Ask question

Ask question

All categories

2 years ago

5

The net below depicts a triangular prism. What is the total surface area of the prism? And please explain how did you get your a

nswer. A.282 cm B. 210 cm C.312 cm D. 624 cm

1 answer:

Mariana [72]2 years ago

6 0

Answer is A - 282

You have (3) rectangles, A=L x W
6x14 = 84 x 3 shapes = 252
2 triangles A= 1/2BxH
1/2 x 6 x 5 = 15 x 2 shapes = 30
252 + 30 = 282

You might be interested in

Suppose a random variable x is best described by a uniform probability distribution with range 22 to 55. Find the value of a tha

const2013 [10]

Answer:

(a) The value of <em>a</em> is 53.35.

(b) The value of <em>a</em> is 38.17.

(c) The value of <em>a</em> is 26.95.

(d) The value of <em>a</em> is 25.63.

(e) The value of <em>a</em> is 12.06.

Step-by-step explanation:

The probability density function of <em>X</em> is:

$f_{X}(x)=\frac{1}{55-22}=\frac{1}{33}$

Here, 22 < X < 55.

(a)

Compute the value of <em>a</em> as follows:

$P(X\leq a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.95\times 33=[x]^{a}_{22}\\\\31.35=a-22\\\\a=31.35+22\\\\a=53.35$

Thus, the value of <em>a</em> is 53.35.

(b)

Compute the value of <em>a</em> as follows:

$P(X< a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.49\times 33=[x]^{a}_{22}\\\\16.17=a-22\\\\a=16.17+22\\\\a=38.17$

Thus, the value of <em>a</em> is 38.17.

(c)

Compute the value of <em>a</em> as follows:

$P(X\geq a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.85=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.85\times 33=[x]^{55}_{a}\\\\28.05=55-a\\\\a=55-28.05\\\\a=26.95$

Thus, the value of <em>a</em> is 26.95.

(d)

Compute the value of <em>a</em> as follows:

$P(X\geq a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.89=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.89\times 33=[x]^{55}_{a}\\\\29.37=55-a\\\\a=55-29.37\\\\a=25.63$

Thus, the value of <em>a</em> is 25.63.

(e)

Compute the value of <em>a</em> as follows:

$P(1.83\leq X\leq a)=\int\limits^{a}_{1.83} {\frac{1}{33}} \, dx \\\\0.31=\frac{1}{33}\cdot \int\limits^{a}_{1.83} {1} \, dx \\\\0.31\times 33=[x]^{a}_{1.83}\\\\10.23=a-1.83\\\\a=10.23+1.83\\\\a=12.06$

Thus, the value of <em>a</em> is 12.06.

7 0

3 years ago

If f(x)=x/2-2 and g(x)=2x^2+x-3, find (f+g)(x)

Lelu [443]

Hello :
<span>If f(x)=x/2-2 and g(x)=2x^2+x-3, find (f+g)(x)
</span>(f+g)(x)= f(x) +g(x) = x/2-2 +2x²+x-3 = x/2 +2x²+x-5
(f+g)(x)= (x+4x²-2x-10) /2
(f+g)(x)= (4x²+3x-10)/2 = 2x²+3/2 x -5

4 0

3 years ago

Read 2 more answers

At 11 hours, 42 minutes and 5 hours, 46 minutes

butalik [34]

Answer

https://zoom.us/j/99385592884?pwd=ajVOcXZSbGcyT3N3N0kxQm5Zek81Zz09er:

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

What’s the answer because I’m confused

Scilla [17]

Answer:

i hate Maths

7 0

3 years ago

A spinner has 6 equally divided sections labeled A, B, C, D, E and F. After spinning the spinner, a number cube is rolled. What

bagirrra123 [75]

Answer: $\dfrac{2}{9}$

Step-by-step explanation:

Given

The spinner has 6 equal sections A, B, C, D, E, and F

The two events occurred one after the other. So, the probability is the product of the probabilities of each event.

There are two consonants in the given 6 letter

The probability of landing over a consonant is

$P_1=\dfrac{2}{6}$

The number greater than 2 in a cube are 3,4,5,6

So, probability corresponding to it

$P_2=\dfrac{4}{6}$

So, the required probability is

$\Rightarrow P=P_1\times P_2\\\Rightarrow P=\dfrac{2}{6}\times \dfrac{4}{6}\\\\\Rightarrow P=\dfrac{8}{36}\\\\\Rightarrow P=\dfrac{2}{9}$

8 0

3 years ago