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2 years ago

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A file that is 256 megabytes is being downloaded. If the download is 18.6% complete, how many megabytes have been downloaded? Ro

und your answer to the nearest tenth.

1 answer:

GenaCL600 [577]2 years ago

6 0

<span>If A file that is 256 megabytes is being downloaded and the download is 18.6% complete, then you just need to multiply the 256 megabytes with 18.6%

The calculation would be: 256 megabytes x 0,186= 47.616 megabytes.
If you round up the answer to nearest tenth it will be 47.6 megabytes</span>

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Two machines are used for filling plastic bottles with a net volume of 16.0 ounces. The filling processes can be assumed to be n

CaHeK987 [17]

Answer:

a) Null hypothesis: $\mu_{1}-\mu_{2}=0$

Alternative hypothesis: $\mu_{1} - \mu_{2}\neq 0$

b) $z=\frac{(16.015-16.005)-0}{\sqrt{\frac{0.015^2}{10}+\frac{0.018^2}{10}}}=1.350$

c) Since is a bilateral test the p value would be:

$p_v =2*P(z>1.350)=0.177$

Comparing the p value with the significance level $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the difference between the two groups is NOT significantly different.

d) $0.01-1.96\sqrt{\frac{0.015^2}{10}+\frac{0.018^2}{10}}=-0.0045$

$0.01+1.96\sqrt{\frac{0.015^2}{10}+\frac{0.018^2}{10}}=0.025$

So on this case the 95% confidence interval would be given by $-0.0045 \leq \mu_1 -\mu_2 \leq 0.025[/tex] Step-by-step explanation:For this case we have the following info:Machine 1: 16.03 16.01 16.04 15.96 16.05 15.98 16.05 16.02 16.02 15.99Machine 2: 16.02 16.03 15.97 16.04 15.96 16.02 16.01 16.01 15.99 16.00We can calculate the sample mean with the following formula:[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}$

$\bar X_{1}=16.015$ represent the mean for sample 1

$\bar X_{2}=16.005$ represent the mean for sample 2

$\sigma_{1}=0.015$ represent the population standard deviation for 1

$\sigma_{2}=0.018$ represent the sample standard deviation for 2

$n_{1}=10$ sample size for the group 2

$n_{2}=10$ sample size for the group 2

Significance level provided

z would represent the statistic (variable of interest)

Concepts and formulas to use

Part a

We need to conduct a hypothesis in order to check if the difference in the population means, the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2}=0$

Alternative hypothesis: $\mu_{1} - \mu_{2}\neq 0$

We have the population standard deviation's, so for this case is better apply a z test to compare means, and the statistic is given by:

$z=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

Part b

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

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

$z=\frac{(16.015-16.005)-0}{\sqrt{\frac{0.015^2}{10}+\frac{0.018^2}{10}}}=1.350$

Part c

P value

Since is a bilateral test the p value would be:

$p_v =2*P(z>1.350)=0.177$

Comparing the p value with the significance level $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the difference between the two groups is NOT significantly different.

Part d

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}}$ (1)

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =16.015-16.005=0.01$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

Confidence interval

Now we have everything in order to replace into formula (1):

$0.01-1.96\sqrt{\frac{0.015^2}{10}+\frac{0.018^2}{10}}=-0.0045$

$0.01+1.96\sqrt{\frac{0.015^2}{10}+\frac{0.018^2}{10}}=0.025$

So on this case the 95% confidence interval would be given by $-0.0045 \leq \mu_1 -\mu_2 \leq 0.025$

5 0

2 years ago

A jogger runs around a circular track of radius 55 ft. Let (x,y) be her coordinates, where the origin is the center of the track

bagirrra123 [75]

Answer:

11.25 ft/sec

Step-by-step explanation:

Given that a jogger runs around a circular track of radius 55 ft.

. Let (x,y) be her coordinates, where the origin is the center of the track.

Then we know x,y satisfies the equation of the circle as

$x^2+y^2 = 55^2$

Let us differentiate this implicit function with respect to t

$2x \frac{dx}{dt} +2y \frac{dy}{dt} =0$

At this point, dx/dt 15 and x=33. y =44

Substitute to have

$2(33)(15) +2(44) \frac{dy}{dt}\\=0\\\frac{dy}{dt}=11.25$

dy/dt = 11.25 ft /s

6 0

3 years ago

Please Help! 5.1 Algebra

sasho [114]

Answer:

B

Step-by-step explanation:

Since x is in an absolute value, we know that no matter what value we put in for x, the result will always be positive. B fits this description.

4 0

2 years ago

Read 2 more answers

An open box is formed from a rectangular piece of cardboard, whose length is 3 inches more than it's width, by cutting 3 inch sq

Feliz [49]

3+264-2=265 inches

Hope this helps!!!!

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2 years ago

3/4(x + 3) = 9<br><br><br> How do I solve this

Sonbull [250]

Answer:

x = 10

Step-by-step explanation:

so first distribute the 3/4 to everything on the inside

3/4x + 1 1/2 = 9

now subtract 1 1/2 on both sides

3/4x = 7 1/2

Now divide both sides by 3/4

x = 7 1/2 divided by 3/4

x = 15/2 divided by 3/4

x = 15/2 times 4/3

x = 5/1 times 2/1

x = 10

4 0

2 years ago