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3 years ago

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(HELPPPPPP)This graph relates the volumes of triangular pyramids with a base area of 12 square inches and their heights. Volumes

of Triangular Pyramids With Base Areas of 12 inches squared
According to the graph, what is the volume of a triangular prism with a base area of 12 square inches and a height of 10 inches?

(A)2.5 inches cubed
(B)40 inches cubed
(C)60 inches cubed
(D)120 inches cubed

1 answer:

aleksklad [387]3 years ago

7 0

Answer:

b

Step-by-step explanation:

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PLEASE HELP 8th GRADE MATH QUESTION OVER HERE!!!

patriot [66]

The side AB measures option 2. $\sqrt{20}}$ units long.

Step-by-step explanation:

Step 1:

The coordinates of the given triangle ABC are A (4, 5), B (2, 1), and C (4, 1).

The sides of the triangle are AB, BC, and CA. We need to determine the length of AB.

To calculate the distance between two points, we use the formula $d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}.$

where ( $x_{1},y_{1}$ ) are the coordinates of the first point and ( $x_{2},y_{2}$ ) are the coordinates of the second point.

Step 2:

For A (4, 5) and B (2, 1), ( $x_{1},y_{1}$ ) = (4, 5) and ( $x_{2},y_{2}$ ) = (2, 1). Substituting these values in the distance formula, we get

$d=\sqrt{\left(2-4\right)^{2}+\left(1-5}\right)^{2}} = \sqrt{\left(2\right)^{2}+\left(4}\right)^{2}}=\sqrt{20}}.$

So the side AB measures $\sqrt{20}}$ units long which is the second option.

6 0

3 years ago

J² - jk +k²= 7<br>J⁴ +j²k² + k⁴ = 133

LekaFEV [45]

Answer: You can search it up...

Step-by-step explanation: Have a nice day!

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3 years ago

The score on an exam from a certain MAT 112 class, X, is normally distributed with μ=78.1 and σ=10.8.

salantis [7]

a) $X$

b) 0.1539

c) 0.1539

d) 0.6922

Step-by-step explanation:

a)

In this problem, the score on the exam is normally distributed with the following parameters:

$\mu=78.1$ (mean)

$\sigma = 10.8$ (standard deviation)

We call X the name of the variable (the score obtained in the exam).

Therefore, the event "a student obtains a score less than 67.1) means that the variable X has a value less than 67.1. Mathematically, this means that we are asking for:

$X$

And the probability for this to occur can be written as:

$p(X$

b)

To find the probability of X to be less than 67.1, we have to calculate the area under the standardized normal distribution (so, with mean 0 and standard deviation 1) between $z=-\infty$ and $z=Z$ , where Z is the z-score corresponding to X = 67.1 on the s tandardized normal distribution.

The z-score corresponding to 67.1 is:

$Z=\frac{67.1-\mu}{\sigma}=\frac{67.1-78.1}{10.8}=-1.02$

Therefore, the probability that X < 67.1 is equal to the probability that z < -1.02 on the standardized normal distribution:

$p(X$

And by looking at the z-score tables, we find that this probability is:

$p(z$

And so,

$p(X$

c)

Here we want to find the probability that a randomly chosen score is greater than 89.1, so

$p(X>89.1)$

First of all, we have to calculate the z-score corresponding to this value of X, which is:

$Z=\frac{89.1-\mu}{\sigma}=\frac{89.1-78.1}{10.8}=1.02$

Then we notice that the z-score tables give only the area on the left of the values on the left of the mean (0), so we have to use the following symmetry property:

$p(z>1.02) =p(z$

Because the normal distribution is symmetric.

But from part b) we know that

$p(z$

Therefore:

$p(X>89.1)=p(z>1.02)=0.1539$

d)

Here we want to find the probability that the randomly chosen score is between 67.1 and 89.1, which can be written as

$p(67.1$

Or also as

$p(67.1$

Since the overall probability under the whole distribution must be 1.

From part b) and c) we know that:

$p(X$

$p(X>89.1)=0.1539$

Therefore, here we find immediately than:

$p(67.1$

7 0

3 years ago

A company introduces a new product for which the number of units sold S is

KonstantinChe [14]

(a) The "average value" of a function over an interval [a,b] is defined to be

(1/(b-a)) times the integral of f from the limits x= a to x = b.

Now S = 200(5 - 9/(2+t))

The average value of S during the first year (from t = 0 months to t = 12 months) is then:

(1/12) times the integral of 200(5 - 9/(2+t)) from t = 0 to t = 12

or 200/12 times the integral of (5 - 9/(2+t)) from t= 0 to t = 12

This equals 200/12 * (5t -9ln(2+t))

Evaluating this with the limits t= 0 to t = 12 gives:

708.113 units., which is the average value of S(t) during the first year.

(b). We need to find S'(t), and then equate this with the average value.

Now S'(t) = 1800/(t+2)^2

So you're left with solving 1800/(t+2)^2 = 708.113

<span>I'll leave that to you</span>

6 0

3 years ago

Who can answer this first POINTS AND BRAINLIEST

Thepotemich [5.8K]

Answer:1.45

Step-by-step explanation:

Divide.

7 0

3 years ago