Suppose a survey of 500 people age 18 to 34 indicated that 32.2% of them live with one or both of their parents. Calculate and interpret a confidence interval estimate for the true proportion of all people age 18 to 34 who live with one or both parents. Use a 94% confidence level. ____________________________________________________________________________________________________________________________ CHAPTER 8: FLOW CHART VIEW OF FORMULAS FOR CONFIDENCE INTERVAL

Answer:

The Manufacturing cycle​ efficiency = 69 %

Step-by-step explanation:

The waiting time is  = 72 minute

The manufacturing lead time is ( [tex]T_{lead}[/tex] )= 160 minutes

Total cycle time ( [tex]T_{c}[/tex] ) = 160 + 72 = 232 minutes

Manufacturing cycle​ efficiency =  [tex]\frac{T_{lead} }{T_{c} }[/tex]

⇒ Manufacturing cycle​ efficiency = [tex]\frac{160}{232}[/tex]

⇒ Manufacturing cycle​ efficiency = 69 %

Therefore, the Manufacturing cycle​ efficiency = 69 %

Answer:

The probability that the plane is oveloaded is P=0.9983.

The pilot should take out the baggage and send it in another plain or have less passengers in the plain to not overload.

Step-by-step explanation:

The aircraft will be overloaded if the mean weight of the passengers is greater than 163 lb.

If the plane is full, we have 41 men in the plane. This is our sample size.

The weights of men are normally distributed with a mean of 180.5 lb and a standard deviation of 38.2.

So the mean of the sample is 180.5 lb (equal to the population mean).

The standard deviation is:

[tex]\sigma=\frac{\sigma}{\sqrt{N}} =\frac{38.2}{\sqrt{41}}=\frac{38.2}{6.4} =5.97[/tex]

Then, we can calculate the z value for x=163 lb.

[tex]z=\frac{x-\mu}{\sigma}=\frac{163-180.5}{5.97}=\frac{-17.5}{5.97}= -2.93[/tex]

The probability that the mean weight of the men in the airplane is below 163 lb is P=0.0017

[tex]P(\bar X<163)=P(z<-2.93)=0.00169[/tex]

Then the probability that the plane is oveloaded is P=0.9983:

[tex]P(overloaded)=1-P(X<163)=1-0.0017=0.9983[/tex]

The pilot should take out the baggage or have less passengers in the plain to not overload.

Step-by-step explanation:

<y+60+58=180 (angle sum property)

<y+118=180

<y=180-118=62

<ACB+<ACD=180(degree of a line )

58+x=180

x=122

(btw u cant get a 100 points by answering one question)

Answer: there are 70 dimes and 40 quarters.

Step-by-step explanation:

A dime is 10 cents

A nickel is 5 cents

A quarter is 25 cents

Let x represent the number of dimes at the bottom.

Let y represent the number of quarters at the bottom.

The city park employee collected 1,850 cents in nickels, dimes, and quarters at the bottom of a wishing well. If there were 30 nickels, it means that

10x + 25y + 30 × 5 = 1850

10x + 25y = 1850 - 150

10x + 25y = 1700- - - - - - - - -1

There is a combined total of 110 dimes and quarters. It means that

x + y = 110

Substituting x = 110 - y into equation 1, it becomes

10(110 - y) + 25y = 1700

1100 - 10y + 25y = 1700

1100 + 15y = 1700

15y = 1700 - 1100

15y = 600

y = 600/15

y = 40

x = 110 - y = 110 - 40

x = 70

The question is not complete and the full question says;

Calculate the (a) range, (b) arithmetic mean, (c) mean deviation, and (d) interpret the values. Dave’s Automatic Door installs automatic garage door openers. The following list indicates the number of minutes needed to install a sample of 10 door openers: 28, 32, 24, 46, 44, 40, 54, 38, 32, and 42.

Answer:

A) Range = 30 minutes

B) Mean = 38

C) Mean Deviation = 7.2

D) This is well written in the explanation.

Step-by-step explanation:

A) In statistics, Range = Largest value - Smallest value. From the question, the highest time is 54 minutes while the smallest time is 24 minutes.

Thus; Range = 54 - 24 = 30 minutes

B) In statistics,

Mean = Σx/n

Where n is the number of times occurring and Σx is the sum of all the times occurring

Thus,

Σx = 28 + 32 + 24 + 46 + 44 + 40 + 54 + 38 + 32 + 42 = 380

n = 10

Thus, Mean(x') = 380/10 = 38

C) Mean deviation is given as;

M.D = [Σ(x-x')]/n

Thus, Σ(x-x') = (28-38) + (32-38) + (24-38) + (46-38) + (44-38) + (40-38) + (54-38) + (38-38) + (32-38) + (42-38) = 72

So, M.D = 72/10 = 7.2

D) The range of the times is 30 minutes.

The average time required to open one door is 38 minutes.

The number of minutes the time deviates on average from the mean of 38 minutes is 7.2 minutes

The average time to install an automatic garage door opener, based on the provided data, is approximately 38 minutes, though individual times can vary.

The subject of this question is Mathematics, specifically statistics. The question provides a series of data points representing the time, in minutes, that it took to install 10 automatic garage door openers. To find the average installation time, you would add up all of the times and then divide by the number of data points, which in this case is 10.

The data points are: 28, 32, 24, 46, 44, 40, 54, 38, 32, and 42. So, average installation time = (28+32+24+46+44+40+54+38+32+42) / 10 = 38 minutes.

This suggests that on average, it takes about 38 minutes to install an automatic garage door opener. But remember, this is an average. Individual installation times can vary greatly, as seen in the range of times provided.

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Answer:

-b/a and line is falling

Step-by-step explanation:

Let (-a,0) and (0,-b) be point 1 and 2, respectively. The slope passing through point (-a,0) and (0,-b) can be calculated using the following equation

[tex]m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-b - 0}{0 - (-a)} = -b/a[/tex]

As b and a are positive real number, then the slope m = -b/a is a negative real number. The line passing through these 2 points is falling.

The slope of the line passing through points (-a, 0) and (0, -b) is calculated using the formula for slope, resulting in a slope of -b/a, which indicates that the line falls as it moves from left to right.

To calculate the slope (m), we use the formula m = (y2 - y1) / (x2 - x1), with (x1, y1) = (-a, 0) and (x2, y2) = (0, -b). Substituting the values, we get m = (-b - 0) / (0 - (-a)) = -b/a. Since both a and b are positive real numbers, the sign of the slope depends on the negative numerator, indicating the line falls as it goes from left to right. If a and/or b were zero, revealing a vertical or horizontal line, a different methodology would apply, such as indicating an undefined slope for a vertical line or a zero slope for a horizontal line.

Answer:

The future value of $1,720 in 14 years assuming an interest rate of 7.25 percent compounded semiannually is $4,661.61

Step-by-step explanation:

The compound interest formula is given by:

[tex]A = P(1 + \frac{r}{n})^{nt}[/tex]

Where A is the amount of money, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per unit t and t is the time the money is invested or borrowed for.

In this problem, we have that

Semianually is twice a year, so [tex]n = 2[/tex].

Also, [tex]P = 1720, t = 14, r = 0.0725[/tex]

So

[tex]A = P(1 + \frac{r}{n})^{nt}[/tex]

[tex]A = 1720*(1 + \frac{0.0725}{2})^{2*14}[/tex]

[tex]A = 4661.61[/tex]

The future value of $1,720 in 14 years assuming an interest rate of 7.25 percent compounded semiannually is $4,661.61

The future value of $1,720 in 14 years, given a 7.25% interest rate compounded semiannually, can be calculated using the formula for compound interest: FV = PV * (1 + rate/n)^(nt). Plugging in the values, the future value comes out to be $3,490.91.

We can find the future value using the formula: Future Value = Present Value * (1 + rate/ n)^(n*t), where the Present Value is $1,720, the rate is 7.25% (or 0.0725 in decimal form), n is 2 (since it's compounded semiannually), and t, the time period, is 14 years.

So Future Value  =  $1,720 * (1 + 0.0725 / 2) ^ (2 * 14) .

Therefore, the future value of $1,720 in 14 years, with an interest rate of 7.25 percent compounded semiannually, would be $3,490.91

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Answer:

Mean and standard deviation of the sample mean are 225.8 and 3.56 respectively.

Step-by-step explanation:

The mean (μₓ) and standard deviation of the sample mean (σₓ) are related to the mean (μ) and standard deviation of the population (σ) through the following relationship

μₓ = μ = 225.8

σₓ = σ/√n = 17.8/√25 = 17.8/5 = 3.56

Answer:

Incomplete question

Complete question: in the U.S. 71.3% of those students replied that, yes, they believe that same-sex couples should have the right to legal marital status. Suppose that you randomly select freshman from the study until you find one who replies "yes." You are interested in the number of freshmen you must ask. What is the probability that you will need to ask fewer than three freshmen?

Answer: P = 0.08656

Step-by-step explanation:

Using the concept of geometric distribution

P(X=x) = p(1-p)^1-x

Where x is the number of freshmen asked and p is the probability of success

Given probability of success as 71.3%

Therefore, P = 0.713

Probability that you'll need to ask less than 3 freshmen is given as

P = p(1-p)^1-x

x = 3 and p = 0.713

P = 0.713(1-0.713)^1-3

P = 0.713(0.287)^-2

P =0.713×12.140

P = 8.656%

P = 0.08656

Answer:

[tex]z=\frac{0.45 -0.508}{\sqrt{\frac{0.508(1-0.508)}{50}}}=-0.82[/tex]  

[tex]p_v =P(z<-0.82)=0.206[/tex]  

So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to  FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of newborn boys babies is not significantly lower than 0.508

Step-by-step explanation:

Data given and notation

n=50 represent the random sample taken

[tex]\hat p=0.45[/tex] estimated proportion of newborn boys babies

[tex]p_o=0.508[/tex] is the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level

Confidence=95% or 0.95  (asumed)

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 assume that we want to check if the true proportion is less than 0.508.

Null hypothesis:[tex]p\geq 0.508[/tex]  

Alternative hypothesis:[tex]p < 0.508[/tex]  

When we conduct a proportion test we need to use the z statisitc, 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  

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

[tex]z=\frac{0.45 -0.508}{\sqrt{\frac{0.508(1-0.508)}{50}}}=-0.82[/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=0.05[/tex]. The next step would be calculate the p value for this test.  

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

[tex]p_v =P(z<-0.82)=0.206[/tex]  

So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to  FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of newborn boys babies is not significantly lower than 0.508

It is concluded that the null hypothesis H₀ is not rejected. There is not enough evidence to claim that the population proportion p is less than 0.508, at the α =0.05 significance level.

Suppose that we have:

Then, the z test statistic for one sample proportion is:

[tex]Z = \dfrac{\hat{p} - p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}[/tex]

Making the hypothesis and performing the test, assuming that we want to test if the population mean proportion is < 0.508 at the level of significance 0.05.

The following null and alternative hypotheses for the population proportion needs to be tested:

[tex]H_0: p \geq 0.508 \\\\ H_a: p & < & 0.508 \end{array}[/tex]

This corresponds to a left-tailed test, for which a z-test for one population proportion will be used.

Based on the information provided, the significance level is [tex]\alpha = 0.05[/tex], and the critical value for a left-tailed test is [tex]z_c = -1.64[/tex]

The rejection region for this left-tailed test is [tex]R = \{z: z < -1.645\}[/tex]

The z-statistic is computed as follows:

[tex]z & = & \displaystyle \frac{\hat p - p_0}{\sqrt{ \displaystyle\frac{p_0(1-p_0)}{n}}} \\\\& = & \displaystyle \frac{0.45 - 0.508}{\sqrt{ \displaystyle\frac{ 0.508(1 - 0.508)}{50}}} \\\\ & = & -0.82[/tex]

Since it is observed that [tex]z = -0.82 \ge z_c = -1.645[/tex],  it is then concluded that the null hypothesis is not rejected.

Thus, it is concluded that the null hypothesis H₀ is not rejected. Therefore, there is not enough evidence to claim that the population proportion p is less than 0.508, at the α =0.05 significance level.

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Answer:

r(t) = (e^t +1, -cos(t) + 3, tan(t) + 2)

Step-by-step explanation:

A primitive of e^t is e^t+c, since r(0) has 2 in its first cooridnate, then

e^0+c = 2

1+c = 2

c = 1

Thus, the first coordinate of r(t) is e^t + 1.

A primitive of sin(t) is -cos(t) + c (remember that the derivate of cos(t) is -sin(t)). SInce r(0) in its second coordinate is 2, then

-cos(0)+c = 2

-1+c = 2

c = 3

Therefore, in the second coordinate r(t) is equal to -cos(t)+3.

Now, lets see the last coordinate.

A primitive of sec²(t) is tan(t)+c (you can check this by derivating tan(t) = sin(t)/cos(t) using the divition rule and the property that cos²(t)+sin²(t) = 1 for all t). Since in its third coordinate r(0) is also 2, then we have that

2 = tan(0)+c = sin(0)/cos(0) + c = 0/1 + c = 0

Thus, c = 2

As a consecuence, the third coordinate of r(t) is tan(t) + 2.

As a result, r(t) = (e^t +1, -cos(t) + 3, tan(t) + 2).

To find the function r(t) that satisfies the given condition, integrate each component of r'(t) and solve for the constants using the initial condition.

To find the function r(t) that satisfies the condition r'(t) = (e^t, sin t, sec^2 t) and r(0) = (2, 2, 2), we need to integrate each component of r'(t) with respect to t.

Combining these integrals, we get r(t) = (e^t + C_1, -cos t + C_2, tan t + C_3). Substituting the initial condition r(0) = (2, 2, 2), we can solve for the constants and obtain the function r(t) = (e^t + 2, -cos t + 2, tan t + 2).

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The question given is incomplete, I googled and got the complete question as below:

You are a waterman daily plying the waters of Chesapeake Bay for blue crabs (Callinectes sapidus), the best-tasting crustacean in the world. Crab populations and commercial catch rates are highly variable, but the fishery is under constant pressure from over-fishing, habitat destruction, and pollution. These days, you tend to pull crab pots containing an average of 2.4 crabs per pot. Given that you are economically challenged as most commercial fishermen are, and have an expensive boat to pay off, you’re always interested in projecting your income for the day. At the end of one day, you calculate that you’ll need 7 legal-sized crabs in your last pot in order to break even for the day. Use these data to address the following questions. Show your work.

a. What is the probability that your last pot will have the necessary 7 crabs?

b. What is the probability that your last pot will be empty?

Answer:

a. Probability = 0.0083

b. Probability = 0.0907

Step-by-step explanation:

This is Poisson distribution with parameter λ=2.4

a)

The probability that your last pot will have the necessary 7 crabs is calculated below:

P(X=7)=  {e-2.4*2.47/7!} = 0.0083

b)

The probability that your last pot will be empty is calculated as:

P(X=0)=  {e-2.4*2.40/0!} = 0.0907

The probability of getting 7 crabs in the last pot is approximately 0.0082, while the probability of the last pot being empty is roughly 0.0907.

To solve this problem, we need to use the Poisson distribution. The Poisson distribution is a probability distribution that can be used to predict the number of events occurring within a fixed interval of time or space. Given that the average (λ) number of crabs per pot is 2.4, we can proceed to solve for the probabilities.

a. Probability of having 7 crabs in the last pot:

b. Probability of the last pot being empty:

The complete question is

You are a waterman daily plying the waters of Chesapeake Bay for blue crabs (Callinectes sapidus), the best-tasting crustacean in the world. Crab populations and commercial catch rates are highly variable, but the fishery is under constant pressure from overfishing, habitat destruction, and pollution. These days, you tend to pull crab pots containing an average of 2.4 crabs per pot. Given that you are economically challenged as most commercial fishermen are, and have an expensive boat to pay off, you’re always interested in projecting your income for the day. At the end of one day, you calculate that you’ll need 7 legal-sized crabs in your last pot in order to break even for the day

Use these data to address the following questions. Show your work.

a. What is the probability that your last pot will have the necessary 7 crabs?

b. What is the probability that your last pot will be empty?

Answer:

[tex] 28x^2 [/tex]

Step-by-step explanation:

[tex]area \:o f \: \triangle = \frac{1}{2} \times 8x \times 7x \\ = 4x \times 7x \\ = 28 {x}^{2} [/tex]

Answer:

16.35% probability their combined weight exceeds 46000 pounds.

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sample means with size n of at least 30 can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]

In this problem, we have that:

[tex]\mu = 1135, \sigma = 97, n = 40, s = \frac{97}{\sqrt{40}} = 15.34[/tex]

Find the probability their combined weight exceeds 46000 pounds.

This is 1 subtracted by the pvalue of Z when [tex]X = \frac{46400}{40} = 1150[/tex]. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{1150 - 1135}{15.34}[/tex]

[tex]Z = 0.98[/tex]

[tex]Z = 0.98[/tex] has a pvalue of 0.8365

1 - 0.8365 = 0.1635

16.35% probability their combined weight exceeds 46000 pounds.

Final answer:

The probability that the combined weight of 40 randomly selected one-year-old Hereford bulls exceeds 46,000 pounds is approximately 16.35%. This calculation uses the Z-score related to the sample mean weight exceeding 1,150 pounds.

Explanation:

To find the probability that the combined weight of 40 randomly selected one-year-old Hereford bulls exceeds 46,000 pounds, we first need to determine if the average weight per bull exceeds 1,150 pounds (since 46,000 ÷ 40 = 1,150). The mean weight of a Hereford bull is given as 1,135 pounds with a standard deviation of 97 pounds. Using the Central Limit Theorem, we can find the mean and standard deviation for the sample mean weight of 40 bulls.

The standard deviation of the sample mean (σ_x-bar) is σ ÷ √{n} = 97 pounds ÷ √{40}, which equals approximately 15.34 pounds. To find the Z-score, we use the formula Z = (X - μ) ÷ σ_x-bar, where X is 1,150 pounds, μ (the mean) is 1,135 pounds, and σ_x-bar is 15.34 pounds, yielding a Z-score of about 0.98.

Using the standard normal distribution table, a Z-score of 0.98 corresponds to a probability of approximately 0.8365. However, since we want the probability that the sample mean is greater than 1,150 pounds, we need to look at the upper tail, which is 1 - 0.8365 = 0.1635. Therefore, there's a 16.35% chance the combined weight of 40 randomly selected one-year-old Hereford bulls exceeds 46,000 pounds.

Answer:

0.6844 is the required probability.        

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = $1,250

Standard Deviation, σ = $125

We are given that the distribution of daily sales is a bell like shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

We have to find

P(sales less than $1,310)

[tex]P( x < 1310) = P( z < \displaystyle\frac{1310 - 1250}{125}) = P(z < 0.48)[/tex]

Calculation the value from standard normal z table, we have,  

[tex]P(x < 1310) =0.6844= 68.44\%[/tex]

0.6844 is the probability that sales on a given day at this store are less than $1,310.

The probability that sales on a given day at this store are less than $1,310 is [tex]68.44\%[/tex]

It is given that, mean [tex]\mu=1250[/tex] and deviation [tex]\sigma=125[/tex]

The z- score is given as,

                   [tex]z-score=\frac{x-\mu}{\sigma} \\\\z=\frac{1310-1250}{125}=0.48 \\\\[/tex]

We have to find probability that the sales on a given day at this store are less than $1,310

                [tex]P(x < 1310)=P(z < 0.48)[/tex]

From z- value table.

           [tex]P(x < 1310)=68.44\%[/tex]

The probability that sales on a given day at this store are less than $1,310 is [tex]68.44\%[/tex]

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The probability that all the answers are the same in the section of the exam is 1/8 or 12.5%.

To find the probability that all the answers are the same, we need to consider the number of ways in which all the answers can be the same, divided by the total number of possible outcomes. Since there are only two possible answers (True or False) for each question, there are a total of 2^4 = 16 possible outcomes.

Out of these 16 outcomes, there are only two ways in which all the answers can be the same: either all True or all False.

Therefore, the probability that all the answers are the same is 2/16 = 1/8 = 0.125, or 12.5%.

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Answer:

The average contents of all bottles of juice in the population

[tex]\mu = 473\text{ mL}[/tex]

Step-by-step explanation:

We are given the following in the question:

Population mean, μ =  473 mL

Sample mean, [tex]\bar{x}[/tex] = 472 mL

Sample size, n = 30

Sample standard deviation, s = 0.2

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 473\text{ mL}\\H_A: \mu < 473\text{ mL}[/tex]

Representation of [tex]\mu[/tex]

The average juice a bottle contain is the mean value of the juice.

[tex]\mathbf{\mu }[/tex] is the average content in all bottles of juice in the population, which is 472mL.

The given parameters are:

[tex]\mathbf{n = 30}[/tex] --- the sample size

[tex]\mathbf{\sigma = 0.2}[/tex] --- the standard deviation

[tex]\mathbf{\bar x = 472}[/tex] --- the sample mean

[tex]\mathbf{\mu = 473}[/tex] --- the population mean

The above highlights means that:

The parameter [tex]\mathbf{\mu }[/tex] represents the population mean

This means that:

[tex]\mathbf{\mu }[/tex] is the average content in all bottles of juice in the population.

From the question, the value is given as: 473

Hence, the true statement is:

[tex]\mathbf{\mu }[/tex] is the average content in all bottles of juice in the population, which is 473mL.

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Answer:

Step-by-step explanation:

a) We would apply the periodic interest rate formula which is expressed as

P = a/[{(1+r)^n]-1}/{r(1+r)^n}]

Where

P represents the monthly payments.

a represents the amount of the loan

r represents the annual rate.

n represents number of monthly payments. Therefore

a = $22000

r = 0.065/12 = 0.0054

n = 12 × 3 = 36

Therefore,

P = 22000/[{(1+0.0054)^36]-1}/{0.0054(1+0.0054)^36}]

22000/[{(1.0054)^36]-1}/{0.0054(1.0054)^36}]

P = 22000/{1.214 -1}/[0.0054(1.214)]

P = 22000/(0.214/0.0065556)

P = 22000/32.64

P = $674

b) The total amount that he would pay in 3 years is

674 × 36 = 24265

total interest paid over the life of the loan is

24265 - 22000 = $2265

Answer:

a)Number of terms =15

Value of sum =

(b)Number of terms =6

Value of sum  

Step-by-step explanation:

A.

nth term,

a=5, r=

Simce the bases are the same, the powers are equal.

n-1=14

n=14+1=15

Number of terms =

15

Value of sum =

B.

nth term,

a=,

Simce the bases are the same, the powers are equal.

4+n-1=9

n=9+1-4=6

Number of terms =

6

Value of sum  

NOTE: For the sum, we desire an expression that gives the exact value, rather than entering an approximation

The ODE is separable:

[tex]\dfrac{\mathrm dV}{\mathrm dt}=2\sqrt V\implies\dfrac{\mathrm dV}{2\sqrt V}=\mathrm dt[/tex]

Integrate both sides to get

[tex]\sqrt V=t+C[/tex]

Assuming the bubble starts with zero volume, so that [tex]V(0)=0[/tex], we find

[tex]\sqrt0=0+C\implies C=0[/tex]

Then the volume of the bubble at time [tex]t[/tex] is

[tex]V(t)=t^2[/tex]

and so the time it take for the bubble to reach a volume of 629 cubic cm is

[tex]t^2=529\implies t=23[/tex]

or 23 seconds after the teen first starts blowing the bubble.

The probability that the market research predicts an unsuccessful market for the product and the product is actually unsuccessful is 24 percent. This is calculated using the concept of conditional probability and multiplying the probability of the product actually failing (30%) with the probability of market research correctly predicting a failure (80%).

To calculate the probability that the market research indicates an unsuccessful market for the product and the product is actually unsuccessful, we first need to understand the concept of conditional probability, given certain conditions.

Here, there is a 30 percent chance that a product will fail. The market research incorrectly predicts success for failed products 20 percent of the time. Therefore, the market research predicts failure for failed products 80 percent of the time (100% - 20%).

By multiplying these probabilities together, we can find the joint probability that a product is unsuccessful and market research also predicts it to be unsuccessful.

The result is 0.30 x 0.80 = 0.24. Therefore, there is a 24 percent chance that the market research will predict a failure and the product will also actually be a failure.

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The probability that market research indicates an unsuccessful market and the product is unsuccessful is 24%.

To calculate the probability that the results indicate an unsuccessful market for the product and that the product is unsuccessful, we need to use Bayes' Theorem. Initially, we have the probability of a product being successful, P(Success) = 0.70, and the probability of a product being a failure, P(Failure) = 0.30. Furthermore, if the product is successful, market research predicts it successful 90% of the time. If the product is a failure, market research incorrectly predicts it as a success 20% of the time, meaning it correctly predicts failure 80% of the time (100% - 20%).

We are interested in P(Market Research Failure and Actual Failure). This is the probability that the market research indicates a failure and the product indeed fails. Using the probabilities provided:

Therefore, P(Market Research Failure and Actual Failure) can be calculated as:

P(Market Research Failure and Actual Failure) = P(Market Research Failure | Failure) x P(Failure) = 0.80 x 0.30 = 0.24 or 24%.

Answer:

C = 4T

Step-by-step explanation:

The question is not completed, the complete question and the solution is attached in the file below

Answer:

Marcus rents the car for 5 days

Step-by-step explanation:

For 5 days:

Plan A would equal $150 (30x5)

Plan B would equal $140 (125+15)

Plan b is cheaper only when Marcus rents the car for 5 days.

Answer:

[tex] t=(26-1) [\frac{0.18}{0.25}]^2 =12.96[/tex]

What is the critical value for the test statistic at an α = 0.01 significance level?

Since is a two tailed test the critical zone have two zones. On this case we need a quantile on the chi square distribution with 25 degrees of freedom that accumulates 0.005 of the area on the left tail and 0.995 on the right tail.  

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.005,25)". And our critical value would be [tex]\Chi^2 =10.520[/tex]    

And the right critical value would be :  [tex]\Chi^2 =46.927[/tex]

And the rejection zone would be: [tex] \chi^2 < 10.52 \cup \chi^2 >46.927[/tex]

Since our calculated value is NOT in the rejection zone we FAIL to reject the null hypothesis.

Step-by-step explanation:

Previous concepts and notation

The chi-square test is used to check if the standard deviation of a population is equal to a specified value. We can conduct the test "two-sided test or a one-sided test".

n = 26 sample size

s= 0.18

[tex]\sigma_o =0.25[/tex] the value that we want to test

[tex]p_v [/tex] represent the p value for the test

t represent the statistic

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

State the null and alternative hypothesis

On this case we want to check if the population standard deviation is equal to 0.25, so the system of hypothesis are:

H0: [tex]\sigma =0.25[/tex]

H1: [tex]\sigma \neq 0.25[/tex]

In order to check the hypothesis we need to calculate the statistic given by the following formula:

[tex] t=(n-1) [\frac{s}{\sigma_o}]^2 [/tex]

This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

[tex] t=(26-1) [\frac{0.18}{0.25}]^2 =12.96[/tex]

What is the critical value for the test statistic at an α = 0.01 significance level?

Since is a two tailed test the critical zone have two zones. On this case we need a quantile on the chi square distribution with 25 degrees of freedom that accumulates 0.005 of the area on the left tail and 0.995 on the right tail.  

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.005,25)". And our critical value would be [tex]\Chi^2 =10.520[/tex]

And the right critical value would be :  [tex]\Chi^2 =46.927[/tex]

And the rejection zone would be: [tex] \chi^2 < 10.52 \cup \chi^2 >46.927[/tex]

Since our calculated value is NOT in the rejection zone we FAIL to reject the null hypothesis.

At the 0.01 significance level, we fail to reject the null hypothesis, meaning there is not enough evidence to refute the claim that the standard deviation is 0.25.

To test the claim that the standard deviation of all precipitation amounts is equal to 0.25 using a 0.01 significance level, we use the chi-square test for standard deviation.

The null hypothesis (H0) states that the population standard deviation (σ) is 0.25. The alternative hypothesis (Ha) states that σ is not equal to 0.25.

The test statistic for a chi-square test is calculated using the formula:

Chi-square (χ²) = (n - 1)s² / σ₀²

Where:

Substituting the values:

χ² = (26 - 1)(0.18)² / (0.25)²

Calculating this:

χ² = 25 * 0.0324 / 0.0625 = 12.96

This χ² value is compared to the critical values from the chi-square distribution table with (n-1) or 25 degrees of freedom at a 0.01 significance level. The two-tailed critical values for 25 degrees of freedom at 0.01 significance level are approximately 10.85 and 46.93.

Since 10.85 < 12.96 < 46.93, we fail to reject the null hypothesis.

Hence, at the 0.01 significance level, there is not enough evidence to reject the claim that the standard deviation of all precipitation amounts is 0.25.

Answer:

We conclude that there is not enough evidence to support the claim that company is spending too much time on email.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 96 minutes

Sample mean, [tex]\bar{x}[/tex] = 91 minutes

Sample size, n = 100

Alpha, α = 0.05

Sample standard deviation, s = 25 minutes

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu \leq 96\text{ minutes}\\H_A: \mu > 96\text{ minutes}[/tex]

We use one-tailed t test to perform this hypothesis.

Formula:

[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{91 - 96}{\frac{25}{\sqrt{100}} } = -2[/tex]

Now, [tex]t_{critical} \text{ at 0.05 level of significance, 99 degree of freedom } = 1.6603[/tex]

Since,                        

The calculated test statistic is less than the critical value, we fail to reject the null hypothesis and accept it.

Thus, we conclude that there is not enough evidence to support the claim that company is spending too much time on email.

Answer:

[tex]P(\bar X >1650)=P(Z>\frac{1650-1637.52}{\frac{623.16}{\sqrt{400}}}=0.401)[/tex]

And we can use the complement rule and we got:

[tex]P(Z>0.401) =1-P(Z<0.401) = 1-0.656= 0.344[/tex]

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the bank account balances of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(1637.52,623.16)[/tex]  

Where [tex]\mu=1637.52[/tex] and [tex]\sigma=623.16[/tex]

Since the distribution of X is normal then the distribution for the sample mean [tex]\bar X[/tex] is given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

And we can use the z score formula given by:

[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

And using this formula we got:

[tex]P(\bar X >1650)=P(Z>\frac{1650-1637.52}{\frac{623.16}{\sqrt{400}}}=0.401)[/tex]

And we can use the complement rule and we got:

[tex]P(Z>0.401) =1-P(Z<0.401) = 1-0.656= 0.344[/tex]

Answer: the probability is 0.49

Step-by-step explanation:

Since the account balances at the large bank are normally distributed.

we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = account balances.

µ = mean account balance.

σ = standard deviation

From the information given,

µ = $1,637.52

σ = $623.16

We want to find the probability that a simple random sample of 400 accounts has a mean that exceeds $1,650. It is expressed as

P(x > 1650) = 1 - P(x ≤ 1650)

For x = 1650,

z = (1650 - 1637.52)/623.16 = 0.02

Looking at the normal distribution table, the probability corresponding to the z score is 0.51

P(x > 1650) = 1 - 0.51 = 0.49

Answer:

(a) The probability of catching 5 individuals in the pond is 0.1429.

(b) The probability of catching at least 2 individuals in the pond is 0.8838.

Step-by-step explanation:

Let X = number of water fleas caught by sweeping the water a single time.

The random variable X follows a Poisson distribution with parameter λ = 3.7.

The probability mass function of the Poisson distribution is:

[tex]P(X=x)=\frac{e^{-3.7}3.7^{x}}{x!}[/tex]

(a)

Compute the value of P (X = 5) as follows:

[tex]P(X=5)=\frac{e^{-3.7}3.7^{5}}{5!}=\frac{17.1443}{120} =0.142869\approx0.1429[/tex]

Thus, the probability of catching 5 individuals in the pond is 0.1429.

(b)

Compute the value of P (X ≥ 2) as follows:

P (X ≥ 2) = 1 - P (X < 2)

              = 1 - P (X = 0) - P (X = 1)

              [tex]=1-\frac{e^{-3.7}3.7^{0}}{0!}-\frac{e^{-3.7}3.7^{1}}{1!}\\=1-0.0247-0.0915\\=0.8838[/tex]

Thus, the probability of catching at least 2 individuals in the pond is 0.8838.

Answer: the distance between the the police station and the fire station is 4 miles.

Step-by-step explanation:

Let x represent the distance between the library and the police station.

Let y represent the distance between the the police station and the fires station.

In the city, the distance between the library and the police station is 3 miles less than twice the distance between the police station and the fire station. This is expressed as

x = 2y - 3

The distance between the library and the police station is 5 miles. This means that

5 = 2y - 3

2y = 5 + 3 = 8

y = 8/2 = 4

Answer:

3/20

Step-by-step explanation:

[tex]\frac{3}{5}[/tex]  ÷  4    =     [tex]\frac{3}{5}[/tex] [tex]* \frac{1}{4}[/tex]    =      [tex]\frac{3 * 1}{5 * 4}[/tex]     =     [tex]\frac{3}{20}[/tex]

To calculate the fraction of the class that earned an A, multiply 3/5 by 1/4, which equals 3/20. Thus, 3/20 of the class earned an A. This is the correct answer among the given options.

Calculating the Fraction of a Class That Earned an A

To find out what fraction of the class earned an A, we start with the fact that 3/5 of the class earned an A or a B.

We also know that 1/4 of those students earned an A.

To determine the fraction of the entire class that earned an A, we multiply these two fractions together:

Fraction of class that earned an A = (3/5) × (1/4)

This multiplication gives us:

(3 × 1) / (5 × 4) = 3/20

Therefore, 3/20 of the class earned an A.

This fraction cannot be reduced further and it is also one of the options provided, confirming that it is the correct answer.

Answer:

[tex](0.28-0.18) - 1.96 \sqrt{\frac{0.28(1-0.28)}{200} +\frac{0.18(1-0.18)}{200}}=0.0181[/tex]  

[tex](0.28-0.18) - 1.96 \sqrt{\frac{0.28(1-0.28)}{200} +\frac{0.18(1-0.18)}{200}}=0.182[/tex]  

And the 95% confidence interval would be given (0.0181;0.181).  

We are confident at 95% that the difference between the two proportions is between [tex]0.0181 \leq p_B -p_A \leq 0.182[/tex]

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Solution to the problem

[tex]p_A[/tex] represent the real population proportion for telephone  

[tex]\hat p_A =0.28[/tex] represent the estimated proportion for telephone

[tex]n_A=200[/tex] is the sample size required for telephone

[tex]p_B[/tex] represent the real population proportion for mail

[tex]\hat p_B =0.18[/tex] represent the estimated proportion for mail

[tex]n_B=200[/tex] is the sample size required for mail

[tex]z[/tex] represent the critical value for the margin of error  

The population proportion have the following distribution  

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]  

The confidence interval for the difference of two proportions would be given by this formula  

[tex](\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}[/tex]  

For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.  

[tex]z_{\alpha/2}=1.96[/tex]  

And replacing into the confidence interval formula we got:  

[tex](0.28-0.18) - 1.96 \sqrt{\frac{0.28(1-0.28)}{200} +\frac{0.18(1-0.18)}{200}}=0.0181[/tex]  

[tex](0.28-0.18) - 1.96 \sqrt{\frac{0.28(1-0.28)}{200} +\frac{0.18(1-0.18)}{200}}=0.182[/tex]  

And the 95% confidence interval would be given (0.0181;0.181).  

We are confident at 95% that the difference between the two proportions is between [tex]0.0181 \leq p_B -p_A \leq 0.182[/tex]

The 95% confidence interval for the difference in the proportions of people who make donations if contacted by telephone or first class mail is between 1.14% and 18.86%.

To calculate a 95% confidence interval for the difference in the proportions of people who make donations if contacted by telephone or first class mail, we use the formula for comparing two proportions.

We are given that 28% of the 200 contacted by telephone donated, which is a proportion of p1 = 0.28, and 18% of the 200 contacted by first class mail donated, which is a proportion of p2 = 0.18. Each group size is n1 = n2 = 200.

The formula for the standard error of the difference between two independent proportions is:

SE = sqrt(p1(1-p1)/n1 + p2(1-p2)/n2)

So, the standard error (SE) for our example is:

SE = sqrt(0.28(1-0.28)/200 + 0.18(1-0.18)/200) = sqrt(0.001248 + 0.000792) = sqrt(0.00204)

The Z-score for a 95% confidence interval is approximately 1.96. Therefore, the margin of error (MOE) is calculated as:

MOE = 1.96 * SE

We then compute the confidence interval as:

(p1 - p2) ± MOE = (0.28 - 0.18) ± 1.96 * sqrt(0.00204)

Calculate the MOE:

MOE = 1.96 * sqrt(0.00204) ≈ 1.96 * 0.0452 ≈ 0.0886

Then the 95% confidence interval is:

(0.10 ± 0.0886) which is (0.0114, 0.1886)

This means that we are 95% confident that the true difference in the proportion of people who would donate when contacted by telephone versus first class mail is between 1.14% and 18.86%.