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2 years ago

12

Lindsey wants to paint the wooden cone pictured below. What is the closest to the amount of paint she needs to paint the entire

surface of the cone? Please answer ASAP ^^

1 answer:

Bad White [126]2 years ago

6 0

The answer is 203in^2

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A researcher compares the effectiveness of two different instructional methods for teaching electronics. A sample of 138 student

yan [13]

Answer:

The 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for the difference between population means is:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

The information provided is as follows:

$n_{1}= 138\\n_{2}=156\\\bar x_{1}=61\\\bar x_{2}=64.6\\\sigma_{1}=18.53\\\sigma_{2}=13.43$

The critical value of <em>z</em> for 98% confidence level is,

$z_{\alpha/2}=z_{0.02/2}=2.326$

Compute the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 as follows:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

$=(61-64.6)\pm 2.326\times\sqrt{\frac{(18.53)^{2}}{138}+\frac{(13.43)^{2}}{156}}\\\\=-3.6\pm 4.4404\\\\=(-8.0404, 0.8404)\\\\\approx (-8.04, 0.84)$

Thus, the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

5 0

2 years ago

Mr. Parker wants to rent a cargo van for a day. It will cost the daily fee of $50.00 plus $0.35 per mile driven.

Fittoniya [83]

Answer:

50d * 0.35m

Step-by-step explanation:

so I don't normally do it like this it would normally be 50 * 0.35x. but the days also had a fee unless she doesn't have it for 2 days and only has it one but d is the days she has it and m is the miles driven the * is times

you would have to find out the days she has it unless you don't have that is would be the answer above (if I am wrong I'm sorry)

6 0

3 years ago

Read 2 more answers

PLZ help picture included

cupoosta [38]

Answer:

4 miles

Step-by-step explanation:

Let

x ----> the time to returned back biking in hours

d ---> the distance in miles

s ---> speed in miles per hour

Remember that

The speed or rate is equal to divide the distance by the time

$s=\frac{d}{t}$

The distance is equal to multiply the speed by the time

$d=st$

we know that

$18\ min=\frac{18}{60}\ hours$

The distance to his friend's house and the distance of return is the same

so

$10x=4(x+\frac{18}{60})$

solve for x

$10x=4x+\frac{18}{15}$

subtract 4x both sides

$10x-4x=\frac{18}{15}$

$6x=\frac{18}{15}$

$x=\frac{1}{5}\ hours$

The total distance of the round trip is

$y=10x+4x+\frac{18}{15}$

substitute the value of x

$y=10(\frac{1}{5})+4(\frac{1}{5})+\frac{18}{15}$

$y=2+2=4\ miles$

6 0

3 years ago

|-5x-5|=45 what is the solutions

Paha777 [63]

Answer:-10

Step-by-step explanation:

45 +5=-5x

-5x=50

X=-10

7 0

3 years ago

Read 2 more answers

Help me answer this math question please!!

marusya05 [52]

Hope this helps you!

6 0

3 years ago