a consumer magazine counts the number of tissues per box in a random sample of 15 boxes of No- Rasp facial tissues. The sample standard deviation of the number of tissues per box is 97. Assume that the population is normally distributed. What is the 95% confidence interval for the population variance of the number of tissues per box?

Given:

Given that the graph OACE.

The coordinates of the vertices OACE are O(0,0), A(2m, 2n), C(2p, 2r) and E(2t, 0)

We need to determine the midpoint of EC.

Midpoint of EC:

The midpoint of EC can be determined using the formula,

[tex]Midpoint=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})[/tex]

Substituting the coordinates E(2t,0) and C(2p, 2r), we get;

[tex]Midpoint=(\frac{2t+2p}{2},\frac{0+2r}{2})[/tex]

Simplifying, we get;

[tex]Midpoint=(\frac{2(t+p)}{2},\frac{2r}{2})[/tex]

Dividing, we get;

[tex]Midpoint=(t+p,r)[/tex]

Thus, the midpoint of EC is (t + p, r)

Hence, Option A is the correct answer.

Answer:q(x)=-5(x+10)2-1

t(x)=-2x2-4x-3

Step-by-step explanation:

The functions that have a maximum and are transformed to the left and down of the parent function, f(x) = x2 include:

It should be noted that a function simply means a rule the shows the relationship between the variables. The variables are the dependent and the independent variables.

In order to determine whether the function will have a minimum or a maximum depending on the coefficient of the x² term. When the x² coefficient is positive, the function has a minimum and when it is negative, the function has a maximum.

In this case, the above functions have a maximum and are transformed to the left and down of the parent function, f(x) = x2.

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Answer:

.183 give me brainliest

Step-by-step explanation:

Answer:

Yes. There is enough evidence to support the claim that the GSR for all scholarship athletes in this division differs from 62%.

Step-by-step explanation:

We have to perform a hypothesis test on a proportion.

The claim is that the GSR for all scholarship athletes in this division differs from 62%. Then, the null and alternative hypothesis are:

[tex]H_0: \pi=0.62\\\\H_a:\pi<0.62[/tex]

The significance level, named here as Type I error rate, is 0.05.

The sample size is n=500.

The sample proportion is:

[tex]p=X/n=285/500=0.57[/tex]

The standard deviation of the proportion is:

[tex]\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.62(1-0.62)}{500}}=\sqrt{0.0004712}=0.022[/tex]

The z-statistic is then:

[tex]z=\dfrac{p-\pi+0.5/n}{\sigma_p}=\dfrac{0.57-0.62+0.5/500}{0.022}=\dfrac{-0.049}{0.022} = -2.227[/tex]

The P-value for this left tailed test is:

[tex]P-value=P(z<-2.227)=0.013[/tex]

The P-value is smaller than the significance level, so the effect is significant. The null hypothesis is rejected.

There is enough evidence to support the claim that the GSR for all scholarship athletes in this division differs from 62%.

Answer:

[tex]z=\frac{0.57 -0.62}{\sqrt{\frac{0.62(1-0.62)}{500}}}=-2.303[/tex]  

[tex]p_v =2*P(z<-2.303)=0.0213[/tex]  

So the p value obtained was a very low 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 reject the null hypothesis, and we can said that at 5% of significance the proportion of student athletes who graduate within 6 years is significantly different from 0.62 or 62%

Step-by-step explanation:

Data given and notation

n=500 represent the random sample taken

X=285 represent the student athletes who graduate within 6 years

[tex]\hat p=\frac{285}{500}=0.57[/tex] estimated proportion of student athletes who graduate within 6 years

[tex]p_o=0.62[/tex] is the value that we want to test

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

Confidence=95% or 0.95

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 true proportion differs from 0.62.:  

Null hypothesis:[tex]p=0.62[/tex]  

Alternative hypothesis:[tex]p \neq 0.62[/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  

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

[tex]z=\frac{0.57 -0.62}{\sqrt{\frac{0.62(1-0.62)}{500}}}=-2.303[/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 provided [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

[tex]p_v =2*P(z<-2.303)=0.0213[/tex]  

So the p value obtained was a very low 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 reject the null hypothesis, and we can said that at 5% of significance the proportion of student athletes who graduate within 6 years is significantly different from 0.62 or 62%

Answer:

D

Step-by-step explanation:

If you work backwards and add everything up it works.

Answer:

D

Step-by-step explanation:

-11/30

Step-by-step explanation: Create a common denominator (30) and then subtract

Answer:

Dilation

Step-by-step explanation:

The type of transformation that will produce a similar, but not congruent figure is a dilation. A dilation is a transformation , with center O and a scale factor of k that is not zero, that maps O to itself and any other point P to P'.

Affine Transformation and Similarity Transformation are essential in creating images that are similar but not congruent. Linear transformations play a role in maintaining the properties of lines and parallelism in geometric transformations.

Affine Transformation is a type of transformation that can result in an image that is similar but not congruent to the pre-image. It involves accommodating differences in scale, rotation, and offset along each dimension of the coordinate systems.

A similarity transformation can also be used, which involves a rotation with an angle, scale change, and translation. It preserves the shape but not necessarily the size.

Linear transformations, as in the case of similar transformations, are essential in transforming lines into lines and preserving parallel lines. These transformations play a crucial role in mathematical concepts related to geometry and spatial transformations.

Answer:

a) 0.3012 = 30.12% probability that the demand will exceed 120 cfs during the early afternoon on a randomly selected day.

b) 240.79 cfs.

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

[tex]f(x) = \mu e^{-\mu x}[/tex]

In which [tex]\mu = \frac{1}{m}[/tex] is the decay parameter.

The probability that x is lower or equal to a is given by:

[tex]P(X \leq x) = \int\limits^a_0 {f(x)} \, dx[/tex]

Which has the following solution:

[tex]P(X \leq x) = 1 - e^{-\mu x}[/tex]

The probability of finding a value higher than x is:

[tex]P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}[/tex]

In this problem, we have that:

[tex]m = 100, \mu = \frac{1}{100} = 0.01[/tex]

a. Find the probability that the demand will exceed 120 cfs during the early afternoon on a randomly selected day.

This is [tex]P(X > 120)[/tex]

[tex]P(X > 120) = e^{-0.01*120} = 0.3012[/tex]

0.3012 = 30.12% probability that the demand will exceed 120 cfs during the early afternoon on a randomly selected day.

b. What water-pumping capacity, in cubic feet per second, should the station maintain during early afternoons so that the probability that demand will exceed capacity on a randomly selected day is only 0.09?

We want x for which

[tex]P(X > x) = 0.09[/tex]

So

[tex]e^{-0.01x} = 0.09[/tex]

[tex]\ln{e^{-0.01x}} = \ln{0.09}[/tex]

[tex]-0.01x = \ln{0.09}[/tex]

[tex]0.01x = -\ln{0.09}[/tex]

[tex]x = -\frac{\ln{0.09}}{0.01}[/tex]

[tex]x = 240.79[/tex]

So 240.79 cfs.

The probability that the demand will exceed 120 cfs is approximately 30.12%. To ensure that the demand won't exceed capacity on 91% of early afternoons, the water-pumping station should maintain a capacity of approximately 230 cfs.

The mean (λ) of the exponential distribution equals the rate (1/λ), which in this case is 100 cfs. To find the probability that the demand will exceed 120 cfs, we need to calculate the cumulative distribution function (CDF) for 120 cfs and subtract it from 1. The formula for the CDF is F(x) = 1 - e^(-λx). Replacing x with 120 and λ with 1/100, we get: F(120) = 1 - e^(-120/100) = 1 - e^-1.2. The value of e^-1.2 is approximately 0.3012. Thus, F(120) = 1 - 0.3012 = 0.6988. Therefore, the probability that the demand will exceed 120 cfs is 0.3012 or 30.12%, rounded to four decimal places.

We want to find the volume of water (x) such that the probability that the demand will exceed x is 0.09. To do this, we set F(x) = 1 - 0.09 (or 0.91), and use the CDF formula: F(x) = 1 - e^(-λx). Solving the equation 0.91 = 1 - e^(-x/100) for x yields x = -100ln(1 - 0.91) cfs, which when calculated equals 230 cfs, rounded to two decimal places. Therefore, the water-pumping capacity that should be maintained during early afternoons is approximately 230 cfs.

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Parameter:

Null hypothesis:  μ = 1.5 (the machines work as needed)

Alternative hypothesis: μ ≠ 1.5 (The machines don't work properly)

Since we don't know the population deviation, we will apply a t-test to compare the actual mean to the reference value

Conditions:

Simple random sample: The problem states the sample was chosen at random.

Independence: You can assume there are more than 10(200) = 2000 screws.

Normality: (200 ≥ 30)  the sample is large enough for sampling distribution to assume Normality

Calculations:

Since the conditions are met we will carry out a T-test using a calculator for μ≠μ0

μ = population mean = 1.5

σ= standard dev = 0.204

xbar = sampe mean = 1.521

n = sample size= 200

After adding all of this data into the calculator in the T-test program we get a p-value of 0.147

Conclusion:

We will assume a 0.05 sig level for our conclusion.

Since 0.147 > 0.05 we will fail to reject the null hypothesis meaning that we have enough evidence to show that the machines work as needed.

Given the average length and standard deviation of the manufactured bolts, the machine might require recalibration since the lengths produced are slightly larger than desired, and there's a wide spread in lengths. Application of the empirical rule can provide further insight about the need for machine repair. A larger sample size might give a more accurate assessment.

Given that the machine is set to manufacture bolts of 1.5 inches and a random sample of 200 bolts showed an average length of 1.521 inches with a standard deviation of 0.204 inches, it seems the machine may not be calibrated correctly. The average length is slightly larger than the desired length, a factor that may be important depending on the tolerances required for these bolts. The standard deviation is also relatively high, implying that there is a wide spread in the lengths of bolts being produced.

One way to determine if the machine needs to be repaired or not is to apply the empirical rule, also known as the 68-95-99.7 rule, which says approximately 68% of data falls within one standard deviation from the mean, 95% falls within two standard deviations, and 99.7% falls within three standard deviations in a normal distribution. In this case, it means 95% of the bolts should fall between 1.113 inches (1.521 - 2*0.204) and 1.929 inches (1.521 + 2*0.204). If these lengths are acceptable for the operation, the machine can continue working. If not, it might need to be shut down for repair.

Also, it's also worth noting that a larger sample size could provide a more accurate assessment of whether the machine is working correctly or not. While a sample size of 200 is decent, a larger sample size would reduce the margin of error from sample to population.

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Answer:

35 miles

Step-by-step explanation:

70miles per hour x1/2 hour =35distance the car travels

Answer:

Correct option: (b).

Step-by-step explanation:

Effect size (η) is a statistical measure that determines the strength of the association (numerically) between two variables.  For example, if we have data on the weight of male and female candidates and we realize that, on average, males are heavier than females, the difference between the weight of male and the weight of female candidates is known as the effect size.

The larger the effect size, the larger the weight difference between male and female will be.  

Statistic effect size helps us in analyzing if the difference is factual or if it is affected by a change of factors.  

In hypothesis testing, effect size, power, sample size, and critical significance level are related to each other.

The effect size formula for a hypothesis test of mean difference is:

[tex]\eta =\frac{\bar x_{1}-\bar x_{2}}{\sqrt{s^{2}}}[/tex]

The denominator s² is combined sample variance.

[tex]s^{2}=\frac{n_{1}s_{1}^{2}+n_{2}s_{2}^{2}}{n_{1}+n_{2}-2}[/tex]

The effect size is affected by two components:

As the sample mean difference is directly proportional to the effect size, on increasing the sample mean difference value the effect size will also increase.

Ans the sample variance is inversely proportional to the the effect size, on decreasing the sample variance value the effect size will increase and vice-versa.

Thus, the correct option is (b).

Given:

Given that the graph of the system of equations.

We need to determine the solution to the system of equation.

Solution:

The solution to the system of equations is the point of intersection of these two lines.

The point of intersection of the two lines in the graph is the point at which the two lines meet.

From the graph, it is obvious that the two lines intersect at a common point.

Thus, the common point is the point of intersection of the two lines.

Hence, the point of intersection is (4,2)

Thus, the solution to the system of equation is (4,2)

Therefore, Option B is the correct answer.

Answer:

its B (4,2)

Step-by-step explanation:

sin S = [tex]$\frac{48}{73}[/tex] is the ratio found.

Step-by-step explanation:

It is given that m∠U = 90°

TS is the hypotenuse = 73 units

UT is the adjacent side of the right angle = 48 units

SU is the base of the triangle = 55 units

Now we have to find the ratio as,

sin S = [tex]$\frac{opp}{hyp}[/tex]

sin S = [tex]$\frac{UT}{TS}[/tex]

Plugin the values, we will get,

sin S = [tex]$\frac{48}{73}[/tex]

So the ratio was found.

Answer:

73/48

Step-by-step explanation:

he put it backwards

Answer:

i assume 20 dollars per haircut

Step-by-step explanation:

i do not know what the bottom numbers stand for because there is no label

Answer:

The Answer is B. $20 per haircut.

Step-by-step explanation:

Answer:

No, the citrus grower shouldn't spend the $5000 and thereby reduce the profit to $95,000 as the expected profit from doing nothing to protect the citrus plants ($96,000) is more than the profit that'll be available if $5,000 is spent on protection.

Step-by-step explanation:

First of, we compute the probability distribution of X.

X represents the profit if the citrus grower does nothing to protect the citrus fruits.

If the citrus grower does nothing, there are two possibilities as to what will happen.

1) The temperatures can drop below freezing point at a chance of 10% and the profit plummets to $60,000

2) The temperature can remain mild at a chance of 90% (100%-10%) and the profit stays at $100,000.

The probability distribution will then be

X ||||||||||||||| P(X)

60,000 ||| 0.10

100,000 | 0.90

The expected value of any probability distribution is given as

E(X) = Σ xᵢpᵢ

xᵢ = each variable

pᵢ = probability of each variable

E(X) = (60,000×0.10) + (100,000×0.90)

= 6,000 + 90,000 = $96,000

The expected amount of profits from doing nothing to protect the citrus fruits = $96,000

The expected amount of profits expected from spending $5,000 to protect the citrus fruits = $95,000

$96,000 > $95,000

Hence, the citrus grower is better off doing nothing to protect the citrus fruits.

Hope this Helps!!!

Answer:

the answer will be 0.8

Step-by-step explanation:

hard to explain

Answer:

C. 0.8

Step-by-step explanation:

Answer:

8, 7.92, 2.81

Step-by-step explanation:

For each Social Security recipient, there are only two possible outcomes. Either they are too young to vote, or they are not. The probability of a Social Security recipient is independent of any other Social Security recipient. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The variance of the binomial distribution is:

[tex]V(X) = np(1-p)[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

In this problem, we have that:

[tex]n = 800, p = 0.01[/tex]

So

Mean:

[tex]E(X) = np = 800*0.01 = 8[/tex]

The variance of the binomial distribution is:

[tex]V(X) = np(1-p) = 800*0.01*0.99 = 7.92[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{800*0.01*0.99} = 2.81[/tex]

Formatted answer: 8, 7.92, 2.81

Answer:

(a) The 95% confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football is (0.34, 0.40).

(b) Not reasonable.

Step-by-step explanation:

The information provided is:

n = 1000

[tex]\hat p[/tex] = 0.37

(a)

The (1 - α)% confidence interval for the population proportion p is:

[tex]CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]

Here,

[tex]\hat p[/tex] = sample proportion

n = sample size

[tex]z_{\alpha/2}[/tex] = critical value of z.

Compute the critical value of z for 95% confidence interval as follows:

[tex]z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96[/tex]

*Use a z-table for the value.

Compute the 95% confidence interval for the population proportion p as follows:

[tex]CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]

     [tex]=0.37\pm 1.96\times\sqrt{\frac{0.37(1-0.37)}{1000}}[/tex]

     [tex]=0.37\pm 0.03\\=(0.34, 0.40)[/tex]

Thus, the 95% confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football is (0.34, 0.40).

(b)

Now we need to determine whether it is reasonable to believe that the actual percent of people in the United States whose favorite sport to watch on television is American football is 33%.

The hypothesis can be defined as:

H₀: The percentage of people in the United States whose favorite sport to watch on television is American football is 33%, i.e. p = 0.33.

Hₐ: The percentage of people in the United States whose favorite sport to watch on television is American football is different from 33%, i.e. p ≠ 0.33

The hypothesis can be tested based on a confidence interval.

The decision rule:

If the (1 - α)% confidence interval includes the null value of the test then the null hypothesis will not be rejected. And if the (1 - α)% confidence interval includes the null value of the test then the null hypothesis will be rejected.

The 95 confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football is (0.34, 0.40).

The confidence interval does includes the null value of p, i.e. 0.33.

So, the null hypothesis will be rejected.

Hence, concluding that is is not reasonable to believe that 33% is the actual percent of people in the United States whose favorite sport to watch on television is American football.

95% confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football is (0.34, 0.40).

Concluding that is is not reasonable to believe that 33% is the actual percent of people in the United States whose favorite sport to watch on television is American football

Given that,

A recent survey collected information on television viewing habits from a random sample of 1,000 people in the United States.

Of those sampled, 37 percent indicated that their favorite sport to watch on television was American football.

We have to determine,

Construct and interpret a 95 percent confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football.

According to the question,

Sample proportion p = 37% = 0.37

Sample space n = 1000

[tex]C.I. = P \pm Z_\frac{\alpha}{2} \sqrt{\dfrac{p(1-p)}{n} }[/tex]

To compute the critical value of z for 95% confidence interval as follows:

[tex]z_\frac{ \alpha}{2} = z_\frac{0.05}{2} = 1.96[/tex]

By using a z-table for the value.

Compute the 95% confidence interval for the population proportion p as follows:

[tex]C.I. = p\pm Z_\frac{\alpha}{2} \sqrt{\dfrac{p(1-p)}{n} }\\\\C.I. = 0.37\pm 1.96 \sqrt{\dfrac{0.43(1-0.34)}{1000} }\\\\C.I.= 0.03 \pm 0.09\\\\C.I. = (0.34, \ 0.40)[/tex]

Hence,  95% confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football is (0.34, 0.40).

H₀: The percentage of people in the United States whose favorite sport to watch on television is American football is 33%, i.e. p = 0.33.

Hₐ: The percentage of people in the United States whose favorite sport to watch on television is American football is different from 33%, i.e. p ≠ 0.33

The hypothesis can be tested based on a confidence interval.

The (1 - α)% confidence interval includes the null value of the test then the null hypothesis will not be rejected.

And if the (1 - α)% confidence interval includes the null value of the test then the null hypothesis will be rejected.

The 95 confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football is (0.34, 0.40).

The confidence interval does includes the null value of p, i.e. 0.33.

So, the null hypothesis will be rejected.

Hence, Concluding that is is not reasonable to believe that 33% is the actual percent of people in the United States whose favorite sport to watch on television is American football

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Answer:

Cross section

Step-by-step explanation:

Cross section refers to the new two dimensional face exposed when we slice through a three dimensional objects.

It can also be the surface or shape exposed by making a straight cut through something, especially at right angles to an axis.

Cross section is the plane surface(two-dimensional objects) formed by cutting across a solid shape (three-dimensional shape) especially perpendicular to its longest axis.

Given:

Given that AC and BD are chords of the circle.

The two chords intersect at the point E which makes an angle 93°

The measure of arc BC is 161°

We need to determine the measure of arc AD.

Measure of arc AD:

The measure of arc AD can be determined using the property that "if two chords intersect in the interior of the circle, then the measure of each angle is half the sum of the arcs intercepted by the angles and its vertical angle".

Thus, applying the above theorem, we have;

[tex]m \angle E=\frac{1}{2}(m \widehat{BC}+m \widehat{AD})[/tex]

Substituting the values, we have;

 [tex]93^{\circ}=\frac{1}{2}(161^{\circ}+m \widehat{AD})[/tex]

[tex]186^{\circ}=161^{\circ}+m \widehat{AD}[/tex]

 [tex]25^{\circ}=m \widehat{AD}[/tex]

Thus, the measure of arc AD is 25°

Arc angle AD is 25 degrees

Secant are lines that intersect a circle at two points.

Secant AC intersect secant BD at angle 93 degree.

Using secant rule , in circle theorem,

Therefore,

93° = 1 / 2(AD + 161)

93 = AD / 2 + 161 / 2

93 =  AD + 161/ 2

cross multiply

186 = AD + 161

AD = 186 - 161

arc AD = 25°

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Answer:

it 1595

Step-by-step explanation:

For the reduced image with a scale factor of 1/25, the model height is 58 blocks; with a 1/50 scale, it's 29 blocks.

To fill in the missing information, we can use the scale factor to calculate the model height for the reduced image.

For the reduced image with a scale factor of [tex]\( \frac{1}{25} \)[/tex], we can calculate the model height by dividing the actual height by the scale factor:

[tex]\[ \text{Model Height} = \frac{\text{Actual Height}}{\text{Scale Factor}} \][/tex]

[tex]\[ \text{Model Height} = \frac{1450}{25} = 58 \text{ blocks} \][/tex]

For the reduced image with a scale factor of [tex]\( \frac{1}{50} \)[/tex], we repeat the calculation:

[tex]\[ \text{Model Height} = \frac{1450}{50} = 29 \text{ blocks} \][/tex]

Now, the completed table looks like this:

|             | Original Image | Reduced Image |

|-------------|----------------|---------------|

| Actual Height (in feet) | 1,450 | 1,450 |

| Model Height (in blocks) | - | 58 (1/25 scale) |

|                          | - | 29 (1/50 scale) |

Thus, the missing information in the table has been filled in using the scale factor and calculations based on the actual height of the Empire State Building.

Answer:

4×107 = 428

428-(5×4) =

428-20= 408

Answer:

20

Step-by-step explanation:

as u have to divide the answr and convert it

Answer:

62 is the answer because

15x4=60 + 1/2x4 = 62

Answer:

a) 95% confidence interval estimate of the mean of the population of differences between hospital admissions = (1.69, 11.91)

b) This confidence interval shows there is indeed a significant difference between the number of hospital admissions from motor vehicle crashes on Friday the 13th and the number of hospital admissions from motor vehicle crashes on Friday the 6th as the interval obtained doesn't contain a zero-value of difference.

Hence, the claim that when the 13th day of a month falls on a​ Friday, the numbers of hospital admissions from motor vehicle crashes are not affected is not true.

Step-by-step explanation:

The missing data from the question

The numbers of hospital admissions from motor vehicle crashes

Friday the 6th || 10 | 8 | 4 | 4 | 2

Friday the 13th | 12 | 10 | 12 | 14 | 14

The differences can then be calculated (number on the 13th - number on the 6th) and tabulated as

Friday the 6th || 10 | 8 | 4 | 4 | 2

Friday the 13th | 12 | 10 | 12 | 14 | 14

Differences ||| 2 | 2 | 8 | 10 | 12

To obtain the confidence interval, we need the sample mean and sample standard deviation.

Mean = (Σx)/N

= (2+2+8+10+12)/5 = 6.80

Standard deviation = σ = √[Σ(x - xbar)²/N]

Σ(x - xbar)² = (2-6.8)² + (2-6.8)² + (8-6.8)² + (10-6.8)² + (12-6.8)² = 84.8

σ = √[Σ(x - xbar)²/N] = √(84.8/5) = 4.12

Confidence Interval for the population's true difference between the number of hospital admissions from motor vehicle crashes on Friday the 6th and Friday the 13th is basically an interval of range of values where the population's true difference can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample true difference) ± (Margin of error)

Sample Mean = 6.8

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error of the sample true difference)

Critical value will be obtained using the t-distribution. This is because there is no information provided for the population mean and standard deviation.

To find the critical value from the t-tables, we first find the degree of freedom and the significance level.

Degree of freedom = df = n - 1 = 5 - 1 = 4.

Significance level for 95% confidence interval

(100% - 95%)/2 = 2.5% = 0.025

t (0.025, 4) = 2.776 (from the t-tables)

Standard error of the mean = σₓ = (σ/√n)

σ = standard deviation of the sample = 4.12

n = sample size = 5

σₓ = (4.12/√5) = 1.84

95% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 6.8 ± (2.776 × 1.84)

CI = 6.8 ± 5.10784

95% CI = (1.69216, 11.90784)

95% Confidence interval = (1.69, 11.91)

b) This confidence interval shows there is a significant difference between the number of hospital admissions from motor vehicle crashes on Friday the 13th and the number of hospital admissions from motor vehicle crashes on Friday the 6th as the interval obtained doesn't contain a difference of 0.

Hope this Helps!!!

Answer:

Step-by-step explanation:

Given

Area (A) = 529[tex]\pi[/tex] square inch

radius(r)  =?

Now

we have the formula

[tex]\pi r^{2} = area[/tex]

[tex]\pi r^{2} = 529\pi[/tex]

Both pie will be cancelled and we get

[tex]r^{2} = 529[/tex]

[tex]r =\sqrt{529}[/tex]

r = 23 inch

Hope it helped:)

"The correct answer is 14 inches.

To find the radius of a circle given its area, one can use the formula for the area of a circle, which is [tex]\( A = \pi r^2 \)[/tex], where[tex]\( A \)[/tex] is the area and[tex]\( r \)[/tex] is the radius.

 Given that the area [tex]\( A \) is \( 529\pi \)[/tex] square inches, we can set up the equation:

 [tex]\[ 529\pi = \pi r^2 \][/tex]

To solve for \( r \), we can divide both sides of the equation by [tex]\( \pi \)[/tex]:

[tex]\[ r^2 = \frac{529\pi}{\pi} \][/tex]

[tex]\[ r^2 = 529 \][/tex]

Taking the square root of both sides gives us the radius:

[tex]\[ r = \sqrt{529} \][/tex]

[tex]\[ r = 23 \][/tex]

Therefore, the radius of the circle is 23 inches. However, the question states that the correct answer is 14 inches. This discrepancy arises because the square root of 529 is actually 23, not 14. It seems there was a mistake in the provided answer. The correct radius, based on the calculation, should indeed be 23 inches, not 14 inches."

Answer:

0.8926 = 89.26% probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable

Step-by-step explanation:

To solve this question, we need 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 normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

[tex]\mu = 975, \sigma = 52, n = 31, s = \frac{52}{\sqrt{31}} = 9.34[/tex]

What is the probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable?

pvalue of Z when X = 975 + 15 = 990 subtracted by the pvalue of Z when X = 975 - 15 = 960. So

X = 990

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

By the Central Limit Theorem

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

[tex]Z = \frac{990 - 975}{9.34}[/tex]

[tex]Z = 1.61[/tex]

[tex]Z = 1.61[/tex] has a pvalue of 0.9463

X = 960

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

[tex]Z = \frac{960 - 975}{9.34}[/tex]

[tex]Z = -1.61[/tex]

[tex]Z = -1.61[/tex] has a pvalue of 0.0537

0.9463 - 0.0537 = 0.8926

0.8926 = 89.26% probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable

Final answer:

The solutions to the three parts of the question use different combinatorial methods: for part (a), the stars and bars method is used; for part (b), permutations are appropriate; and for part (c), combinations with fixed capacities are needed. Additionally, probability concepts are used to calculate the chance of an instructor finding an exam with a grade below C within a certain number of tries.

Explanation:

The student's question revolves around combinatorics, which is a field of mathematics that deals with counting, both as an art and as a science. Let's break down the responses to parts (a), (b), and (c) of the question provided by the student:

Part (a): We need to determine the number of ways to distribute 60 exams among three TAs regardless of which specific exams they receive. This problem can be solved using the concept of partitions of integers or stars and bars method. The formula for distributing n indistinguishable items into k distinguishable bins is (n + k - 1)! / [n!(k - 1)!]. Here, n=60 exams, and k=3 TAs.

Part (b): If it matters which students' exams go to which TAs, we are dealing with permutations. The total ways to distribute the exams in this case is 60!, because each exam is distinct and can be assigned to each TA.

Part (c): With TAs grading at different rates with predetermined numbers of exams (25, 20, 15), we need to use combinations. This is similar to distributing indistinguishable items to distinguishable bins with fixed capacities. The number of ways to distribute the exams in this scenario is the product of combinations: 60C25 for the first TA, then 35C20 for the second TA, and the remaining 15C15 for the third TA.

To answer the other part of the student's multifaceted question related to probability, the instructor looking for an exam graded below a C: If 15% of the students get below a C, then the probability that the instructor needs to look at at least 10 exams can be found using the geometric distribution. The mathematical statement of this probability question is P(X ≥ 10), where X follows a geometric distribution with success probability p = 0.15.

The number of ways to distribute 60 exams to 3 TAs varies based on specific conditions. If only the count of exams per TA matters, there are 1891 ways. If specific exams matter, there are approximately 4.05 × 1028 ways, and if the specific quantity per TA matters, there are about 4.28 × 1016 ways.

Distribution of Exams Among TAs

Let's break down the problem into three parts:

(a) Distribution Based on Number of Exams Each TA Grades

→ This problem can be approached using the stars and bars combinatorial method. We need to distribute 60 → → indistinguishable exams to 3 TAs.

→ The formula for this is:

C(n + r - 1, r - 1) where n = 60 exams and r = 3 TAs.

C(60 + 3 - 1, 3 - 1) = C(62, 2)

→ Calculating this combination:

C(62, 2) = 62! / (2!(60!))

62! / (2! × 60!) = (62 × 61) / (2 × 1)

                       = 1891

Thus, there are 1891 ways to distribute the exams such that only the number of exams per TA matters.

(b) Distribution Where Specific Exams Matter

Now, we are interested in which specific exams go to which TA.

→ This is a permutations problem with repetition. Each of the 60 exams can go to any of the 3 TAs.

3⁶⁰

→ Calculating this value:

3⁶⁰ ≈ 4.0528564 × 10²⁸

Therefore, there are approximately 4.05 × 10²⁸ ways to distribute the specific exams to the TAs.

(c) Distribution with Specific Numbers and Specific Exams

Here, we need to distribute the exams where each TA has a predetermined number of exams (25, 20, and 15).

→ This scenario uses the multinomial coefficient:

C(60, 25, 20, 15)

→ This is calculated as:

60! / (25! 20! 15!)

→ Finding the exact value:

60! is a very large number, but using software/tools to confirm, we get the result.

Thus, there are 60! / (25! 20! 15!) ≈ 4.28 × 10¹⁶ ways to distribute the exams under these conditions.

Answer: he set the factored expressions equal to each other

Step-by-step explanation:

Answer:he set the factored expressions equal to each other.

Step-by-step explanation:

Answer:

59 inches

Step-by-step explanation:

a square has 4 equal sides

if the sum of all sides(perimeter) is 236 you can do the inverse of that and divide by 4

so your answer should be 59

Answer:

59 inches

Step-by-step explanation:

All of the 4 sides in a square are equal, and the perimeter is all the sides added together.

s+s+s+s=236

4s=236

Divide both sides by 4

s=59

Each side is 59 inches