.

(2) The probability of the event X ≤ 1 is 0.009.

(3) The probability of the event X < 6 is 0.6846.

Step-by-step explanation:

Let the random variable X follow a Binomial distribution with parameter n = 8 and p = 0.60.

The probability mass function of the Binomial distribution is:

[tex]P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0, 1, 2,...[/tex]

(1)

Compute the probability of the event X = 2 as follows:

[tex]P(X=2)={8\choose 2}(0.60)^{2}(1-0.60)^{8-2}\\=\frac{8!}{2!(8-2)!}\times (0.60)^{2}\times(0.40)^{6}\\=28\times0.36\times0.004096\\=0.0413[/tex]

Thus, the probability of the event X = 2 is 0.0413.

(2)

Compute the probability of the event X ≤ 1 as follows:

P (X ≤ 1) = P (X = 0) + P (X = 1)

[tex]={8\choose 0}(0.60)^{0}(1-0.60)^{8-0}+{8\choose 1}(0.60)^{1}(1-0.60)^{8-1}\\=\frac{8!}{0!(8-0)!}\times (0.60)^{0}\times(0.40)^{8}+\frac{8!}{1!(8-1)!}\times (0.60)^{1}\times(0.40)^{7}\\=(1\times1\times0.00066)+(8\times0.60\times0.00164)\\=0.008532\approx0.009[/tex]

Thus, the probability of the event X ≤ 1 is 0.009.

(3)

Compute the probability of the event X < 6 as follows:

P (X < 6) = 1 - P (X ≥ 6)

              = 1 - P (X = 6) - P (X = 7) - P (X = 8)

    [tex]=1-{8\choose 6}(0.60)^{6}(1-0.60)^{8-6}+{8\choose 7}(0.60)^{7}(1-0.60)^{8-7}+{8\choose 8}(0.60)^{8}(1-0.60)^{8-8}\\=1-\frac{8!}{6!(8-6)!}\times (0.60)^{6}\times(0.40)^{2}+\frac{8!}{7!(8-7)!}\times (0.60)^{7}\times(0.40)^{1}\\+\frac{8!}{8!(8-8)!}\times (0.60)^{8}\times(0.40)^{0}\\=1-0.2090-0.0896-0.0168\\=0.6846[/tex]

Thus, the probability of the event X < 6 is 0.6846.

The probabilities for a binomial distribution with p=0.6 and n=8 for different scenarios such as P(x=2), P(x≤1), and P(x>6), which involves combinations and powers of success and failure probabilities is equal to 0.2799, 0.0888, and 0.0332 respectively.

1) Upper P(x = 2):

To find P(x = 2), we can use the formula for the binomial probability:

P(x = 2) = (8 choose 2) × (0.6)² × (0.4)⁶= 0.2799

2) Upper P(x ≤ 1):

P(x ≤ 1) = P(x = 0) + P(x = 1) = (8 choose 0)×  (0.6)⁰ ×(0.4)⁸ + (8 choose 1) ×(0.6)¹ ×(0.4)⁷ = 0.0888

3) Upper P(x > 6):

P(x > 6) = 1 - P(x ≤ 6) = 1 - (P(x = 0) + P(x = 1) + ... + P(x = 6)) = 1 - 0.9668 = 0.0332

Answer: The probability of drawing a red marble the sixth time is 1/2

Step-by-step explanation:

Here is the complete question:

A box contains 10 red marbles and 10 green marbles. Sampling at random from the box five times with replacement, you have drawn a red marble all five times. What is the probability of drawing a red marble the sixth time?

Explanation:

Since the sampling at random from the box containing the marbles is with replacement, that is, after picking a marble, it is replaced before picking another one, the probability of picking a red marble is the same for each sampling. Probability, P(A) is given by the ratio of the number of favourable outcome to the total number of favourable outcome.

From the question,

Number of favourable outcome = number of red marbles =10

Total number of favourable outcome = total number of marbles = 10+10= 20

Hence, probability of drawing a red marble P(R) = 10 ÷ 20

P(R) = 1/2

Since the probability of picking a red marble is the same for each sampling, the probability of picking a red marble the sixth time is 1/2

The probability is [tex]\frac{1}{2}[/tex]

First, we have to calculate the probability of drawing a red marble on any given try using the formula of probability:

Probability of an event = [tex]\frac{Number\ of\ favorable\ outcomes}{Total\ number\ of\ outcomes}[/tex]

P(red) = [tex]\frac{Number\ of\ red\ marbles}{Total\ number\ of\ marbles} =\frac{10}{20} = \frac{1}{2}[/tex]

Since we are replacing after every time we pick up a marble, each event of picking a marble is an independent event

Thus, every time a marble is to be picked, the probabilities remain same.

Hence the probability of drawing a red marble the sixth time is = [tex]P(red) = \frac{1}{2}[/tex]

The complete question is:
A box contains 10 red marbles and 10 green marbles. Sampling at random from the box five times with replacement, you have drawn a red marble all five times. What is the probability of drawing a red marble the sixth time?

Answer:

a) [tex]P(X=5)=(9C5)(0.47)^5 (1-0.47)^{9-5}=0.228[/tex]

b) [tex]P(X=0)=(9C0)(0.47)^0 (1-0.47)^{9-0}=0.0033[/tex]

And replacing we got:

[tex]P(X \geq 1)= 1-P(X<1) = 1-P(X=0)=1-0.0033= 0.9967[/tex]

c) [tex]P(X=4)=(9C4)(0.47)^4 (1-0.47)^{9-4}=0.257[/tex]

[tex]P(X=5)=(9C5)(0.47)^5 (1-0.47)^{9-5}=0.228[/tex]

[tex]P(X=6)=(9C6)(0.47)^6 (1-0.47)^{9-6}=0.135[/tex]

And adding we got:

[tex] P(4 \leq X \leq 6)= 0.257+0.228+0.135=0.620 [/tex]

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

[tex]X \sim Binom(n=9, p=0.47)[/tex]

The probability mass function for the Binomial distribution is given as:

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]

Where (nCx) means combinatory and it's given by this formula:

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]

Assuming the following questions:

a. exactly five

For this case we can use the probability mass function and we got:

[tex]P(X=5)=(9C5)(0.47)^5 (1-0.47)^{9-5}=0.228[/tex]

b. at least one

For this case we want this probability:

[tex] P(X \geq 1)[/tex]

And we can use the complement rule and we got:

[tex]P(X \geq 1)= 1-P(X<1) = 1-P(X=0)[/tex]

[tex]P(X=0)=(9C0)(0.47)^0 (1-0.47)^{9-0}=0.0033[/tex]

And replacing we got:

[tex]P(X \geq 1)= 1-P(X<1) = 1-P(X=0)=1-0.0033= 0.9967[/tex]

c. between four and six, inclusive.

For this case we want this probability:

[tex] P(4 \leq X \leq 6)[/tex]

[tex]P(X=4)=(9C4)(0.47)^4 (1-0.47)^{9-4}=0.257[/tex]

[tex]P(X=5)=(9C5)(0.47)^5 (1-0.47)^{9-5}=0.228[/tex]

[tex]P(X=6)=(9C6)(0.47)^6 (1-0.47)^{9-6}=0.135[/tex]

And adding we got:

[tex] P(4 \leq X \leq 6)= 0.257+0.228+0.135=0.620 [/tex]

It seems like your question was cut off at the end. You did not specify the exact number of adults out of the 9 randomly selected ones that have experienced a breakup. However, I can help you understand how to approach this kind of problem generally using the binomial probability formula.
When we randomly select adults and we are interested in those who have experienced a breakup, with a probability of 47% (or 0.47), this can be modeled as a binomial distribution because:
1. There are a fixed number of trials (n = 9).
2. Each trial has only two possible outcomes (experienced a breakup or did not experience a breakup).
3. The probability of experiencing a breakup is the same for each trial (p = 0.47).
4. Each trial is independent of the others.
The probability of exactly x adults (out of 9) having experienced a breakup can be calculated using the binomial probability formula:
P(X = x) = (n choose x) * p^x * (1 - p)^(n - x)
Where:
- "n choose x" is the binomial coefficient C(n, x) = n! / [x!(n - x)!],
- p is the probability of success on any given trial (in this case, 0.47),
- x is the number of successes out of n trials,
- n! denotes the factorial of n,
- x! denotes the factorial of x.
Because you didn't specify the value of x in your question, I can't give you the exact probability. However, if you give me a specific number of adults who have experienced a breakup (x), I can provide the calculation for that scenario.
For now, if you'd like to calculate this probability for a certain number of adults x, you would plug your values for x and n into the formula and calculate accordingly.
Let's say you wanted to find the probability that exactly 4 out of 9 adults have experienced a breakup, for example. You would calculate it as follows:
P(X = 4) = C(9, 4) * (0.47)^4 * (0.53)^(9 - 4)
         = 126 * (0.47)^4 * (0.53)^5
C(9, 4) is the number of combinations of 9 things taken 4 at a time, which is 9! / (4!(9 - 4)!) = 126.
You would then calculate (0.47)^4, (0.53)^5, and multiply these by 126 to get the probability.
For other values of x, you would undergo a similar process, using the appropriate value of x in the formula.

The function that determines the rock's height above the road in terms of the number of seconds since it was thrown is f(t) = -16t² + 24.32t + 57. This quadratic function accounts for the initial upward velocity and the constant downward acceleration due to gravity.

To determine the rock's height above the road as a function of time after it was thrown, we can use the standard equation of motion under constant acceleration due to gravity, which in this case is downward. The formula for the height f(t) of the rock at time t seconds after it is thrown is:

f(t) = -16t² + vt + s

where:

Since the rock reaches its maximum height 0.76 seconds after it is thrown, the initial velocity v can be calculated using the fact that the velocity at the peak height is 0 (the rock stops moving upward for an instant before descending).

At maximum height, the velocity v had decreased by 32 feet/second for 0.76 seconds, which is v - 32(0.76) = 0. Solving for v, we find that v = 32(0.76).

The rock contacts the road 3.15 seconds after it was thrown, meaning it falls 57 feet in that time. Plugging these values into the formula:

f(t) = -16t² + 32(0.76)t + 57

which simplifies to:

f(t) = -16t² + 24.32t + 57

This function f(t) gives the rock's height in feet above the road at any time t measured in seconds since it was thrown, assuming negligible air resistance.

The correct function for the rock's height above the road in terms of the number of seconds [tex]\( t \)[/tex] since the rock was thrown is: [tex]\[ f(t) = -16t^2 + v_0t + 57 \][/tex] where [tex]\( v_0 \)[/tex] is the initial velocity of the rock in feet per second.

To derive this function, we use the kinematic equation for the vertical motion of an object under constant acceleration due to gravity, which is:

[tex]\[ h(t) = h_0 + v_0t - \frac{1}{2}gt^2 \][/tex]

Given that the bridge is 57 feet above the road, [tex]\( h_0 = 57 \)[/tex] feet. The acceleration due to gravity is [tex]\( g = 32 \)[/tex] feet per second squared. Plugging these values into the kinematic equation, we get:

[tex]\[ h(t) = 57 + v_0t - \frac{1}{2}(32)t^2 \][/tex]

Simplifying the equation by multiplying [tex]\( \frac{1}{2} \)[/tex] with 32 gives us:

[tex]\[ h(t) = 57 + v_0t - 16t^2 \][/tex]

This simplifies to the function [tex]\( f(t) \)[/tex] as provided above. Note that [tex]\( v_0 \)[/tex] is still unknown and will need to be determined using additional information given in the problem, such as the time it takes for the rock to reach its maximum height.

To find [tex]\( v_0 \)[/tex], we can use the fact that at the maximum height, the velocity of the rock is zero. The kinematic equation relating velocity and time under constant acceleration is:

[tex]\[ v(t) = v_0 - gt \][/tex]

Setting [tex]\( v(t) = 0 \)[/tex] at the maximum height and solving for [tex]\( v_0 \)[/tex] gives us:

[tex]\[ 0 = v_0 - gt_{\text{max}} \][/tex]

[tex]\[ v_0 = gt_{\text{max}} \][/tex]

Given that the rock reaches its maximum height 0.76 seconds after it is thrown, [tex]\( t_{\text{max}} = 0.76 \)[/tex] seconds. Plugging this into the equation for [tex]\( v_0 \)[/tex], we get:

[tex]\[ v_0 = 32 \times 0.76 \][/tex]

[tex]\[ v_0 = 24.32 \][/tex]

Now we can write the complete function with the known values:

[tex]\[ f(t) = -16t^2 + 24.32t + 57 \][/tex]

This function represents the height of the rock above the road as a function of time since it was thrown.

Answer:

a) B(t)= 64.8e^(-0.01255t), k=-0.0125 b) 56.5 c)2150

Step-by-step explanation:

a) B(t)= Ae^(-kt)

at t=0, B(t)=64.8

A=64.8

at t=6, B(t)=60.1

60.1=64.8e^(-6k)

k=0.0125

b) B(t)=64.8e^(-0.0125×11)

B(t)= 56.5

c) 10=64.8e^(-0.0125t)

0.15432=e^(-0.0125t)

-0.0125t=ln(0.15432)

-0.0125t=-1.869

t=149.5 or 150

Year= 2150

a)k ≈ 0.01267 and the equation is [tex]B(t) = 64.8e^{-0.01267t[/tex]

b)B(11) ≈ 56.38 lb

c) t ≈ 143.86

A beef consumption model follows exponential decay where B(t), the annual beef consumption t years after 2000,

can be modeled by the equation: [tex]B(t) = B_0e^{-kt}.[/tex]

Given that [tex]B_0[/tex] = 64.8 lb in 2000 and B(6) = 60.1 lb in 2006,

we first need to find the decay constant k.

a) Find the value of k, and write the equation:

We use the given data points to solve for k.

[tex]B(6) = 64.8e^{-6k} = 60.1[/tex]

[tex]e^{-6k} = 60.1 / 64.8[/tex]
[tex]e^{-6k}[/tex] ≈ 0.9272
Taking the natural log of both sides:
-6k = ln(0.9272)
k ≈ -ln(0.9272) / 6
k ≈ 0.01267

Therefore, the equation for B(t) is:

[tex]B(t) = 64.8e^{-0.01267t[/tex]

b) Estimate the consumption of beef in 2011:

For t = 11 (since 2011 is 11 years after 2000):

[tex]B(11) = 64.8e^{-0.01267 * 11[/tex]
B(11) ≈ 64.8[tex]e^{-0.13937[/tex]
B(11) ≈ [tex]64.8 * 0.8699[/tex]
B(11) ≈ 56.38 lb

c) In what year (theoretically) will the consumption of beef be 10 lb?

We need to solve for t when B(t) = 10 lb:

[tex]10 = 64.8e^{-0.01267t[/tex]
[tex]e^{-0.01267t} = 10 / 64.8[/tex]
[tex]e^{-0.01267t[/tex] ≈ 0.1543
Taking the natural log of both sides:
-0.01267t = ln(0.1543)
t ≈ -ln(0.1543) / 0.01267
t ≈ 143.86

Thus, the theoretical year is approximately 2000 + 144 = 2144.

Answer:

a) The minimum head breadth that will fit the clientele = 4.105 inches to 3d.p = 4.1 inches to 1 d.p

b) The maximum head breadth that will fit the clientele = 8.905 inches to 3 d.p = 8.9 inches to 1 d.p

Step-by-step explanation:

This is normal distribution problem.

A normal distribution has all the data points symmetrically distributed around the mean in a bell shape.

For this question, mean = xbar = 6.1 inches

Standard deviation = σ = 1 inch

And we want to find the lowermost 2.3% and uppermost 2.3% of the data distribution.

The minimum head breadth that will fit the clientele has a z-score with probability of 2.3% = 0.023

Let that z-score be z'

That is, P(z ≤ z') = 0.023

Using the table to obtain the value of z'

z' = - 1.995

P(z ≤ - 1.995) = 0.023

But z-score is for any value, x, is that value minus the mean then divided by the standard deviation.

z' = (x - xbar)/σ

- 1.995 = (x - 6.1)/1

x = -1.995 + 6.1 = 4.105 inches

The maximum head breadth that will fit the clientele has a z-score with probability of 2.3% also = 0.023

Let that z-score be z''

That is, P(z ≥ z'') = 0.023

Using the table to obtain the value of z''

P(z ≥ z") = P(z ≤ -z")

- z'' = - 1.995

z" = 1.995

P(z ≥ 1.995) = 0.023

But z-score is for any value, x, is that value minus the mean then divided by the standard deviation.

z'' = (x - xbar)/σ

1.995 = (x - 6.1)/1

x = 1.995 + 6.1 = 8.905 inches

Using z-scores for the smallest and largest 2.3% of the normal distribution, and a formula that accounts for the mean and standard deviation of head breadths, we can calculate that the helmets need to accommodate head breadths between 4.1 inches (minimum) and 8.1 inches (maximum).

To determine the minimum and maximum head breadths that will fit the clientele, we need to find the z-scores that correspond to the smallest 2.3% and the largest 2.3% of the normal distribution. Then we can use these z-scores to calculate the specific head breadths.

First, we find the z-scores using a standard z-table or a calculator since the normal distribution table typically provides the area to the left of a z-score. For the smallest 2.3%, we need the z-score that has 0.023 to its left. This value is approximately z = -2. Conversely, for the largest 2.3%, since the normal distribution is symmetric, the z-score will have the same absolute value but be positive, which would be z = 2. However, to be precise, one should use statistical tables or software to find the exact z-scores close to these values.

To convert the z-scores to specific head breadths, we use the formula:

X = μ + (z * σ)

where X is the head breadth, μ is the mean, and σ is the standard deviation.

The minimum head breadth (Xmin) is calculated as follows:

The maximum head breadth (Xmax) is calculated as follows:

Therefore, the minimum head breadth that will fit the clientele is 4.1 inches, and the maximum head breadth is 8.1 inches.

Answer:

a.) 0.0402

b.) 0.00789

Step-by-step explanation:

In between the numbers 1 and 1012,

The number of perfect squares we have is 31, that is all numbers between 1 and 31 inclusive have their squares between the numbers 1 and 1012

The number of perfect cubes we have is 10, that is all numbers between 1 and 10 Inclusive have their perfect cube between the numbers 1 and 1012.

The number of perfect Fourth we have is 5, that is all numbers between 1 and 5 inclusive have their perfect fourth between the numbers 1 and 1012.

The number of perfect Sixth we have is 3, that is all numbers between 1 and 3 inclusive have their perfect sixth between the numbers 1 and 1012.

Hence,

a.) probability of choosing a perfect square = 31/1012

Probability of choosing a perfect cube = 10/1012

Probability of choosing a perfect square or a perfect cube = (31/1012) + (10/1012) - [(31/1012)*(10/1012)]

=41/1012 - 310/1024144

=0.0405 - 0.0003

=0.0402.

b.) Probability of choosing a perfect fourth or a perfect Sixth = (5/1012) + (3/1012) - [(5/1012)*(3/1012)]

= 8/1012 - 15/1024144

= 0.00789.

Answer:

Therefore, we conclude that 2/15  of the team work part-time.

Step-by-step explanation:

We know that a softball team is composed of 15 employees of a furniture store of whom 2 work part-time. We calculate what proportion of the team work part-time.

So, we will divide the number of those people who work part-time, by the number of people employed in  a softball team. We get:

x=2/15

Therefore, we conclude that 2/15  of the team work part-time.

Answer:

[tex]-\frac{1}{2} (-\frac{3}{2} x+6x+1)-3x[/tex] [tex]=-\frac{21x}{4} -\frac{1}{2}[/tex] [tex]=-\frac{1}{2}(\frac{21x}{2} +1)[/tex]

Step-by-step explanation:

Given,

[tex]-\frac{1}{2} (-\frac{3}{2} x+6x+1)-3x[/tex]

Applying distribution law

[tex]=(-\frac{1}{2}) (-\frac{3}{2} x)+(-\frac{1}{2}).6x+(-\frac{1}{2}).1-3x[/tex]

[tex]=\frac{3}{4} x-3x-\frac{1}{2}-3x[/tex]

Combine like terms

[tex]=\frac{3}{4} x-3x-3x-\frac{1}{2}[/tex]

Adding like terms

[tex]=\frac{3x-12x-12x}{4} -\frac{1}{2}[/tex]

[tex]=-\frac{21x}{4} -\frac{1}{2}[/tex]

[tex]=-\frac{1}{2}(\frac{21x}{2} +1)[/tex]

Answer:

(a)y(t)=30exp(-0.0231t)

(b)y(20)=18.9mg

Step-by-step explanation:

At a particular time t, the mass of a radioactive substance like Cesium-137 is governed by the equation:

N=N₀e⁻ᵏᵗ where k=ln 2/half life

(a)Mass that remains after t years

Half Life= 30 years

k= ln2/30=0.0231

Initial Mass, N₀=30mg

Therefore the mass N that remains at time t

N=N₀e⁻ᵏᵗ

N=30exp(-0.0231t)

y(t)=30exp(-0.0231t)

(b)We want to determine how much of the sample remains after 20 years.

At t=20 years

y(t)=30exp(-0.0231t)

y(20)=30exp(-0.0231X20)

=30 X 0.63

y(20)=18.9mg

Final answer:

23.81 mg of the original 30 mg sample remaining, rounded to two decimal places.

Explanation:

The half-life of a radioactive isotope like cesium-137 is time it takes for half of the original amount of the substance to decay. For cesium-137, this period is 30 years. To find out how much of a sample remains after a specific amount of time, such as 20 years in the given question, we use the decay formula.

Step 1: Let y(t) be the mass remaining after t years.
Step 2: The decay formula is y(t) = y(0) · (1/2)^(t/half-life), where y(0) is the initial mass and t is the time in years.

For cesium-137 after 20 years: y(20) = 30 mg · (1/2)^(20/30).
Calculating the remaining mass: y(20) = 30 mg · (1/2)^(2/3) ≈ 30 mg · 0.7937 ≈ 23.81 mg.

Answer:

Probability that the weight of a randomly selected steer is between 939 and 1417 lbs is 0.68389 .

Step-by-step explanation:

We are given that the weights of steers in a herd are distributed normally. The standard deviation is 200 lbs and the mean steer weight is 1300 lbs.

So, Let X = weights of steers in a herd ,i.e.; X ~ N([tex]\mu,\sigma^{2}[/tex])

Here, [tex]\mu[/tex] = population mean = 1300 lbs

         [tex]\sigma[/tex] = population standard deviation = 200 lbs

The z score area distribution is given by;

            Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)

So, Probability that the weight of a randomly selected steer is between 939 and 1417 lbs = P(939 lbs < X < 1417 lbs)

P(939 lbs < X < 1417 lbs) = P(X < 1417) - P(X <= 939)

P(X < 1417) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{1417-1300}{200}[/tex] ) = P(Z < 0.58) = 0.71904

P(X <= 939) = P( [tex]\frac{X-\mu}{\sigma}[/tex] <= [tex]\frac{939-1300}{200}[/tex] ) = P(Z <= -1.81) = 1 - P(Z < 1.81)

                                                       = 1 - 0.96485 = 0.03515

Therefore,  P(939 lbs < X < 1417 lbs) = 0.71904 - 0.03515 = 0.68389 .

Answer:

[tex]X \sim Binom(n=10, p=0.30)[/tex]

And we want to find this probability:

[tex] P(X=3)[/tex]

And using the probability mass function we got:

[tex] P(X=3) = (10C3) (0.3)^3 (1-0.3)^{10-3]= 0.2668[/tex]

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:  

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]  

Where (nCx) means combinatory and it's given by this formula:  

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]  

The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: [tex]P(A)+P(A') =1[/tex]

Solution to the problem

Let X the random variable of interest "workers that take public transportation daily", on this case we now that:

[tex]X \sim Binom(n=10, p=0.30)[/tex]

And we want to find this probability:

[tex] P(X=3)[/tex]

And using the probability mass function we got:

[tex] P(X=3) = (10C3) (0.3)^3 (1-0.3)^{10-3]= 0.2668[/tex]

Answer:

a) FALSE we obtain a NOT significant results after conduct the hypothesis tes.

b) FALSE, the standard error is not associated to a certain % of people included in the study

c) FALSE. if we want to reduce the standard error we need to increase the sample size or data

d) FALSE, always if we have a higher confidence level the confidence interval associated to this level would be wider than for a lower confidence interval

Step-by-step explanation:

Data given and notation n  

n represent the random sample taken

[tex]\hat p=0.52[/tex] estimated proportion of U.S. adult Twitter user get at least some news on Twitter

[tex]p_o=0.5[/tex] is the value that we want to test

[tex]\alpha=0.01[/tex] represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that more than half U.S. adult twitter users get some news throught Twitter:  

Null hypothesis:[tex]p\leq 0.5[/tex]  

Alternative hypothesis:[tex]p > 0.5[/tex]  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

Calculate the statistic  

The standard error is given:

[tex] SE = \sqrt{\frac{p_o (1-p_o)}{n}}=0.024[/tex]

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.52 -0.5}{0.024}=0.833[/tex]  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level assumed is [tex]\alpha=0.01[/tex]. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

[tex]p_v =P(z>0.833)=0.2024[/tex]  

So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.01[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

Now let's idendity the statements

a) FALSE we obtain a NOT significant results after conduct the hypothesis tes.

b) FALSE, the standard error is not associated to a certain % of people included in the study

c) FALSE. if we want to reduce the standard error we need to increase the sample size or data

d) FALSE, always if we have a higher confidence level the confidence interval associated to this level would be wider than for a lower confidence interval

The statement a) is true because the data is statistically significant. Statements b), c) and d) are false because: standard error doesn't reflect the portion of the population included in the study; standard error reduces with more data; and a higher confidence level results in a wider interval, not narrower.

a) This statement is true. The data can be considered statistically significant as long as the percentage is not within the standard error range below 50%. In this case, 52% minus 2.4% (standard error) = 49.6% which is below 50%. However, at a significance level of alpha = 0.01, it would be statistically significant since it is more than one standard error away from 50%.

b) This statement is false. The standard error is not an indication of what portion of the population is included in the study. Instead, it's a measure of statistical accuracy of the estimate.

c) This statement is false. The standard error of the estimate would typically decrease as we collect more data. As more data is collected, estimates are usually more accurate.

d) This statement is false. A 90% confidence interval is narrower than a 99% confidence interval. The larger the confidence level, the wider the interval to factor in more uncertainty.

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Answer:

See answers below

Step-by-step explanation:

# prior

mu0 <- 5

k0 <- 1

s20 <- 4

nu0 <- 2

# read in data

dat <- list()

dat[1] <- read.table("school1.dat")

dat[2] <- read.table("school2.dat")

dat[3] <- read.table("school3.dat")

n <- sapply(dat, length)

ybar <- sapply(dat, mean)

s2 <- sapply(dat, var)

# posterior

kn <- k0 + n

nun <- nu0 + n

mun <- (k0 * mu0 + n * ybar)/kn

s2n <- (nu0 * s20 + (n - 1) * s2 + k0 * n * (ybar - mu0)^2/kn)/nun

# produce a posterior sample for the parameters

s.postsample <- s2.postsample <- theta.postsample <- matrix(0, 10000, 3, dimnames = list(NULL,  

   c("school1", "school2", "school3")))

for (i in c(1, 2, 3)) {

   s2.postsample[, i] <- 1/rgamma(10000, nun[i]/2, s2n[i] * nun[i]/2)

   s.postsample[, i] <- sqrt(s2.postsample[, i])

   theta.postsample[, i] <- rnorm(10000, mun[i], s.postsample[, i]/sqrt(kn[i]))

}

# posterior mean and 95% CI for Thetas (the means)

colMeans(theta.postsample)

## school1 school2 school3  

##   9.294   6.943   7.814

apply(theta.postsample, 2, function(x) {

   quantile(x, c(0.025, 0.975))

})

##       school1 school2 school3

## 2.5%    7.769   5.163   6.152

## 97.5%  10.834   8.719   9.443

# posterior mean and 95% CI for the Standard Deviations

colMeans(s.postsample)

## school1 school2 school3  

##   3.904   4.397   3.757

apply(s.postsample, 2, function(x) {

   quantile(x, c(0.025, 0.975))

})

##            school1 school2 school3

## 2.5%    2.992   3.349   2.800

## 97.5%   5.136   5.871   5.139

The question calls for Bayesian inference with a normal model and a conjugate prior to derive posterior distributions and probabilities concerning means and standard deviations. Statistical software such as R or Python with libraries like PyMC3 is needed to perform the detailed computations, which are not provided here.

The provided question involves advanced statistical analysis, specifically Bayesian inference with prior distributions and the computation of posterior probabilities. Unfortunately, without software or computational methods to process the provided datasets for schools school1.dat, school2.dat, and school3.dat, a comprehensive answer cannot be provided. However, the general approach to tackle this problem would involve:

To complete the analysis, one would typically utilize statistical software capable of Bayesian computation, such as R or Python with appropriate libraries (e.g., PyMC3).

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Answer:

a. 0.692 or 69.2%; b. 0.308 or 30.8%.

Step-by-step explanation:

This is the case of the probability of the sum of two events, which is defined by the formula:

[tex] \\ P(A \cup B) = P(A) + P(B) - P(A \cap B)[/tex] (1)

Where [tex] \\ P(A \cup B)[/tex] represents the probability of the union of both events, that is, the probability of event A plus the probability of event B.

On the other hand, [tex] \\ P(A \cap B)[/tex] represents the probability that both events happen at once or the probability of event A times the probability of event B (if both events are independent).

Notice the negative symbol for the last probability. The reason behind it is that we have to subtract those common results from event A and event B to avoid count them twice when calculating [tex] \\ P(A \cup B)[/tex].

We have to remember that a sample space (sometimes denoted as S) is the set of the all possible results for a random experiment.

From the question, we have two events:

Event A: event subscribers rented a car during the past 12 months for business reasons.

Event B: event subscribers rented a car during the past 12 months for personal reasons.

[tex] \\ P(A) = 45.2\%\;or\;0.452[/tex]

[tex] \\ P(B) = 56\%\;or\;0.56[/tex]

[tex] \\ P(A \cap B) = 32\%\;or\;0.32[/tex]

With all this information, we can proceed as follows in the next lines.

The probability that a subscriber rented a car during the past 12 months for business or personal reasons.

We have to use here the formula (1) because of the sum of two probabilities, one for event A and the other for event B.

Then

[tex] \\ P(A \cup B) = P(A) + P(B) - P(A \cap B)[/tex]

[tex] \\ P(A \cup B) = 0.452 + 0.56 - 0.32[/tex]

[tex] \\ P(A \cup B) = 0.692\;or\;69.2\%[/tex]

Thus, the probability that a subscriber rented a car during the past 12 months for business or personal reasons is 0.692 or 69.2%.

The probability that a subscriber did not rent a car during the past 12 months for either business or personal reasons.

As we can notice, this is the probability for the complement event that a subscriber did not rent a car during the past 12 months, that is, the probability of the events that remain in the sample space. In this way, the sum of the probability for the event that a subscriber rented a car plus the event that a subscriber did not rent a car equals 1, or mathematically:

[tex] \\ P(\overline{A \cup B}) + P(A \cup B)= 1[/tex]

[tex] \\ P(\overline{A \cup B}) = 1 - P(A \cup B)[/tex]

[tex] \\ P(\overline{A \cup B}) = 1 - 0.692[/tex]

[tex] \\ P(\overline{A \cup B}) = 0.308\;or\;30.8\%[/tex]

As a result, the requested probability for a subscriber that did not rent a car during the past 12 months for either business or personal reasons is 0.308 or 30.8%.

We can also find the same result if we determine the complement for each probability in formula (1), or:

[tex] \\ P(\overline{A}) = 1 - P(A) = 1 - 0.452 = 0.548[/tex]

[tex] \\ P(\overline{B}) = 1 - P(B) = 1 - 0.56 = 0.44[/tex]

[tex] \\ P(\overline{A \cup B}) = 1 - P(A \cup B) = 1 - 0.32 = 0.68[/tex]

Then

[tex] \\ P(\overline{A \cup B}) = P(\overline{A}) + P(\overline{B}) - P(\overline{A\cap B})[/tex]

[tex] \\ P(\overline{A \cup B}) = 0.548 + 0.44 - 0.68[/tex]

[tex] \\ P(\overline{A \cup B}) = 0.308[/tex]

Final answer:

The probability of a magazine subscriber renting a car in the past 12 months for business or personal reasons is 69.2%, and the probability of not renting for either reason is 30.8%.

Explanation:

To solve the question: What is the probability that a subscriber rented a car during the past 12 months for business or personal reasons? and What is the probability that a subscriber did not rent a car during the past 12 months for either business or personal reasons?, we can use the principle of inclusion-exclusion in probability.

(a) The probability of renting a car for either reason can be calculated by adding the probabilities of renting for each reason and then subtracting the probability of renting for both reasons (to avoid double counting). Hence, the formula is: P(Business or Personal) = P(Business) + P(Personal) - P(Both).

Substituting the given values: P(Business or Personal) = 45.2% + 56% - 32% = 69.2%.

(b) The probability that a subscriber did not rent a car for either business or personal reasons is the complement of the probability calculated in part (a). Hence, it is calculated as 100% - 69.2% = 30.8%.

Complete Question

The complete question is shown on the first uploaded image

Answer:

The cognitive percentile for someone who has played for 7 years is 78.62

The cognitive percentile for someone who has played for 16 years is 48.56

Step-by-step explanation:

From the question the regression line for predicting percentile from years is given as

                 percentile = [tex]102 - 3.34 \ Years[/tex]

For someone who has played for 7 years his cognitive percentile would be

           [tex]Percentile = 102 - 3.34(7)[/tex]

                             [tex]=78.62[/tex]

For someone who has played for 16 years his cognitive percentile would

be   [tex]Percentile = 102 - 3.34(16)[/tex]

                        [tex]=48.56[/tex]

Final answer:

The predicted cognitive percentile for someone who has played football for 7 years is 78.62, while for someone who has played for 16 years, it is 48.56, using the regression equation Percentile = 102 - 3.34 Years.

Explanation:

To predict the cognitive percentile for a football player who has played for 7 years, we use the given regression equation Percentile = 102 - 3.34 Years. By plugging 7 into the equation for Years, we get:

Percentile = 102 - 3.34 × 7

Percentile = 102 - 23.38

Percentile = 78.62

So, for someone who has played football for 7 years, the predicted cognitive percentile would be 78.62.

Similarly, for someone who has played football for 16 years:

Percentile = 102 - 3.34 × 16

Percentile = 102 - 53.44

Percentile = 48.56

Therefore, the predicted cognitive percentile for someone who has played for 16 years is 48.56.

The phrases you would like to be written as expressions are not listed. I would nevertheless, explain how to write phrases as expressions so that the same approach could be applied to you own question.

Phrases are dynamic, depending on the problem. They do not necessarily take a particular form.

The constant thing about phrases is the operators connecting the words in the phrases. Theses operators are:

Addition (+), Subtraction (-), Division (÷), and Multiplication (×).

In word problems, it is a matter of interpretation, these operators can be written in many ways.

ADDITION

plus

the sum of

increase

grow

add

profit

And so on.

SUBTRACTION

minus

loss

decrease

reduce

subtract

And so on

MULTIPLICATION

times

multiply

triple

And so on

DIVISION

split

share

divide

distribute

And so on.

Examples

(1) 56 is added to a number to give 100

Interpretation: x + 56 = 100

(2)The difference between Mr. A and Mr. B is 5

Interpretation: A - B = 5

(3) This load (L1) is three times heavier than that one (L2)

Interpretation: L1 = 3L2

(4) Share this orange (P) equally between the three children

Interpretation: P/3

To write expressions using the numbers 54 and 67 with operations, you can add (54 + 67), subtract (54 - 67), multiply (54 x 67), or divide (54 ÷ 67). The order of operations is important and parentheses may be used to dictate the sequence of calculations. Practice with expressions can be enhanced by working through problems symbolically and then substituting numbers.

The question asks to write expressions using the numbers 54 and 67 with operations. When creating expressions, operations such as addition (+), subtraction (-), multiplication (x or ×), and division (÷) are typically used. For example, if we want to add 54 and 67, the expression would be 54 + 67.

If we wanted to multiply them, the expression would be 54 x 67 (or 54 × 67). Remember that in writing expressions, the order of operations is important, so if we want to perform different operations, we may need to use parentheses to ensure the correct sequence of calculations.

For more practice with expressions and operations, you can approach problems by covering up one number and solving for it using the remaining information (as mentioned in the given examples). This technique helps you solve various types of problems using the same principle. Moreover, when working with algebraic expressions, it's beneficial to solve the equation symbolically first and then substitute the numbers.

Substituting numbers and parentheses placement can significantly alter the result of the expression, as shown in the examples given in the question. The expressions created through experimentation, such as (74) to the third power, can be solved symbolically or using a calculator to understand how different operations and powers affect the result.

Answer: the population in 5 years is

187600

Step-by-step explanation:

The value of a house is increasing by 1900 per year. This means that it is increasing in an arithmetic progression. The formula for determining the nth term of an arithmetic sequence is expressed as

Tn = a + d(n - 1)

Where

a represents the first term of the sequence.

d represents the common difference.

n represents the number of terms in the sequence.

From the information given,

a = 180000

d = 1900

n = 5

We want to determine the value of the 5th term, T5. Therefore,

T5 = 180000 + 1900(5 - 1)

T5 = 180000 + 7600

T5 = 187600

Answer:

254.34 in^3

Step-by-step explanation:

Volume of a cylinder is V=pi*r^2*h

Since r=6/2=3, and h=3r=3*3=9, then we solve for V:

V=3.14*3^2*9

V=3.14*9*9

V=3.14*81

V=254.34

So the volume of the cylinder is 254.34 in^3

Answer: volume = 254.3 inches³

Step-by-step explanation:

The formula for determining the volume of a cylinder is expressed as

Volume = πr²h

Where

r represents the radius of the cylinder.

h represents the height of the cylinder.

π is a constant whose value is 3.14

From the information given,

Diameter = 6 inches

Radius = diameter/2 = 6/2 = 3 inches

Height = 3 × radius = 3 × 3 = 9 inches

Therefore,

Volume = 3.14 × 3² × 9

Volume = 254.3 inches³ to the nearest tenth

To find cos(1/2 B) given cos B = 3/5, we can use the double angle identity for cosine. The positive value of cos(1/2 B) in simplest radical form with a rational denominator is (2√5) / 5.

If cos B = 3/5, we are looking to find cos(1/2 B) in simplest radical form. To find this, we can use the double angle identity for cosine, which tells us that cos(2θ) = 2cos^2(θ) - 1. If we let θ = 1/2 B, then B = 2θ and the identity becomes cos(B) = 2cos^2(1/2 B) - 1. We can solve for cos(1/2 B) by rearranging into 2cos^2(1/2 B) = cos(B) + 1. Substituting the known value of cos(B) gives us:

2cos^2(1/2 B) = 3/5 + 1 = 8/5

So,

cos^2(1/2 B) = 8/10

cos^2(1/2 B) = 4/5

Taking the positive square root, since cosine is positive in the question:

cos(1/2 B) = √(4/5)

cos(1/2 B) = √4 / √5

cos(1/2 B) = 2 / √sqrt(5)

To rationalize the denominator, we multiply the numerator and denominator by √5:

cos(1/2 B) = (2√5) / 5

Therefore, the positive value of cos(1/2 B), in simplest radical form with a rational denominator, is (2√5) / 5.

The positive value of [tex]$\cos \frac{1}{2}B$[/tex] in simplest radical form with a rational denominator is [tex]$\frac{2}{\sqrt{5}}$[/tex].

To find the value of [tex]$\cos \frac{1}{2}[/tex][tex]B$ given $\cos B[/tex][tex]= \frac{3}{5}$[/tex], we can use the half-angle formula for cosine:

[tex]\[ \cos \frac{1}{2}B = \pm\sqrt{\frac{1 + \cos B}{2}} \][/tex]

Since we are looking for the positive value, we will use the positive square root. Plugging in the given value of [tex]$\cos B$[/tex]:

[tex]\[ \cos \frac{1}{2}B = \sqrt{\frac{1 + \frac{3}{5}}{2}} = \sqrt{\frac{\frac{5}{5} + \frac{3}{5}}{2}} = \sqrt{\frac{\frac{8}{5}}{2}} = \sqrt{\frac{8}{5} \cdot \frac{1}{2}} = \sqrt{\frac{8}{10}} = \sqrt{\frac{4}{5}} \][/tex]

Now, we can simplify the square root of a fraction by taking the square root of the numerator and the denominator separately:

[tex]\[ \cos \frac{1}{2}B = \frac{\sqrt{4}}{\sqrt{5}} = \frac{2}{\sqrt{5}} \][/tex]

This is the simplest radical form with a rational denominator. Note that we do not need to rationalize the denominator since the question only asks for a rational denominator, not necessarily an integer denominator. The answer[tex]$\frac{2}{\sqrt{5}}$[/tex] is already in the required form."

Answer:

The answer to the question is

It would take about 167.021 s  to splash down in the arctic ocean.

Step-by-step explanation:

The gravitational force is given by

[tex]F_G = \frac{G*m_1*m_2}{r^2}[/tex]

Where m₁ mass of the piano and

m₂ = mass of the Earth

r = 3·R where R = radius of the earth

as stated in the question we have varying acceleration due to the inverse square law

Therefore at 3 × Radius of the earth we have

[tex]F_G[/tex] = 109.083 N and the acceleration =1.09083 m/s²

If the body falls from 3·R to 2·R with that acceleration we have

S = u·t +0.5×a·t²  = 0.5×a·t² as u = 0

That is 6371 km = 0.5·1.09083·t²

t₁ = 108.079 s and we have

v₁² = u₁² +2·a₁·s₁ =  2·a₁·s₁ = 117.895 m/s

For the next stage r₂ = 2R

Therefore F = 245.436 N and a₂ = F/m₁ = 2.45436 m/s²

Therefore the time from 2R to R is given by

S₂ =R=u·t+0.5·a₂·t² = v₁·t + 0.5·a₂·t²

or 6371 km = 117.895 m/s × t + 0.5 × 2.45436 × t²

Which gives 1.22718 × t² + 117.895 × t -6371  = 0

Factorizing we have (t+134.631)(t-38.56)×1.22718 = 0

Therefore t = -134.631 s or 38.56 s as we only deal with positive values of time in the present question we have t₂ = 38.56 s and

v₂² = v₁² + 2·a₂·S  = (117.895 m/s)² + 2·2.45436 m/s²×6371 km = 45172.686

v₂ = 212.54 m/s

For final stage we have r = R and

[tex]F_{G3}[/tex] = 981.746 N and a₂ = F/m₁ = 9.81746 m/s²

Therefore the time from R to the arctic ocean  is given by

S₃ =R=v₂·t+0.5·a₂·t² = 212.54·t + 0.5·9.81746·t² = 6371

Which gives

Therefore t₃ = 20.382 s

Therefore, it will take about t₁ + t₂ + t₃ = 108.079 s + 38.56 s + 20.382 s  = 167.021 s  to splash down in the ocean

Use the formula t = sqrt((2 * d) / g), where d = 4R and g = G * (mass of earth) / (distance from center)^2 to calculate the time taken for the piano to hit the ocean from a height 3R above the Earth's surface.

This physics question deals with the concept of

gravity

and the formation of a free fall situation in a vacuum situation where no other forces (apart from gravitation) are considered. In this case, the value of gravity will not be 9.8 m/s^2, as we are not at the surface but at a distance 3 times the radius of Earth. To find how long the piano takes to hit the ocean, we need to use the following formula for the time taken (t) for an object to fall a certain distance under gravity: t = sqrt((2 * d) / g). Here, d is the total distance that the piano would fall, and g is the acceleration due to gravity. We would first need to calculate the distance the piano falls, which will be the sum of the Earth’s radius (R) and the height the piano is dropped from (3R). This equals 4R. Then, calculate the value of gravity at that point, using formula G * (mass of Earth)/(distance from point to center of earth)^2, and then substitute these values back into the equation.

Note

: This is a idealized scenario and other factors like air resistance, influence from moon and sun, and the non-uniform distribution of Earth’s mass etc., are not considered.

https://brainly.com/question/35699425

#SPJ3

Answer:

5278.0

21112

2927

Step-by-step explanation:

P = Po[2^(t/5)]

8000 = Po(2^⅗)

Po = 5278.0

P = 5278(2^(10/5))

P = 21112

P = Po[2^(t/5)]

ln(P/Po) = (t/5)ln2

ln(P) - ln(Po) = (t/5)ln2

1/P . dP/dt = ln2/5

dP/dt = P(ln2)/5

At t = 10, P = 21112

dP/dt = 2927

Answer:

The value of this inheritance is $78,192.28

Step-by-step explanation:

The monthly payments Sara will receive  starting today for next 40 years is $500.

Annual Interest Rate = 7.3%

Monthly Interest Rate = Annual Interest Rate/12

                                    =7.3/12

Monthly Interest Rate = 0.6083%

Present Value = $500 + $500/1.006083 + $500/1.006083^2 + $500/1.006083^3 + ... + $500/1.006083^479

Present Value = $500 * 1.006083 * (1 - (1/1.006083)^480) / 0.006083

Present Value = $500 * 156.39156

Present Value = $78,192.28

Thus, the value of this inheritance is $78,192.28

Answer:

86.46% probability that at least 2 crashes occur in the next month.

Step-by-step explanation:

For each TIE ghter ights, there are only two possible outcomes. Either it crashes, or it does not. The probability of a TIE ghter ights crashing is independent from other TIE ghter ights. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

Every month, there are 1000 independent TIE ghter ights, and each TIE ghter ight crashes with a probability of 0.0035.

This means that [tex]n = 1000, p = 0.0035[/tex]

(a) What is the probability that at least 2 crashes occur in the next month?

Either less than 2 crashes occur, or at least 2 do. The sum of the probabilities of these events is decimal 1. So

[tex]P(X < 2) + P(X \geq 2) = 1[/tex]

We want [tex]P(X \geq 2)[/tex]

So

[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]

In which

[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{1000,0}.(0.0035)^{0}.(0.9965)^{1000} = 0.0300[/tex]

[tex]P(X = 1) = C_{1000,1}.(0.0035)^{1}.(0.9965)^{999} = 0.1054[/tex]

[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.0300 + 0.1054 = 0.1354[/tex]

[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.1354 = 0.8646[/tex]

86.46% probability that at least 2 crashes occur in the next month.