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3 years ago

6

a serving of ice cream contains 1,200 calories. One hundred forty four calories come from fat. What percent of the total calorie

s come from fat?

2 answers:

nekit [7.7K]3 years ago

7 0

12% here is the procedure: (144/1,200)*100

Gre4nikov [31]3 years ago

6 0

12% come from fat, 144/1200*100= 12%

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Majesty Video Production Inc. wants the mean length of its advertisements to be 26 seconds. Assume the distribution of ad length

Paladinen [302]

Answer:

a) By the Central Limit Theorem, approximately normally distributed, with mean 26 and standard error 0.44.

b) s = 0.44

c) 0.84% of the sample means will be greater than 27.05 seconds

d) 98.46% of the sample means will be greater than 25.05 seconds

e) 97.62% of the sample means will be greater than 25.05 but less than 27.05 seconds

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 $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation(also called standard error) $s = \frac{\sigma}{\sqrt{n}}$ .

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:

$\mu = 26, \sigma = 2, n = 21, s = \frac{2}{\sqrt{21}} = 0.44$

a. What can we say about the shape of the distribution of the sample mean time?

By the Central Limit Theorem, approximately normally distributed, with mean 26 and standard error 0.44.

b. What is the standard error of the mean time? (Round your answer to 2 decimal places)

$s = \frac{2}{\sqrt{21}} = 0.44$

c. What percent of the sample means will be greater than 27.05 seconds?

This is 1 subtracted by the pvalue of Z when X = 27.05. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{27.05 - 26}{0.44}$

$Z = 2.39$

$Z = 2.39$ has a pvalue of 0.9916

1 - 0.9916 = 0.0084

0.84% of the sample means will be greater than 27.05 seconds

d. What percent of the sample means will be greater than 25.05 seconds?

This is 1 subtracted by the pvalue of Z when X = 25.05. So

$Z = \frac{X - \mu}{s}$

$Z = \frac{25.05 - 26}{0.44}$

$Z = -2.16$

$Z = -2.16$ has a pvalue of 0.0154

1 - 0.0154 = 0.9846

98.46% of the sample means will be greater than 25.05 seconds

e. What percent of the sample means will be greater than 25.05 but less than 27.05 seconds?"

This is the pvalue of Z when X = 27.05 subtracted by the pvalue of Z when X = 25.05.

X = 27.05

$Z = \frac{X - \mu}{s}$

$Z = \frac{27.05 - 26}{0.44}$

$Z = 2.39$

$Z = 2.39$ has a pvalue of 0.9916

X = 25.05

$Z = \frac{X - \mu}{s}$

$Z = \frac{25.05 - 26}{0.44}$

$Z = -2.16$

$Z = -2.16$ has a pvalue of 0.0154

0.9916 - 0.0154 = 0.9762

97.62% of the sample means will be greater than 25.05 but less than 27.05 seconds

8 0

3 years ago

Rewrite each sum as a product of a gcf as a new sum 9+15

Crazy boy [7]

4 +5 = 9 and 12 + 3 =15 and there is more than this
hope I helped

7 0

3 years ago

Read 2 more answers

Suppose that \nabla f(x,y,z) = 2xyze^{x^2}\mathbf{i} + ze^{x^2}\mathbf{j} + ye^{x^2}\mathbf{k}. if f(0,0,0) = 2, find f(1,1,1).

lesya [120]

The simplest path from (0, 0, 0) to (1, 1, 1) is a straight line, denoted $C$ , which we can parameterize by the vector-valued function,

$\mathbf r(t)=(1-t)(\mathbf i+\mathbf j+\mathbf k)$

for $0\le t\le1$ , which has differential

$\mathrm d\mathbf r=-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$

Then with $x(t)=y(t)=z(t)=1-t$ , we have

$\displaystyle\int_{\mathcal C}\nabla f(x,y,z)\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}\nabla f(x(t),y(t),z(t))\cdot\mathrm d\mathbf r$

$=\displaystyle\int_{t=0}^{t=1}\left(2(1-t)^3e^{(1-t)^2}\,\mathbf i+(1-t)e^{(1-t)^2}\,\mathbf j+(1-t)e^{(1-t)^2}\,\mathbf k\right)\cdot-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$

$\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)(t^2-2t+2)\,\mathrm dt$

Complete the square in the quadratic term of the integrand: $t^2-2t+2=(t-1)^2+1=(1-t)^2+1$ , then in the integral we substitute $u=1-t$ :

$\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)((1-t)^2+1)\,\mathrm dt$

$\displaystyle=-2\int_{u=0}^{u=1}e^{u^2}u(u^2+1)\,\mathrm du$

Make another substitution of $v=u^2$ :

$\displaystyle=-\int_{v=0}^{v=1}e^v(v+1)\,\mathrm dv$

Integrate by parts, taking

$r=v+1\implies\mathrm dr=\mathrm dv$

$\mathrm ds=e^v\,\mathrm dv\implies s=e^v$

$\displaystyle=-e^v(v+1)\bigg|_{v=0}^{v=1}+\int_{v=0}^{v=1}e^v\,\mathrm dv$

$\displaystyle=-(2e-1)+(e-1)=-e$

So, we have by the fundamental theorem of calculus that

$\displaystyle\int_C\nabla f(x,y,z)\cdot\mathrm d\mathbf r=f(1,1,1)-f(0,0,0)$

$\implies-e=f(1,1,1)-2$

$\implies f(1,1,1)=2-e$

3 0

3 years ago

Find the inverse of 6x^2-7

valina [46]

This would be the correct answer

6 0

2 years ago

Read 2 more answers

A man wants to test for diabetes. His family history shows that the probability of getting diabetes is 0.6. He decides to do a h

lesantik [10]

So,

The probability of the man having diabetes is 0.6 or 60%. Because we are figuring the probability BEFORE the test is taken that he has the disease, we can disregard the test and its accuracy rate. That rate is 60%, the probability of him having diabetes.

The correct option is D.

8 0

2 years ago