When testing for current in a cable with eight ​color-coded wires, the author used a meter to test five wires at a time. How many different tests are required for every possible pairing of five ​wires?

Answer:

(a) The probability of the event (X > 84) is 0.007.

(b) The probability of the event (X < 64) is 0.483.

Step-by-step explanation:

The random variable X follows a Poisson distribution with parameter λ = 64.

The probability mass function of a Poisson distribution is:

[tex]P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0, 1, 2, ...[/tex]

(a)

Compute the probability of the event (X > 84) as follows:

P (X > 84) = 1 - P (X ≤ 84)

                [tex]=1-\sum _{x=0}^{x=84}\frac{e^{-64}(64)^{x}}{x!}\\=1-[e^{-64}\sum _{x=0}^{x=84}\frac{(64)^{x}}{x!}]\\=1-[e^{-64}[\frac{(64)^{0}}{0!}+\frac{(64)^{1}}{1!}+\frac{(64)^{2}}{2!}+...+\frac{(64)^{84}}{84!}]]\\=1-0.99308\\=0.00692\\\approx0.007[/tex]

Thus, the probability of the event (X > 84) is 0.007.

(b)

Compute the probability of the event (X < 64) as follows:

P (X < 64) = P (X = 0) + P (X = 1) + P (X = 2) + ... + P (X = 63)

                [tex]=\sum _{x=0}^{x=63}\frac{e^{-64}(64)^{x}}{x!}\\=e^{-64}\sum _{x=0}^{x=63}\frac{(64)^{x}}{x!}\\=e^{-64}[\frac{(64)^{0}}{0!}+\frac{(64)^{1}}{1!}+\frac{(64)^{2}}{2!}+...+\frac{(64)^{63}}{63!}]\\=0.48338\\\approx0.483[/tex]

Thus, the probability of the event (X < 64) is 0.483.

To approximate probabilities for a Poisson distribution with mean 64, one can use the cumulative distribution function of the Poisson distribution. But for large means like 64, using a Normal approximation to the Poisson might be more accurate.

The question is about the Poisson distribution which is a probability distribution used often in statistical modelling where we are counting the number of times an event happens in a fixed interval of time or space. The mean is given as 64. To compute probabilities for values more than or less than a certain value, we use the cumulative distribution function (cdf) of the Poisson distribution.

The cumulative distribution function gives us the probability that the random variable is less than or equal to a certain value. In other words, if we want to find P(X > 84), we can find it by subtracting P(X <= 83) from 1. Likewise, to find P(X < 64), we can directly use the cdf at 63.

Please note however, because of the large mean (64), it may be more accurate to use a Normal approximation to the Poisson with mean 64 and variance 64. In that case, standardize the variables and use the Standard Normal table for the probabilities.

https://brainly.com/question/33722848

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Answer:

  43 months

Step-by-step explanation:

The amortization formula is ...

  A = P(r/n)/(1 -(1 +r/n)^(-nt))

We want to find the value of t for monthly payment A, financed amount P, interest rate r, and compounding monthly (n=12). Filling in the values, we have ...

  75 = 2500(0.135/12)/(1 -(1 +0.135/12)^(-12t))

  1 -(1.01125^(-12t) = 2500·0.135/(12·75) = 0.375

  1 -0.375 = 1.01125^(-12t) . . . . . add 1.01125^(-12t) -0.375

Next, take logarithms and divide by the coefficient of t.

  log(0.625)/(-12log(1.01125)) = t ≈ 3.50106 . . . . years

In months, that is ...

  3.50106×12 ≈ 42.01

The balance will be not quite zero after 42 payments. (It will be about $0.94.) It will take 43 payments to pay off the credit card balance.

Answer:

42

I did the test and got it right hope this helps

Option C: The sum of the infinite geometric series is [tex]\frac{15}{2}[/tex]

Explanation:

The sum of infinite geometric series can be determined using the formula,

[tex]S_{\infty}=\frac{a}{1-r}[/tex]

Substituting the values [tex]a=5[/tex] and [tex]r=\frac{1}{3}[/tex] in the above formula, we have,

[tex]S_{\infty}=\frac{5}{1-\frac{1}{3} }[/tex]

Simplifying the denominator by taking LCM.

Thus, we have,

[tex]S_{\infty}=\frac{5}{\frac{2}{3} }[/tex]

Simplifying, we get,

[tex]S_{\infty}=5\times{\frac{3}{2} }[/tex]

Multiplying, we get,

[tex]S_{\infty}={\frac{15}{2} }[/tex]

Thus, the sum of the infinite geometric series is [tex]\frac{15}{2}[/tex]

Hence, Option C is the correct answer.

Answer:

C

Step-by-step explanation:

Sum = a/(1 - r)

= 5/(1 - ⅓)

= 5/(⅔)

= 5 × 3/2

= 15/2

= 7.5

Answer:

10 bags of chips

Step-by-step explanation:

15(2)+3x=60

30+3x=60

3x=30

x=10

Therefore, 10 bags of chips

Answer:

Which of the following is a quantitative continuous variable?

b. Distance a family travels on vacation

e. Amount spent on the vacation, to the nearest $100.

Step-by-step explanation:

Quantitative variables use numbers to express attributes, there are 2 types:

1. Quantitave continuous variables use numbers to express the attributes but, could have an infinite value between its maximum and minimum values, and are able to be divided into smaller values, such as in options b and e.

2. Quantitative discrete, also use numbers to express attributes, but they may have some, but not all the values, and must be integers, such as in options:

c. Number of vacations a family takes per year

d. Number of consecutive years that a family goes on a vacation (Let a trip of at least 200 miles from home be defined as "vacation").

e. Amount spent on the vacation, to the nearest $100.

And there are the cualitative variables, which use words to express attributes, such as in:

a. Where a family goes on vacation.

Answer:

The median is located at the 2.5th position, which is halfway between the values 52 and 57.

Step-by-step explanation:

The Median is referred to as the Middle term of a set of Data when ordered in Ascending/Descending order.

Consider the numbers 50, 230, 52, and 57

Arranging them in an Ascending Order

50, 52, 57, 230

The number of data in the set is even.

Dividing the data into two halves (50, 52 and 57, 230)

The median is halfway between the two halves.

The median is located at the 2.5th position, which is halfway between the values 52 and 57.

Answer: 0.0935

Step-by-step explanation: p(business major) = 0.17, p(receiving financial aids) = 0.55

The two events are independent that's the occurrence of one does not affect the other.

Hence, the probability of that the student is a business major and receiving financial aid is given below as

p(business major and financial aid) = 0.17×0.55 = 0.0935.

Answer:

The slope is -5/6

Step-by-step explanation:

To find the slope of a line given two points, we use the formula

m= (y2-y1)/(x2-x1)

  = (-1-4)/(4--2)

  =-5/(4+2)

  =-5/6

Answer: the slope of the line between the two points is - 5/6

Step-by-step explanation:

Slope, m = change in value of y on the vertical axis / change in value of x on the horizontal axis.

change in the value of y = y2 - y1

Change in value of x = x2 -x1

y2 = final value of y

y 1 = initial value of y

x2 = final value of x

x1 = initial value of x

The line passes through (- 2, 4) and (4, - 1),

y2 = - 1

y1 = 4

x2 = 4

x1 = - 2

Slope,m = (- 1 - 4)/(4 - - 2) = - 5/6

Answer and Step-by-step explanation:

(a)

E[XY] = 1/4*0*1 + 1/4*1*0 + 1/4*-1*0 +1/4*0*-1 = 0

(b)

E[X] = 0, E[Y] = 0

Thus, Cov(X,Y) = E[XY] - E[X]E[Y] = 0

So, Var(X + Y) = Var(X) + Var(Y) is True

The answer is Yes

(c) No, the converse is not true

(a) The value of E[XY] is zero.

(b) Yes  For this pair of random variables (X, Y) is true for the given relation.

(c) No, the converse is not true.

Variance is the value of the squared variation of the random variable from its mean value, in probability and statistics.

The given data in the problem will be;

(X, Y)  is a random variable

pairs of values is given as  (0,1), (1,0), (−1,0), (0,−1).

E[XY] =?

(a) The value of E[XY] is zero.

[tex]E[XY] = \frac{1}{4}\times 0\times1 +\frac{1}{4}\times 1\times 0 + \frac{1}{4} \times 1\times 0 +\frac{1}{4}\times0\times1 = 0[/tex]

Hence the value of E[XY] is zero.

(b) Yes or this pair of random variables (X, Y) is true for the given relation.

[tex]E[X] = 0, E[Y] = 0[/tex]

[tex]\rm Cov(X,Y) = E[X,Y] - E[X]E[Y] = 0Var(X + Y) = Var(X) + Var(Y)[/tex]

Hence ) Yes  For this pair of random variables (X, Y) is true for the given relation.

(c) No, the converse is not true

X and Y are independent

[tex]Var(X+Y)=Var(X)+Var(Y).[/tex]

Hence ) No, the converse is not true

To learn more about the variance refer to the link;

https://brainly.com/question/7635845

Answer:

0.44290869

Step-by-step explanation:

The Maclaurin series for sin⁻¹(x) is given by

sin⁻¹(x) = x + [tex]x^{\alpha } _{n=1}[/tex] [tex]\frac{1.3.5...(2n - 1)}{2.4.6...(2n)} * \frac{x^{2n + 1} }{2n + 1}[/tex]

Use the first five terms of the Maclaurin series above to approximate sin⁻¹ [tex]\frac{3}{7}[/tex]. (Round your answer to eight decimal places.)

Answer

sin⁻¹(x) = x + [tex]x^{\alpha } _{n=1}[/tex] [tex]\frac{1.3.5...(2n - 1)}{2.4.6...(2n)} * \frac{x^{2n + 1} }{2n + 1}[/tex]

in the above equation [tex]x^{\alpha } _{n=1}[/tex]  summation from n=1 to ∞

we are estimating this for the first 5 terms as follows

sin⁻¹(x) = x +   [tex]\frac{1}{2} * \frac{x^{3} }{3}[/tex]  +  [tex]\frac{1*3}{2*4} * \frac{x^{5} }{5}[/tex]  +  [tex]\frac{1*3*5}{2*4*6} * \frac{x^{7} }{7}[/tex]  +  [tex]\frac{1*3*5*7}{2*4*6*8} * \frac{x^{9} }{9}[/tex]

sin⁻¹(x) = x +  [tex]\frac{x^{3} }{6}[/tex]  +  [tex]\frac{3x^{5} }{40}[/tex]  +[tex]\frac{15x^{7} }{336}[/tex]  +  [tex]\frac{105x^{9} }{3456}[/tex]  

now to get

sin⁻¹([tex]\frac{3}{7}[/tex]) substitute

hence,

sin⁻¹([tex]\frac{3}{7}[/tex]) = [tex]\frac{3}{7} + \frac{\frac{3}{7} ^{3} }{6} + \frac{3 * \frac{3}{7} ^{5} }{40} + \frac{15* \frac{3}{7} ^{7} }{336} + \frac{105* \frac{3}{7} ^{9} }{3456}[/tex]  

sin⁻¹([tex]\frac{3}{7}[/tex]) = 0.42857142 + 0.01311953 + 0.00108437 + 0.00011855 + 0.00001482

           =  0.44290869

Answer:

Option A) 0.012

Step-by-step explanation:

We are given the following information:

We treat customers asking for water as a success.

P(customers ask for water) = [tex]\dfrac{3}{5}[/tex] = 0.6

Then the number of customers follows a binomial distribution, where

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

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 10

We have to evaluate:

[tex]P(x < 3) = P(x = 0) + P(x = 1) + P(x = 2) \\= \binom{10}{0}(0.6)^0(1-0.6)^{10} +\binom{10}{1}(0.6)^1(1-0.6)^{9} + \binom{10}{2}(0.6)^2(1-0.6)^{8}\\= 0.0001 +0.0015 + 0.0106\\=0.012[/tex]

0.012 is the probability that less than 3 customers ask for water with their meal.

The correct answer is option a) 0.012

The probability of success in each trial is 3 out of 5 (or 0.6), and the probability of failure is 2 out of 5 (or 0.4). Denoting the number of customers who ask for water as a binomial random variable X with n = 10 and p = 0.6.

The binomial probability formula is given by:

[tex]P(X=k) ={n \choose k}p^k(1-p)^{n-k}[/tex]

where:

n is the number of trials (10 in this case),
k is the number of successes (in this case, less than 3),
p is the probability of success (0.6), and
1−p is the probability of failure (0.4).

Probability that fewer than 3 customers ask for water needs to be found. This means we need to find:

P(X<3)=P(X=0)+P(X=1)+P(X=2)

Calculating each of these probabilities using the binomial formula:

For k=0:
[tex]P(X=0) ={10 \choose 0}(0.6)^0(0.4)^{10-0} = 1\times1\times0.4^{10} = 0.0001048576[/tex]

For k=1:

[tex]P(X=1) ={10 \choose 1}(0.6)^1(0.4)^{10-1} = 10\times0.6\times0.4^{9} = 0.001572864[/tex]

For k=2:

[tex]P(X=2) ={10 \choose 2}(0.6)^2(0.4)^{10-2} = 45\times0.36\times0.4^{8} = 0.010616832[/tex]

Now, we sum these probabilities to get the total probability that fewer than 3 customers ask for water:

P(X<3) = P(X=0) + P(X=1) + P(X=2)

P(X<3) = 0.0001048576 + 0.001572864 + 0.010616832

P(X<3) = 0.0122945536

Therefore rounding this value results in a probability of approximately 0.012.

Answer:

The probability that the age of a randomly selected CEO will be between 50 and 55 years old is 0.334.

Step-by-step explanation:

We have a normal distribution with mean=56 years and s.d.=4 years.

We have to calculate the probability that a randomly selected CEO have an age between 50 and 55.

We have to calculate the z-value for 50 and 55.

For x=50:

[tex]z=\frac{x-\mu}{\sigma}=\frac{50-56}{4}=\frac{-6}{4}= -1.5[/tex]

For x=55:

[tex]z=\frac{x-\mu}{\sigma}=\frac{55-56}{4}=\frac{-1}{4}=-0.25[/tex]

The probability of being between 50 and 55 years is equal to the difference between the probability of being under 55 years and the probability of being under 50 years:

[tex]P(50<x<55)=P(x<55)-P(x<50)\\\\P(50<x<55)=P(z<-0.25)-P(z<-1.5)\\\\P(50<x<55)=0.40129-0.06681\\\\P(50<x<55)=0.33448[/tex]

Final answer:

The probability that the age of a randomly selected CEO will be between 50 and 55 years old is approximately 0.3345.

Explanation:

To find the probability that the age of a randomly selected CEO will be between 50 and 55 years old, we need to calculate the z-scores for both values and then use the standard normal distribution table.

Step 1: Calculate the z-score for 50 years old:

z = (50 - 56) / 4 = -1.5

Step 2: Calculate the z-score for 55 years old:

z = (55 - 56) / 4 = -0.25

Step 3: Use the standard normal distribution table to find the area to the left of each z-score:

P(z < -1.5) = 0.0668

P(z < -0.25) = 0.4013

Step 4: Calculate the probability between the two z-scores:

P(-1.5 < z < -0.25) = P(z < -0.25) - P(z < -1.5) = 0.4013 - 0.0668 = 0.3345

Therefore, the probability that the age of a randomly selected CEO will be between 50 and 55 years old is approximately 0.3345.

Answer:

(b) 0.251

Step-by-step explanation:

1. Standard deviation equation:

[tex]SD=\sqrt{\frac{\sum\limits^N_i {(x_{i}-X)^{2} } }{N} }[/tex]

Where X is the mean of the data, and N the amount of data. Then, N=5

2. Estimate the Mean:

[tex]X=\frac{0.750+0.750+0.200+0.600+0.200}{5}=\frac{2.5}{5}=0.5\\[/tex]

3. Caclulate Standard deviation:

[tex]SD=\sqrt{\frac{{(0.750-0.500)^{2}+(0.750-0.500)^{2}+(0.200-0.500)^{2}+(0.600-0.500)^{2}+(0.200-0.500)^{2} } }{5} }[/tex]

[tex]SD=\sqrt{\frac{0.315}{5} }=\sqrt{0.063}\\SD=0.251[/tex]

Answer:

Step-by-step explanation:

a) November sales = (Total Uncollected Sales - Uncollected Sales from December) / Collection rate after two months

= ($121,100 - $87,300) / 0.20

November sales = $169, 000.00

b) December sales = Uncollected sales from December / Collection rate of the previous month sales,

Therefore: December sales = $87,300 / 0.45 = $194,000

December sales = $194,000.00

c) Each month's collection for the company are:

Collections for ech month = 0.20(Sales from 2 months ago) + 0.25(Last month's sales) + 0.55 (Current sales)

January collections = 0.20($169000.00) + 0.25($194,000.00) + 0.55($132,000)

January collections = $154,900.00

February collections = 0.20($194,000.00) + 0.25($132,000) + 0.55($149,000)

February collections = $153,750.00

March collections = 0.20($132,000) + 0.25($149,000) + 0.55($164,000)

March collections = $153,850.00

Answer:

76 ≤ B ≤ 89

Step-by-step explanation:

Mean score (μ) = 73.3

Standard deviation (σ) = 9.7

If B Scores are below the top 5% and above the bottom 62%, then:

62% ≤ B ≤ 95%

In a normal distribution, the 62nd percentile has a corresponding z-score of z = 0.305, while the 95th percentile has a corresponding z-score of z = 1.650.

The grades X1 and X2 which are the limits for a B grade are given by:

[tex]z= \frac{X-\mu}{\sigma}\\0.305= \frac{X_1-73.3}{9,7}\\X_1=76.2\\1.650= \frac{X_2-73.3}{9,7}\\X_2=89.3[/tex]

Rounding to the nearest whole number, a B grade is given to grades between 76 and 89.

Final answer:

To calculate the numerical limits for a B grade in a normally distributed set of test scores with a mean of 73.3 and a standard deviation of 9.7, we find the z-scores corresponding to the top 5% and bottom 62%. The B grade range is approximately between 70 and 89.

Explanation:

To find the numerical limits for a B grade in a normally distributed set of test scores, we must determine the z-scores that correspond to the top 5% and the bottom 38% of scores since a B grade is assigned to scores below the top 5% and above the bottom 62% (which is equivalent to being below the top 38%). Given the mean score is 73.3 and the standard deviation is 9.7, we can use z-score formulas to convert these percentages to raw scores.

For the top 5% cutoff (z = 1.645), the score is calculated as follows:

Top 5% cutoff score = mean + (z x standard deviation)

Top 5% cutoff score = 73.3 + (1.645 x 9.7)

Top 5% cutoff score = 73.3 + 15.9565

Top 5% cutoff score ≈ 89

For the top 38% cutoff (z = -0.31), which is the bottom 62%, the score is calculated as:

Bottom 62% cutoff score = mean + (z x standard deviation)

Bottom 62% cutoff score = 73.3 + (-0.31 x 9.7)

Bottom 62% cutoff score = 73.3 + (-2.997)

Bottom 62% cutoff score ≈ 70

Therefore, a B grade on this test would be for scores roughly between 70 and 89.

Answer:

6

Step-by-step explanation:

2x² + 4xy − 48y²

2 (x² + 2xy − 24y²)

2 (x − 4y) (x + 6y)

Answer:

solution attached below

Step-by-step explanation:

To find the sinusoids in the given equations, we can use the derivative and integral properties of phasors. For the first equation, υ2(t) = 1/100 dυ1(t)/dt + 20υ1(t), you can find the phasor representation of υ1(t) using the given equation υ1(t) = 10cos(100t + 90°) V and apply the derivative property of phasors. For the second equation, i2(t) = 10∫i1(t)dt - 3i1(t), you can find the phasor representation of i1(t) using the given equation i1(t) = -4cos(5t) A and apply the integral property of phasors. Therefore, the sinusoids in the given equations are υ2(t) ≈ 10cos(100t + 90°) + 2000cos(100t + 90°) V and i2(t) ≈ (4/5)cos(5t + 180°) + 12cos(5t + 180°) A·s.

To find the sinusoids in the given equations, we can use the derivative and integral properties of phasors.

For the first equation, υ2(t) = 1/100 dυ1(t)/dt + 20υ1(t), we can start by finding the phasor representation of υ1(t) using the given equation υ1(t) = 10cos(100t + 90°) V:

υ1(t) = 10e^(j(100t + 90°))

Next, we differentiate υ1(t) with respect to t to find dυ1(t)/dt:

dυ1(t)/dt = -1000sin(100t + 90°) V/s

Using the derivative property of phasors (multiplying by jω), we have:

jωυ1(t) = j(100)(10)e^(j(100t + 90°)) = 1000ej(100t + 90°) V/s

Now, we can substitute these phasor representations back into the original equation:

υ2(t) = 1/100 (1000ej(100t + 90°) V/s) + 20(10e^(j(100t + 90°)) V

Simplifying, we get:

υ2(t) ≈ 10ej(100t + 90°) + 2000ej(100t + 90°) V

For the second equation, i2(t) = 10∫i1(t)dt - 3i1(t), we can find the phasor representation of i1(t) using the given equation i1(t) = -4cos(5t) A:

i1(t) = -4e^(j(5t + 180°))

Next, we can integrate i1(t) with respect to t to find ∫i1(t)dt:

∫i1(t)dt = (-4/5)e^(j(5t + 180°)) A·s

Using the integral property of phasors (dividing by jω), we have:

(1/jω)i1(t) = (-1/(j5))(4)e^(j(5t + 180°)) = (4/5)ej(5t + 180°) A·s

Substituting these phasor representations back into the original equation, we get:

i2(t) ≈ (4/5)ej(5t + 180°) - 3(-4)e^(j(5t + 180°)) A·s

Therefore, the sinusoids in the given equations are υ2(t) ≈ 10cos(100t + 90°) + 2000cos(100t + 90°) V and i2(t) ≈ (4/5)cos(5t + 180°) + 12cos(5t + 180°) A·s.

Answer:

note:

solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment

Answer:

The answer to your question is Domain (-∞, ∞) Range  [-4, ∞)

Step-by-step explanation:

The Domain is the set of all possible values of the independent variable (x).

The Range is the set of all the possible values of the dependent variable when substitute the domain in the function.

On the graph, we find the domain looking at the x-axis

On a graph, we find the range, looking at all the y-axis

In this graph, x has values from -infinite to infinite, then, the domain is (-∞, ∞).

In this graph, y has values from -4 to infinite, then, the range is [-4, ∞)

Answer:

Expected value: 3/7

Change landing on purple to -1.5

Step-by-step explanation:

Expected value

=3(1/7) + 1(2/7) + 0(2/7) - 1(2/7) = 3/7

Fair game: Expected value = 0

Change the value of 0 to -1.5

Answer:

3/7

change 0 to -1.5

Step-by-step explanation:

Got a 100

Answer:

[tex]P(x \geq 9)=P(X=9)+P(X=10)[/tex]

[tex]P(X=9)=(10C9)(0.9)^9 (1-0.9)^{10-9}=0.387[/tex]

[tex]P(X=10)=(10C10)(0.9)^{10} (1-0.9)^{10-10}=0.349[/tex]

And adding we got:

[tex]P(x \geq 9)=P(X=9)+P(X=10)=0.387+0.349=0.736[/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=10, p=1-0.1=0.9)[/tex]

Where 1-p = 1-0.1=0.9 represent the probability of being NOT defective

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]

And we want to find this probability:

[tex]P(x \geq 9)=P(X=9)+P(X=10)[/tex]

[tex]P(X=9)=(10C9)(0.9)^9 (1-0.9)^{10-9}=0.387[/tex]

[tex]P(X=10)=(10C10)(0.9)^{10} (1-0.9)^{10-10}=0.349[/tex]

And adding we got:

[tex]P(x \geq 9)=P(X=9)+P(X=10)=0.387+0.349=0.736[/tex]

Answer: Third answer

Step-by-step explanation:

Answer:

[tex] SE= \frac{\sigma}{\sqrt{n}} = \frac{47}{\sqrt{30}}= 8.58[/tex]

[tex] ME= 1.64 *\frac{\sigma}{\sqrt{n}} = 1.64*\frac{47}{\sqrt{30}}= 14.073[/tex]

[tex] \bar X - ME = 237- 14.073 = 222.927[/tex]

[tex] \bar X + ME = 237+ 14.073 = 251.073[/tex]

[tex]n=(\frac{1.640(47)}{5})^2 =237.65 \approx 238[/tex]

So the answer for this case would be n=238 rounded up to the nearest integer

Null hypothesis:[tex]\mu \geq 247[/tex]  

Alternative hypothesis:[tex]\mu <247[/tex]  

[tex]z=\frac{237-247}{\frac{47}{\sqrt{30}}}=-1.165[/tex]  

[tex]p_v =P(z<-1.165)=0.122[/tex]  

If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that the true mean is lower than 247 at 10% of significance.

Step-by-step explanation:

For this case we have the following data given:

[tex] \bar X =237[/tex] represent the sample mean

[tex]\sigma = 47[/tex] represent the population deviation

[tex] n =30[/tex] represent the sample size selected

[tex]\mu_0 = 247[/tex] represent the value that we want to test.

The standard error for this case is given by:

[tex] SE= \frac{\sigma}{\sqrt{n}} = \frac{47}{\sqrt{30}}= 8.58[/tex]

For the 90% confidence the value of the significance is given by [tex] \alpha=1-0.9 = 0.1[/tex] and [tex] \alpha/2 = 0.05[/tex] so we can find in the normal standard distribution a quantile that accumulates 0.05 of the area on each tail and we got:

[tex] z_{\alpha/2}= 1.64[/tex]

And the margin of error would be:

[tex] ME= 1.64 *\frac{\sigma}{\sqrt{n}} = 1.64*\frac{47}{\sqrt{30}}= 14.073[/tex]

The confidence interval for this case would be given by:

[tex] \bar X - ME = 237- 14.073 = 222.927[/tex]

[tex] \bar X + ME = 237+ 14.073 = 251.073[/tex]

The margin of error is given by this formula:

[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]    (a)

And on this case we have that ME =5 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex]   (b)

Replacing into formula (b) we got:

[tex]n=(\frac{1.640(47)}{5})^2 =237.65 \approx 238[/tex]

So the answer for this case would be n=238 rounded up to the nearest integer

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is lower than 247 pounds, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \geq 247[/tex]  

Alternative hypothesis:[tex]\mu <247[/tex]  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

[tex]z=\frac{237-247}{\frac{47}{\sqrt{30}}}=-1.165[/tex]  

P-value  

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

[tex]p_v =P(z<-1.165)=0.122[/tex]  

Conclusion  

If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that the true mean is lower than 247 at 10% of significance.

Answer:

1) n = 39916800

2) n = 1663200

3) n = 330

Step-by-step explanation:

1) If the blue balls are distinguishable as are the red balls

Then you can arrange these balls in the following ways, we must use a permutation

In totally we have 11 balls, then

n = 11P11

[tex]n = \frac{11!}{(11-11)!} = 11! = 39916800\\[/tex]  

2) If Blue balls are distinguishable, but the red balls are identical

In this case, we need to do a correction due to the red balls are identical and we cannot identify the difference when we interchange two red balls

[tex]n = \frac{11!}{4!} = \frac{39916800}{24} = 1663200[/tex]

3) If the balls of each color are indistinguishable

We proceed equal to the before case but we include a correction due to blue balls also

[tex]n = \frac{11!}{4!*7!} = \frac{39916800}{24*5040} = 330[/tex]

Complete Question

Experiments on learning in animals sometimes measure how long it takes a gerbil to nd its way through a maze. The mean time is 14 seconds for one particular maze. A researcher thinks that playing soothing music will cause the gerbils to complete the maze slower. She measures how long each of 34 gerbils takes with a noise stimulus.

She measures how long each of 34 gerbils takes with a noise stimulus. The sample mean is x = 16.5 seconds.

The alternative hypothesis for the significance test is

a. Ha:μ≠14

b. Ha:μ=16.5.

c. Ha:μ>14

Answer:

Option C is correct.

The alternative hypothesis for the significance test is Ha:μ>14

Step-by-step explanation:

The researcher sets up a maze and notes that the mean time to complete the maze is 14 seconds for gerbils.

She then theorizes that playing soothing music for the gerbils while they complete the maze slows the gerbils down.

Using a sample of 34 gerbils, she tests her hypothesis, and truly, the sample mean, when she plays soothing noise in the background, rises from 14 to 16.5 seconds; indicating that the noise stimulus indeed, slows down the gerbils.

Alternative hypothesis theorizes that there is a relationship between two variables. A relationship that is significant enough statistically, that when the hypothesis is introduced, it affects the result of the experiment just in the way that the conductors of the experimemt have predicted.

So, if the alternative hypothesis for this 'gerbils completing maze experiment' is represented by Ha, just like the researcher theorizes, when Ha is introduced, the gerbils become slower and the mean time for completing the maze is evidently higher than the mean.

Mathematically, it can be written as

Ha:μ>14

Again, this is interpreted as, when Ha is introduced (soothing music), the mean time for the gerbils completing the maze is higher than the original mean (14).

Hope this Helps!!!

Answer:

Margin of error = 0.01344

Step-by-step explanation:

Margin of error = critical value × standard deviation.

critical value for 95% confidence interval = 1.96

Standard deviation = √[(p)(q)/n]

p = 0.4, q = 1 - p = 1 - 0.4 = 0.6, n = 5100

Standard deviation = √[(0.6×0.4)/5100] = 0.00686

Margin of error = 1.96 × 0.00686 = 0.0134

Answer:

0.8753

Step-by-step explanation:

Total Population=194+142+89=425

Population of those who smoke=48+33+20=101

Population of non smokers=425-101=324

Population of those in 18-22 bracket=194

Population of nonsmoker's in 18-22 age bracket=194-48=146

Probability of getting someone who is age​ 18-22 = [TeX]\frac{194}{425}[/TeX]

Probability of getting someone who does not smoke = [TeX]\frac{324}{425}[/TeX]

Let the event that the person is in age bracket 18-22 be A

Let the event that the person does not smoke be B

Therefore, the probability of getting someone who is age​ 18-22 or does not smoke

=Probability of getting someone who is age​ 18-22 + Probability of getting someone who does not smoke - Probability of those who do not smoke and are in the age bracket 18-22

P(A or B)=P(A)+P(B)-P(A and B)

=[TeX]\frac{194}{425}+\frac{324}{425}-\frac{146}{425}[/TeX]

[TeX]=\frac{194+324-146}{425}=\frac{372}{425}=0.8753[/TeX]

In simplifying complex fractions, first, put the individual fractions in simplest terms. Solve the equations based on the order of operations, simplify the fractions, find a common denominator, and combine the fractions. Your skill in Algebraic operations, particularly with polynomials and fractions, play a significant role in solving the problem.

When it comes to simplifying complex fractions, you need to first ensure that the individual fractions are in their simplest terms. For the fraction n-4/n², consider factoring out n from the denominator, while for 2n-15/n +1/n +3, the expression can be simplified to (2n-15+n+3)/n or (3n-12)/n.

Following an order of operations, evaluate the expressions and perform the indicated operations for a simple and correct answer. Briefly, your steps should include: simplifying the complex fractions, finding a common denominator, and combining the fractions. Once you have done this, continue to simplify until you are left with one fraction.

Also note that a key aspect to solving this is applying your knowledge of Algebraic Operations, specifically with polynomials and fractions. It's important to take care not to make any mistakes in signs or operations as they can significantly alter your responses.

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The simplified form is (3n² - n + 11)/n².

To simplify the complex fraction (n - 4)/(n²) - (2n - 15)/n + 1/n + 3, follow these steps:

Find a common denominator for all the fractions involved. Here, the common denominator will be n².

1. Rewrite each term with the common denominator:

(n - 4)/n²

(2n - 15)/n becomes (2n - 15)n/n²

1/n becomes n/n²

3 becomes 3n²/n²

2. Combine all terms to have the same denominator:

(n - 4 - 2n + 15 + n + 3n²)/n²

3. Simplify the numerator:

(n - 4 - 2n + 15 + n + 3n²)

Combine like terms: (3n² + n - 2n + n - 4 + 15)

4. Simplifies to: (3n² - n + 11)

5. Rewrite the fraction:

(3n² - n + 11)/n²

So, the simplified fraction is (3n² - n + 11)/n².

Answer:

yes, he was effectively a .400 hitter

Step-by-step explanation:

The mean rate of success ( ÿ ) is 0.394.

The total number of hits was 419.

Under table t the  value at 5% confidence interval is 1.96.  

Here se the asymptotic standard deviation is given as follows

se = [tex]\sqrt{\frac{y(1 - y)}{n} }[/tex]

     = [tex]\sqrt{\frac{0.394(1 - 0.394)}{419} }[/tex]

    = 0.02387  

The confidence interval is calculated as follows:  

y ± 1.96 * se

0.394 ± 1.96  * 0.02387  

the answer is  -0.347 or  0.4407)  

meaning that the mean lies between -0.347 to 0.4407.  

There is strong evidence that 0.4 lies between the above confidence interval.

This means that  Gwynn was effectively a 0.4 hitter

Tony Gwynn's batting average was .394 with 165 hits in 419 at bats. He would have needed approximately 3 more hits to officially reach a .400 average. Despite his impressive performance, it is factually incorrect to label him a .400 hitter for that season.

To test the hypothesis that Tony Gwynn was a .400 hitter in 1994 before the strike ended the season, we need to see if his actual batting average could be considered close to .400 even though it was .394.

Gwynn had 165 hits in 419 at bats, which gives a batting average of 165/419, or approximately .394 when rounded to three decimal places. To have a .400 average, he would need to have 419 * .400 hits, which equals 167.6. Therefore, he needed approximately 3 more hits (rounding up from 2.6) to achieve a .400 average with the same number of at bats.

Given that batting averages are not rounded up after the season ends, stating that Gwynn was a .400 hitter would be incorrect. However, his achievement of .394 is remarkably high, and some may argue that rounding up for discussion purposes could informally suggest he was 'effectively' a .400 hitter.

Answer:

D.

The IQR for Week 1 is 31, and the IQR for Week 2 is 25.

Step-by-step explanation:

Answer: D.The IQR for week 1 is 31 and the IQR for week 2 is 25

Step-by-step explanation: When you divid a data set in groups of 4 and measure the bulk of the values, this is called Interquatile Range (IQR).

It's calculated as follows:

1) Put the data in order;

week 1 : 39 50 58 63 72 81 104

week 2 : 34 54 62 68 72 79 110

2) Find the median of each data set:

*Note: Median is the middle value of a set.

week 1 : Median 63

week 2 : Median 68

3) Find Q1, which is the median in the lower half of the set:

week 1: the lower set is 39 50 58.

The middle value is 50.

So Q1 = 50.

week 2: the lower set is 34 54 62

The middle value is 54.

So Q1 = 54.

4) Find Q3, which is the middle value of the upper half:

week 1 : the upper half is 72 81 104

Q2 = 81

week 2 :  the upper half is 72 79 110

Q2 = 79

5) To determine IQR, subtract Q1 and Q3:

week 1: 81 - 50 = 31

week 2 : 79 - 54 = 25

In conclusion, the IQR for week 1 is 31 and for week 2 is 25.