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bagirrra123 [75]
3 years ago
9

Use a net to find the surface area of the right triangular prism shown below: A. 104 square feet

Mathematics
2 answers:
Phoenix [80]3 years ago
6 0
I'd say by rounding and similar techniques the answer is 468, though it doesn't say all the needed numbers.
MakcuM [25]3 years ago
6 0

Answer:  The correct option is (C) 468 sq. ft.

Step-by-step explanation:  We are given to use a net to find the surface area of the right-triangular prism shown in the figure.

Let us divide the net of the prism in three parts, a rectangle ABCD and two congruent right-angled triangles T_1 and T_2 as shown in the attached figure below.

From the figure, we note that

the length and width of the rectangle ABCD are as follows :

AD = 15 + 9 +12 = 36 ft

and

AB = 10 ft.

So, the area of the rectangle ABCD is given by

a_r=AD\times AB=36\times10=360~\textup{sq. ft}.

Now, the base of each right-angled triangle is 9 ft and altitude is 12 ft.

So, the total area of both the right-angled triangle will be

a_t=2\times\dfrac{1}{2}\times 9\times12=108~\textup{sq. ft}.

Therefore, the total surface area of the given right triangular prism is given by

A_s=a_r+a_t=360+108=468~\textup{sq ft}.

Thus, the surface area of the given right-triangular prism is 468 sq. ft.

Option (C) is CORRECT.

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2 years ago
Data on oxide thickness of semiconductors are as follows: 426, 433, 415, 420, 420, 438, 417, 410, 430, 434, 423, 426, 412, 434,
Luda [366]

Answer:

a) 423.458

b) 9.53

c) 1.94

d) 424.5

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Step-by-step explanation:

We are given the following in the question:

426, 433, 415, 420, 420, 438, 417, 410, 430, 434, 423, 426, 412, 434, 435, 432, 409, 426, 409, 436, 422, 430, 411, 415

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Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{10163}{24} = 423.458

b) point estimate of the standard deviation of oxide thickness

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Sum of squares of differences = 2089.958

s = \sqrt{\dfrac{2089.958}{23}} = 9.53

c) standard error of the point estimate

Standard error =

=\dfrac{\sigma}{\sqrt{n}} = \dfrac{9.53}{\sqrt{24}} = 1.94

d) point estimate of the median oxide thickness

Sorted data: 409, 409, 410, 411, 412, 415, 415, 417, 420, 420, 422, 423, 426, 426, 426, 430, 430, 432, 433, 434, 434, 435, 436, 438

Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}

Median =

\dfrac{12^{th}+13^{th}}{2} = \dfrac{423+426}{2} = 424.5

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Answer:

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Simplifying first:

2(7) = 2 × 7 = 14

Our answer is 14 So answer is OA) 14

Expanding first then simplifying:

2(8-1) = 16-2 = 14

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