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3 years ago

What is the surface area of a square prism with sides that measure 8 units?

Mathematics

1 answer:

eimsori [14]3 years ago

5 0

Answer:

384 square units

Step-by-step explanation:

Square prism = Cube

Formula for finding the surface area of a cube = L x L (6)

Surface area = 8 x 8 (6)

Surface area = 64(6)

Surface area = 384 square units

You might be interested in

Which equation has a solution of k = 6.5?

olga_2 [115]

Answer:

-7k = -45.5

Explanation:

You can solve this by plugging in the “k” to the problems that are listed.
If you were to plug 6.5 in for all those problems, you’d get:

-3(6.5) = -19.5

-1 + 6.5 = 5.5

-7(6.5) = -45.5

-2 + 6.5 = 4.5

So, in this case, looking at the ones I answered and looking at the problems listed, the correct answer would have to be -7k = -45.5.

4 0

3 years ago

Read 2 more answers

Heather earns $16 for each hour she works as a lifeguard. Choose the equation that models the situation, where h represents the

lord [1]

Answer:

I'm pretty sure the answer is m = 16h

Hope this helps!

Please mark as brainliest!

5 0

3 years ago

Which pyramids with base area B and height h have equal volumes?

Pachacha [2.7K]

Option A:

V = 1/3 * b^2 * h

V = 1/3 * 10.5^2 * 6

V = 1/3 * 110.25 * 6

V = 220.5

Option B:

V = 1/3 * b^2 * h

V = 1/3 * 3.6^2 * 8

V = 1/3 * 12.96 * 8

V = 34.56

Option C:

V = 1/3 * b^2 * h

V = 1/3 * 4.2^2 * 12

V = 1/3 * 17.64 * 12

V = 70.56

Option D:

V = 1/3 * b^2 * h

V = 1/3 * 6^2 * 8.4

V = 1/3 * 36 * 8.4

V = 100.8

I guess it's 'None of the Above'

3 0

3 years ago

4. A random variable X has a mean of 10 and a standard deviation of 3. If 2 is added to each value of X, what will the new mean

Ede4ka [16]

Adding 2 to each value of the random variable $X$ makes a new random variable $X+2$ . Its mean would be

$E[X+2]=E[X]+E[2]=E[X]+2$

since expectation is linear, and the expected value of a constant is that constant. $E[X]$ is the mean of $X$ , so the new mean would be

$E[X+2]=10+2=12$

The variance of a random variable $X$ is

$V[X]=E[X^2]-E[X]^2$

so the variance of $X+2$ would be

$V[X+2]=E[(X+2)^2]-E[X+2]^2$

We already know $E[X+2]=12$ , so simplifying above, we get

$V[X+2]=E[X^2+4X+4]-12^2$

$V[X+2]=E[X^2]+4E[X]+4-12^2$

$V[X+2]=(V[X]+E[X]^2)+4E[X]-140$

Standard deviation is the square root of variance, so $V[X]=3^2=9$ .

$\implies V[X+2]=(9+10^2)+4(10)-140=9$

so the standard deviation remains unchanged at 3.

NB: More generally, the variance of $aX+b$ for $a,b\in\mathbb R$ is

$V[aX+b]=a^2V[X]+b^2V[1]$

but the variance of a constant is 0. In this case, $a=1$ , so we're left with $V[X+2]=V[X]$ , as expected.

5 0

3 years ago

Find all values of k so that the trinomial x^2 + kx - 35 can be factored using integers

Korvikt [17]

$(x-r_1)(x-r_2)=x^2-(r_1+r_2)x+r_1r_2$

So the trinomial $x^2+kx-35$ can be factored as long as

$\begin{cases}r_1+r_2=-k\\r_1r_2=-35\end{cases}$

has integer solutions for $r_1,r_2$ . Clearly, both have to be factors of -35, which leaves only a handful of cases:

$(r_1,r_2)=(1,-35)\implies r_1+r_2=-34$
$(r_1,r_2)=(-1,35)\implies r_1+r_2=34$
$(r_1,r_2)=(5,-7)\implies r_1+r_2=-2$
$(r_1,r_2)=(-5,7)\implies r_1+r_2=2$

So the possible values of $k$ are $\pm34$ and $\pm2$ .

8 0

3 years ago