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3 years ago

find the surface area of the rectangular prism that is 16 inches long, 12 inches wide, and 5 inches in length

2 answers:

7 0

<span>Surface Area of a closed rectangular Prism equals to twice the area of the base plus the perimeter of the base multiplied by the height.

In your question:

Area of the base = (16)(12) = 192

Twice the Area of the base = 2(192) = 384

Perimeter of the base = 2(16) + 2(12) = 56

Surface Area = Twice the Area of the base + the Perimeter of the base x height

S.A. = 384 + 56 x 5
</span>
The surface area is 664 <span>in². Hope that helps.</span>
.

Afina-wow [57]3 years ago

4 0

960 square unit

length x width x height

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Need help asap please

Ira Lisetskai [31]

Answer:

858 the next one is 947

Step-by-step explanation:

I hope it's right bye

3 0

3 years ago

Find the measure of angle 4 AND...… angle 7

zaharov [31]

Answer:

<4 = 126

<7 = 55

Step-by-step explanation:

Remark

There are only 2 answers to this question. <4 has one of them and <7 has the other one. <7 is acute. <2 is obtuse.

Givens

<2 = 55 degrees

Solution

<4 is the supplement of <2

Therefore <4 + <2 = 180 A straight line has an angle of 180°

<4 + 55 = 180

<4 = 180 - 55

<4 = 125

So <7 is either 55 or 125. Since <7 is acute, it is 55 degrees but we have another way of proving it.

<4 + <6 = 180 degrees Interior angles on the same side of the transversal are supplementary

125 + <6 = 180

<6 = 180 - 125

<6 = 55

<7 =<6 Vertically opposite angles are equal.

<7 = 55 degrees.

4 0

3 years ago

What is the value of X?

zhenek [66]

Tough sorry wish I could help u can’t too hard but I think it is equivalent to the other triangle so I think it’s 31? I THINK

7 0

3 years ago

Read 2 more answers

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

marusya05 [52]

Answer:

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

$\hat p =\frac{955-812}{955}= 0.150$

$0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131$

$0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169$

And the 90% confidence interval would be given (0.131;0.169).

Step-by-step explanation:

We have the following info given:

$n= 955$ represent the sampel size slected

$x = 812$ number of students who read above the eighth grade level

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

$\hat p =\frac{955-812}{955}= 0.150$

The confidence interval for the proportion would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 90% confidence interval the significance is $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution and we got.