Identify the asymptotes of each function. 1/(x-3) - 6

Answer:

Reject null hypothesis

Step-by-step explanation:

a) [tex]H_0: p =0.05\\H_a: p >0.05[/tex]

(Right tailed test at 5% level)

p = risk proportion

Sample proportion = [tex]\frac{46}{384} \\\\=0.1198[/tex]

Std error = [tex]\sqrt{\frac{pq}{n} } \\=\sqrt{\frac{0.05(0.95)}{384} } \\=0.0111[/tex]

p difference = [tex]0.1198-0.05=0.0698[/tex]

b) Sampling proportion is Normal with mean = 0.05 and std error = 0.0698

c) Z = test statistic = p diff/std error

= 6.29

p value <0.05

d) Since p < alpha we reject null hypothesis.

e) The probability that null hypothesis is rejected when true is negligible =p value

Answer:

Step-by-step explanation:

Reject null hypothesis

The question is about calculating the mean and standard deviation of a discrete random variable representing the size of the freezer model a customer buys, and then the mean and standard deviation of the price paid, which is a function of the freezer size.

The random variable X in this scenario represents the size of the freezer model (in cubic feet) that the next customer buys from the appliance dealer. This is a discrete random variable, with possible values being the sizes of the three available freezer models: 14.5, 16.9, and 19.1 cubic feet.

To calculate the mean and standard deviation of X, we can use the formula for the mean of a discrete random variable and the formula for the standard deviation respectively, which are given as follows:
Mean (expected value), E(X) = Σ [x * p(x)]
Standard Deviation, σ(X) = sqrt{Σ [(x - μ)² * p(x)]}

The mean value of the second variable (Price) can be calculated by plugging the expected value of X into the given Price = 25X - 8.5 equation, and the standard deviation by applying the transformation rule for standard deviations (σ_Y = |b| * σ_X, where b is the coefficient of X in the original expression for Y, which in this case is 25).

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Answer:

Step-by-step explanation:

Convert 7.2 to grams then divide the grams evenlly to the 24 people then you get 300

The question involves calculating the probabilities of different combinations of females and males being selected for positions using the hypergeometric distribution, a concept in mathematics related to probability without replacement from two distinct groups.

The question is related to the concept of probability in mathematics, specifically the use of the hypergeometric distribution, which is appropriate when sampling without replacement from a finite population divided into two groups. To calculate the probability of various combinations of female and male applicants selected for the positions, we need to use the hypergeometric probability formula:

These calculations involve combinations (denoted as C(n, k)) and the formula for hypergeometric probability, which is:

Hypergeometric Probability = [C(group1 size, group1 subset size) * C(group2 size, group2 subset size)] / C(total population size, total subset size)

Answer:

0.1587

Step-by-step explanation:

Let X be the random variable that represents a return on long-term government bond. We know that X has a mean of 6.0 and a standard deviation of 9.9, in order to compute the approximate probability that your return on these bonds will be less than -3.9 percent in a given year, we should compute the z-score related to -3.9, i.e., (-3.9-6.0)/9.9 =  -1. Therefore, we are looking for P(Z < -1) = 0.1587

Final answer:

To calculate the 99% confidence interval for the population mean commute time, we use the formula CI = x± t * (s/√n), where x is the sample mean, t is the t-value, s is the sample standard deviation, and n is the sample size. By substituting the given values into the formula, we find that the confidence interval is approximately (29.4, 36.2) and the margin of error is 1.7. This means we are 99% confident that the population mean commute time lies within the interval, and the margin of error represents the possible deviation of the sample mean from the population mean.

Explanation:

To construct a 99% confidence interval for the population mean μ, we can use the formula:

CI = x± t * (s/√n)

Where CI is the confidence interval, x is the sample mean, t is the t-value, s is the sample standard deviation, and n is the sample size. In this case, x = 32.8, s = 7.2, and n = 27.

To find the t-value, we need to determine the degrees of freedom (df), which is equal to n - 1. So, df = 27 - 1 = 26. Looking up the t-value for a 99% confidence level with 26 degrees of freedom in a t-table or using calculator software, we find it to be approximately 2.787.

Substituting these values into the formula, we get:

CI = 32.8 ± 2.787 * (7.2/√27)

Calculating this, we find that the confidence interval is approximately (29.4, 36.2).

The margin of error, also known as the error bound for a population mean (EBM), is half the width of the confidence interval. So, the margin of error in this case is (36.2 - 32.8)/2 = 1.7.

Interpreting the results, we can say that we are 99% confident that the population mean μ lies within the range of 29.4 to 36.2 minutes. Furthermore, the margin of error of 1.7 indicates how much the sample mean could vary from the population mean, taking into account the variability in the data.

Good evening ,

Answer:

x=2

Step-by-step explanation:

Look at the photo below for the details.

:)

Answer:  (0.3332, 0.33341)

Step-by-step explanation:

Formula to find the confidence interval for population mean[tex](\mu)[/tex] :

[tex]\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}[/tex]

, where n= Sample size

[tex]\overline{x}[/tex] = sample mean.

[tex]z^*[/tex] = Critical z-value (two-tailed)

[tex]\sigma[/tex] = population standard deviation.

As per given , we have

n= 53

[tex]\sigma=0.000507\ mm [/tex]

[tex]\overline{x}=0.3333\ mm[/tex]

The critical values for 90% confidence interval : [tex]z^*=\pm1.645[/tex]

Now , the 90 percent confidence interval for the true mean metal thickness:

[tex]0.3333\pm (1.645)\dfrac{0.000507}{\sqrt{53}}\\\\=0.3333\pm(1.645)(0.0000696)\approx0.3333\pm0.00011456\\\\=(0.3333-0.00011456,\ 0.3333+0.00011456)\\\\=(0.33318544,\ 0.33341456)\approx(0.3332,\ 0.3334)[/tex]

Hence, the 90 percent confidence interval for the true mean metal thickness. : (0.3332, 0.3334)

Answer: the fraction of the whole pie that Rebecca ate is 1/15

Step-by-step explanation:

Let x represent the size or area of the whole pie. Maria ate 1/3 of the pie. This means that the amount of pie that maria ate is 1/3 × x = x/3

Her sister,Rebecca ate 1/5 of that. This means that Rebecca ate 1/5 of the amount that Maria ate. The amount that Rebecca ate will be expressed as 1/5 × x/3 = x/15

To determine the fraction of the whole pie that Rebecca ate, we will divide the amount that Rebecca ate by the size of the whole pie. It becomes

x/15/x = x/15 × 1/x

= 1/15

Answer:

Part (A): The total number of ways are [tex]4^{100}[/tex]

Part (B): The total number of ways are [tex]1.6122075076\times10^{57}[/tex]

Step-by-step explanation:

Consider the provided information.

Part(A) a) How many ways are there for her to plan her schedule if there are no restrictions on the number of days she studies each of the four subjects?

She plans on studying four different subjects.

That means she has 4 choices for each day also there is no restriction on the number of days.

Hence, the total number of ways are:

[tex]4\times 4\times4\times4\times......4=4^{100}[/tex]

Part (B) How many ways are there for her to plan her schedule if she decides that the number of days she studies each subject will be the same?

She wants that the number of day she studies each subject will be the same that means she studies for 25 days each subject.

Therefore, the number of ways are:

[tex]^{100}C_{25}\times ^{75}C_{25}\times ^{50}C_{25}\times ^{25}C_{25}=1.6122075076\times10^{57}[/tex]

In the case of no restrictions, there are 4^100 possible study schedules. If each subject is studied the same number of days, the number of possible schedules is found using a permutation with repetitions of multiset formula, resulting in 100! / (25! * 25! * 25! * 25!).

This question focuses on combinatorics which is a branch of mathematics that deals with combinations of objects belonging to a finite set. It features the tasks of counting different objects as per certain restrictions, arrangement of objects with regard to considered conditions, and selection of objects satisfying specific criteria.

For part (a), if there are no restrictions, then she could pick any of the four subjects to study each day. For each day, there are 4 choices, so for 100 days, that results in 4^100 possible schedules.

For part (b), if she wants to study each subject the same number of days, then each subject would be studied for 100/4=25 days. This situation is a permutation with repetitions of multiset, which can be found through the formula n!/(r1! * r2! * ... * rk!) where n is total number of items, and r1, r2, ..., rk are numbers of same elements of multiset. Using this equation, the number of ways would be 100! / (25! * 25! * 25! * 25!).

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Answer:B

Step-by-step explanation: The answer is B. 1/6^7. Your trying to find the reciprocal of 6^-7. The way you do that is put a 1 over the 6 with a power of -7 like this (6^-7)/(1) and that's how you get your answer!

[tex]\huge\text{Hey there!}[/tex]

[tex]\huge\textbf{This equation should be somewhat}\\\huge\textbf{easy because it has like THREE steps}\\\huge\textbf{to solve it.}[/tex]

[tex]\huge\textbf{But....}[/tex]

[tex]\huge\textbf{Your equation should look like:}[/tex]

[tex]\mathsf{6^{-7}}[/tex]

[tex]\huge\textbf{Now, let's get to the steps:}[/tex]

[tex]\mathsf{6^{-7}}\\\\\mathsf{\approx \dfrac{1}{6\times6\times6\times6\times6\times6\times6}}\\\\\mathsf{\approx \dfrac{1}{6^7}}\\\\\mathsf{\approx \dfrac{1}{279,936}}[/tex]

[tex]\huge\textbf{Therefore, the closet answer to your}\\\huge\textbf{question should be:}[/tex]

[tex]\huge\boxed{\frak{Option\ B. \rm{\dfrac{1}{6^7}}}}\huge\checkmark[/tex]

[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]

~[tex]\frak{Amphitrite1040:)}[/tex]

Answer:

We conclude that  the mean number of residents in the retirement community household is less than 3.36 persons.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 3.36

Sample mean, [tex]\bar{x}[/tex] = 2.71

Sample size, n =  25

Alpha, α = 0.10

Sample standard deviation, s = 1.10

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 3.36\text{ residents per household}\\H_A: \mu < 3.36\text{ residents per household}[/tex]

We use One-tailed(left) 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{2.71 - 3.36}{\frac{1.10}{\sqrt{25}} } = -2.95[/tex]

Now, [tex]t_{critical} \text{ at 0.10 level of significance, 24 degree of freedom } =-1.31[/tex]

Since,                  

[tex]t_{stat} < t_{critical}[/tex]

We reject the null hypothesis and fail to accept it.

Thus, we conclude that  the mean number of residents in the retirement community household is less than 3.36 persons.

The null hypothesis for testing if the mean number of residents in Arizona retirement community households is less than the national average is that the mean is equal to or greater than 3.36 residents. The alternate hypothesis is that it is less than 3.36 residents. A one-sample t-test would typically be used to test these hypotheses.

Conducting a hypothesis test to determine if the mean number of residents in retirement community households in Arizona is significantly less than the national average reported by the Census Bureau, which is 3.36 residents per household. To address this,

The null hypothesis (H0) is that the mean number of residents per household in the Arizona retirement communities is equal to or greater than the national average, = 3.36 residents.

The alternate hypothesis (Ha) is that the mean number of residents per household in the Arizona retirement communities is less than the national average, < 3.36 residents.

To test this hypothesis at a 0.10 significance level, we would typically use a one-sample t-test to compare the sample mean (2.71 residents) against the national average (3.36 residents) considering the sample standard deviation (1.10 residents) and sample size (25 households).

The hypothesis test would determine if the observed difference is statistically significant, implying that the mean number of residents in retirement community households is indeed less than 3.36 persons.

Final answer:

The experimental results support the claim that less than 0.2 percent of the company’s batteries will fail during the advertised time period.

Explanation:

To test if the experimental results support the claim that less than 0.2 percent of the company’s batteries will fail during the advertised time period, we can conduct a hypothesis test.

H0 (null hypothesis): The proportion of batteries that fail during the advertised time period is equal to or greater than 0.2 percent.

H1 (alternative hypothesis): The proportion of batteries that fail during the advertised time period is less than 0.2 percent.

We can use a one-sample proportion test to compare the proportion of batteries that fail in the sample to the claimed proportion. The test statistic for this hypothesis test is the z-test statistic.

The critical value(s) for a one-sided test with α = 0.01 is z = -2.33.

The critical (rejection) region is z < -2.33.

The p-value is calculated by finding the probability of observing 15 or fewer failures out of 5000 batteries assuming that the null hypothesis is true. Using a normal approximation, we can calculate the p-value as the probability of observing x ≤ 15, where x is the number of failures. We find the p-value is less than 0.01.

Since the p-value is less than the significance level of 0.01, we reject the null hypothesis. We have sufficient evidence to support the claim that less than 0.2 percent of the company’s batteries will fail during the advertised time period.

Answer: a) 0.079589 b) 0.079656

Step-by-step explanation:

Since we have given that

Number of times a coin is flipped = 100 times

Number of times he get exactly head = 50

Probability of getting head = [tex]\dfrac{1}{2}[/tex]

We will use "Binomial distribution":

Probability would be

[tex]^{100}C_{50}(\dfrac{1}{2})^{50}(\dfrac{1}{2})^50\\\\=0.079589[/tex]

Using "Normal approximation":

n = 100

p = 0.5

So, mean = [tex]np=100\times 0.5=50[/tex]

Standard deviation is given by

[tex]\sqrt{np(1-p)}\\\\=\sqrt{50(1.05)}\\\\=\sqrt{50\times 0.5}\\\\=\sqrt{25}\\\\=5[/tex]

So,

[tex]P(X<x)=P(Z<\dfrac{\bar{x}-\mu}{\sigma})\\\\So, P(X=50)=P(49.5<X<50.5)\\\\=P(\dfrac{49.5-50}{5}<Z<\dfrac{50.5-50}{5})\\\\=P(-0.1<Z<0.1)\\\\=P(Z<0.1)-P(Z<-0.1)\\\\=0.539828-0.460172\\\\=0.079656[/tex]

Hence, a) 0.079589 b) 0.079656

The subject of this question is Physics and the grade level is High School. The article describes a factorial experiment designed to determine factors in a high-energy electron beam process that affect hardness in metals. The experiment focuses on two factors: travel speed and beam current. The aim of the experiment is to determine how these factors affect hardness in metals.

In the given article, the subject of the question is physics, specifically the experimentation involving a high-energy electron beam process that affects hardness in metals. The experiment focuses on two factors: travel speed and beam current. The travel speed is measured in mm/s and has three levels: 10, 20, and 30. The beam current is a continuous variable that can be adjusted and is a factor of two. The experiment aims to determine how these factors affect hardness in metals.

Answer:

-1/3

Step-by-step explanation:

rise / run = slope

4-5/ 1-(-2)

= -1/3

Answer:

Since p value <0.1 accept the claim that oven I repair costs are more

Step-by-step explanation:

The data given for two types of ovens are summarised below:

Group   Group One     Group Two  

Mean 85.7900 78.6700

SD 15.1300 17.8400

SEM 1.9533 2.3840

N 60       56      

Alpha = 10%

[tex]H_0: \mu_1 - \mu_2 =0\\H_a: \mu_1 - \mu_2> 0[/tex]

(Right tailed test)

The mean of Group One minus Group Two equals 7.1200

df = 114

 standard error of difference = 3.065

 t = 2.3234

p value = 0.0219

If p value <0.10 reject null hypothesis

4) Since p value <0.1 accept the claim that oven I repair costs are more

Answer:

\mu = 14.5\\

\sigma = 5.071\\

k = 1.084

Step-by-step explanation:

given that a  statistician uses Chebyshev's Theorem to estimate that at least 15 % of a population lies between the values 9 and 20.

i.e. his findings with respect to probability are

[tex]P(9<x<20) \geq 0.15\\P(|x-14.5|<5.5) \geq 0.15[/tex]

Recall Chebyshev's inequality that

[tex]P(|X-\mu |\geq k\sigma )\leq {\frac {1}{k^{2}}}\\P(|X-\mu |\leq k\sigma )\geq 1-{\frac {1}{k^{2}}}\\[/tex]

Comparing with the Ii equation which is appropriate here we find that

[tex]\mu =14.5[/tex]

Next what we find is

[tex]k\sigma = 5.5\\1-\frac{1}{k^2} =0.15\\\frac{1}{k^2}=0.85\\k=1.084\\1.084 (\sigma) = 5.5\\\sigma = 5.071[/tex]

Thus from the given information we find that

[tex]\mu = 14.5\\\sigma = 5.071\\k = 1.084[/tex]

Answer: 98% confidence interval is (0.06,0.11).

Step-by-step explanation:

Since we have given that

Number of men = 550

Number of women = 2400

Number of men have color blindness = 48

Number of women have color blindness = 5

So, [tex]p_M=\dfrac{48}{550}=0.087\\\\p_F=\dfrac{5}{2400}=0.0021[/tex]

So, at 98% confidence interval, z = 2.326

so, interval would be

[tex](p_M-P_F)\pm z\sqrt{\dfrac{p_M(1-p_M)}{n_M}+\dfrac{p_F(1-p_F)}{n_F}}\\\\=(0.087-0.0021)\pm 2.326\sqrt{\dfrac{0.087\times 0.912}{550}+\dfrac{0.0021\times 0.9979}{2400}}\\\\=0.0849\pm 2.326\times 0.012\\\\=(0.0849-0.027912, 0.0849+0.027912)\\\\=(0.06,0.11)[/tex]

Hence, 98% confidence interval is (0.06,0.11).

No, there does not appears to be a significant difference.

Answer:

its 114

Step-by-step explanation:

180-66=114 so that's it

Answer:

I guess they rounded, but the answer is .5000 or 50% chance

Step-by-step explanation:

Count the total chances. There is a chance it will be 10oz, 11oz, 12oz, 13oz, 14oz, 15oz, or 16oz. The question asked 13 or more. That is 13, 14, 15, and 16.

That is a 4/7 chance or probability it will be more than 13oz. Divide 4 and 7 to get your answer.

0.57 or 57%

Answer:

C i think i already answerd this

Step-by-step explanation:

The music professor attacking the test is validity. College courses in music-related subjects, such as voice, instruments, music appreciation, theory, and performance, are taught by music professors.

College courses in music-related subjects, such as voice, instruments, music appreciation, theory, and performance, are taught by music professors. They might instruct sophisticated, highly specialized courses or wide, basic ones. They design curricula, teach classes, and evaluate students' work.

Other names for music teachers include: music educator, band teacher, choir teacher, piano teacher, violin teacher, guitar teacher, elementary school music teacher, high school music teacher, and vocal teacher.

The highest academic position at a college, university, or postsecondary institution is professor. Professors are well-known, competent academics who are frequently regarded as authorities in their fields of study. In addition to graduate courses, a professor teaches upper-level undergraduate courses.

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Answer:

70th percentile for the amount of time between customers entering is 30.10

Step-by-step explanation:

given data

mean =  25

to find out

What is the 70th percentile for the amount of time between customers entering Clay's store

solution

we know that here mean is

mean = [tex]\frac{1}{\lambda}[/tex]   ..................1

so here 25 =  [tex]\frac{1}{\lambda}[/tex]

and we consider time value corresponding to 70th percentile = x

so we can say

P(X < x) = 1 - [tex]e^{- \lambda *x}[/tex]      

P(X < x) = 70 %

1 - [tex]e^{\frac{-x}{25}}[/tex]  = 70 %

1 - [tex]e^{\frac{-x}{25} }[/tex]  = 0.70

[tex]e^{\frac{- x}{25} }[/tex]  = 0.30

take ln both side

[tex]\frac{-x}{25}[/tex] = ln 0.30

[tex]\frac{x}{25}[/tex]  = 1.203973

x = 30.10

70th percentile for the amount of time between customers entering is 30.10

The 70th percentile for the amount of time between customers entering Clay's store is 30

The mean of an exponential distribution is:

[tex]E(x) = \frac 1\lambda[/tex]

The mean is 25.

So, we have:

[tex]\frac 1\lambda = 25[/tex]

Solve for [tex]\lambda[/tex]

[tex]\lambda = \frac 1{25}[/tex]

[tex]\lambda = 0.04[/tex]

An exponential function is represented as:

[tex]P(x < x) = 1 - e^{-\lambda x}[/tex]

For the 70th percentile, we have:

[tex]1 - e^{-\lambda x} = 0.70[/tex]

This gives

[tex]e^{-\lambda x} = 1 - 0.70[/tex]

[tex]e^{-\lambda x} = 0.30[/tex]

Substitute [tex]\lambda = 0.04[/tex]

[tex]e^{-0.04x} = 0.30[/tex]

Take the natural logarithm of both sides

[tex]-0.04x = \ln(0.30)[/tex]

This gives

[tex]-0.04x = -1.20[/tex]

Divide both sides by -0.04

[tex]x = 30[/tex]

Hence, the 70th percentile for the amount of time between customers entering Clay's store is 30

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Answer:

0.1724 or 17.24%

Step-by-step explanation:

There are  

10+13+7 = 30 people

so there are C(30,5) combinations of 30 elements taken 5 at a time ways to select groups of 5 people.

[tex]C(30,5)=\displaystyle\binom{30}{5}=\frac{30!}{5!25!}=142,506[/tex]

There are C(10,2) combinations of 10 elements taken 2 at a time ways to select the 2 Democrats

[tex]C(10,2)=\displaystyle\binom{10}{2}=\frac{10!}{2!8!}=45[/tex]

There are C(13,2) combinations of 13 elements taken 2 at a time ways to select the 2 Republicans

[tex]C(13,2)=\displaystyle\binom{13}{2}=\frac{13!}{2!11!}=78[/tex]

There are C(7,1) combinations of 7 elements taken 1 at a time ways to select the Independent

[tex]C(7,1)=\displaystyle\binom{7}{1}=\frac{7!}{1!6!}=7[/tex]

By the Fundamental Principle of Counting, there are  

45*78*7 = 24,570

ways of making the group of 5 people with 2 Democrats, 2 Republicans and 1 independent, and the probability of forming the group is

[tex]\displaystyle\frac{24,570}{142,506}=0.1724[/tex] or 17.24%

Answer:

A recursive rule for the sequence is f(1) = -8; f(n) = -4 (n – 1) for all n ≥ 2 is "FALSE"

An explicit rule for the sequence is f(n) = -8 + 4 (n – 1) is "TRUE"

The tenth term is 28 is "TRUE"

Step-by-step explanation:

Statement (1)

While the first part [f(1) = –8] is TRUE, the second part [f(n) = –4 (n – 1) for all n ≥ 2] would only be true if the sequence ends at the second term.

Check: Since the fifth term of the sequence is 8, then f(5) = 8

From the statement,

f(5) = –4 (5 – 1)

f(5) = –4 × 4 = –16

:. f(5) ≠ 8

Statement (2)

f(n) = –8 + 4 (n – 1) is TRUE

Check: The fifth term of the sequence is 8 [f(5) = 8]

From the statement,

f(5) = –8 + 4 (5 – 1)

f(5) = –8 + 4 (4)

f(5) = –8 + 16 = 8

:. f(5) = 8

Statement (3)

f(10) = 28 is TRUE

Since the explicit rule is TRUE, use to confirm if f(10) = 28:

f(10) = –8 + 4 (10 – 1)

f(10) = –8 + 4 (9)

f(10) = –8 + 36

f(10) = 28

:. f(10) = 28

Answer:

Step-by-step explanation:

Hello!

To be able to resolve this kind of exercise nice and easy the first step is to determine your study variable and population parameter.

The variable is

X: daily dietary allowance for zinc for a male 50 years old or older. (mg/day)

The parameter of interest is the population mean. (μ)

You need to test if the average zinc intake is bellowed the recommended allowance, symbolically μ < 15. This will be the alternative hypothesis, it's complement will be the null hypothesis. (easy tip to detect the null hypothesis fast, it always carries the = sign)

The hypothesis is:

H₀: μ ≥ 15

H₁: μ < 15

α: 0,05

Now you have no information about the variable distribution. You need the variable to be normally distributed to study the population mean. Since the sample is large enough, you can apply the Central Limit Theorem and approximate the distribution of the sample mean to normal, this way, you can use the approximate Z to make the test.

X[bar]≈N(μ;σ²/n)

Z=  X[bar]-μ ≈N(0;1)

√(σ²/n)

Z=  11.2 - 15  = -6.17

6.58/√114

The critical region of this test is one-tailed (left) the critical value is:

[tex]Z_{\alpha } = Z_{0.05} = -1.64[/tex]

If Z ≤ -1.64, you reject the null hypothesis.

If Z > -1.64, you support the null hypothesis

The calculated value is less than the critical value, so the decision is to reject the null hypothesis.

This means that the average daily zinc intake of males age 65 - 74 is below the recommended allowance.

I hope it helps!

Answer:

0.682 is the probability that one newborn baby will have a weight within one standard deviation of the​ mean.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  8 pounds

Standard Deviation, σ = 0.6 pounds

We are given that the distribution of weights of​ full-term newborn babies is a bell shaped distribution that is a normal distribution.

Formula:

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

P(weight between 7.4 and 8.6 ​pounds)

[tex]P(7.4 \leq x \leq 8.6) = P(\displaystyle\frac{7.4 - 8}{0.6} \leq z \leq \displaystyle\frac{8.6-8}{0.6}) = P(-1 \leq z \leq 1)\\\\= P(z \leq 1) - P(z < -1)\\= 0.841 - 0.159 = 0.682 = 68.2\%[/tex]

[tex]P(7.4 \leq x \leq 8.6) = 6.82\%[/tex]

This could also be found with the empirical formula.

0.682 is the probability that one newborn baby will have a weight within 0.6 pounds of the mean.

Answer:38 min

Step-by-step explanation:

Given

Cutting time [tex]t_1=14 min[/tex]

Assembly Line time [tex]t_2=24 min[/tex]

[tex]t_3=18[/tex] to move a batch of rugs from cutting to assembly

Value-added refers to a process or phase in a system that turns raw materials or work in progress into much more desirable goods and services for downstream consumers.

thus value added time is [tex]t_1+t_2=14+24 =38 min[/tex]                

Answer: 90% confidence interval would be (6.022,6.986).

Step-by-step explanation:

Since we have given that

Number of students = 361

Sample mean = 6.504

Sample standard deviation = 5.584

Standard error of the mean = 0.294

At 90% confidence level,

So, α = 0.01

So, z = 1.64

Margin of error is given by

[tex]z\times \text{Standard error}\\\\=1.64\times 0.294\\\\=0.48216[/tex]

So, Lower limit would be

[tex]\bar{x}-0.482\\\\=6.504-0.482\\\\=6.022[/tex]

Upper limit would be

[tex]\bar{x}+0.482\\\\=6.504+0.482\\\\=6.986[/tex]

So, 90% confidence interval would be (6.022,6.986).

Answer:

Which is the output of the formula =AND(12>6;6>3;3>9)?

A.

TRUE

B.

FALSE

C.

12

D.

9

The test statistic for the main effect for factor A cannot be computed from the provided information because the mean square for error (MSE) is not given, which is required for the F statistic calculation.

The test statistic for determining whether there is a main effect for factor A is calculated as follows.

First, the mean square for factor A (MSA) is found by dividing SSA by its degrees of freedom, which for factor A is a-1. In this case, with a = 6, degrees of freedom for A is 6-1, which is 5.

Therefore, MSA = SSA / dfA = 1.05 / 5 = 0.21.

The mean square for error (MSE) is not provided, which is necessary to calculate the F statistic for factor A.

To obtain the test statistic for factor A, we need to divide MSA by MSE. Without SSE or MSE, the test statistic cannot be computed. Hence, the correct answer is c.

The test statistic cannot be computed because SSE is not given.