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3 years ago

Would the base be the 12 in??

Mathematics

1 answer:

3241004551 [841]3 years ago

4 0

To find volume use the formula L × W × H = V

L = length
W = width
H = height
V = volume

Hope this helps

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A rectangular page is to contain 24 sq. in. of print. The margins at the top and bottom of the page are each 1.5 inches. The mar

suter [353]

Answer:

Dimensions of page should be width of 6 inches and height of 9 inches

Step-by-step explanation:

Let x be the width of the printed part in inches

Let y be height of the printed part in inches.

Thus, Area of printed part; A = xy

And area of printed part is given as 24.

Thus, xy = 24

Making y the subject, we have;

y = 24/x

Now, the question says the top and bottom margins are 1.5 inches.

Thus, width of page = x + 1 + 1 = x + 2

And also the margins on each side are both 1m in length, thus the height of page will be:

y + 1.5 + 1.5 = y + 3

So area of page will now be;

A = (x + 2)•(y+3)

From earlier, we got y = 24/x

Thus,plugging this into area of page, we have;

A = (x + 2)•((24/x)+3)

A = 24 + 3x + 48/x + 6

A = 30 + 3x + 48/x

For us to find the minimum dimensions, we have to find the derivative of A and equate to zero

Thus,

dA/dx = 3 - 48/x²

Thus, dA/dx = 0 will be

3 - 48/x² = 0

Multiply through by x²:

3x² - 48 = 0

Thus,

3x² = 48

x² = 48/3

x = √16

x = 4 inches

Plugging this into y = 24/x,we have;

y = 24/4 = 6 inches

We want dimensions of page at x = 4 and y = 6.

From earlier, width of page = x + 2.

Thus,width = 4 + 2 = 6 inches

Height = y + 3 = 6 + 3 = 9 inches

So dimensions of page should be width of 6 inches and height of 9 inches

4 0

3 years ago

A manager records the repair cost for 4 randomly selected TVs. A sample mean of $88.46 and standard deviation of $17.20 are subs

WINSTONCH [101]

Answer:

a) The critical value is $z = 2.575$ .

b) The 99% confidence interval for the mean repair cost for the TVs is ($66.31, $110.61).

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ . This is our critical value

So it is z with a pvalue of $1-0.005 = 0.995$ , so $z = 2.575$

a. Find the critical value that should be used in constructing the confidence interval.

The critical value is $z = 2.575$ .

b. Construct the 99% confidence interval. Round your answer to two decimal places.

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample. So, in this problem

$M = 2.575*\frac{17.20}{\sqrt{4}} = 22.15$

The lower end of the interval is the mean subtracted by M. So it is 88.46 - 22.15 = $66.31

The upper end of the interval is the mean added to M. So it is 88.46 - 22.15 = $110.61

The 99% confidence interval for the mean repair cost for the TVs is ($66.31, $110.61).

5 0

3 years ago

-68C <br> 57C<br> How many degrees separate these two temperatures

mafiozo [28]

Answer:

wait what the letter look weird ?

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

PLEASE HELP!!!!!!!!!!!!!!!!

TEA [102]

I AM TOO YOUNG FRO THIS

3 0

3 years ago

Plz help me understand

faltersainse [42]

So what I have gotten from this is the dates 1 Jan -8 Jan they where gone 7 days so look at the prices of the per night I hope it helps

5 0

4 years ago