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3 years ago

Look at the triangular prism below. Each triangular face or the

Mathematics

2 answers:

Goryan [66]3 years ago

7 0

Answer:

The volume of the triangular prism is $V=72\ cm^3$

Step-by-step explanation:

we know that

The volume of the triangular prism is equal to

$V=BL$

where

B is the area of the triangular face

L is the length of the prism

step 1

Find the area of the triangular face

The area of a triangle is equal to

$B=\frac{1}{2} bh$

we have

$b=3\ cm\\h=4\ cm$

substitute

$B=\frac{1}{2}(3)(4)=6\ cm^2$

step 2

Find the volume

we have

$B=6\ cm^2$

$L=12\ cm$

substitute in the formula of volume

$V=6(12)$

$V=72\ cm^3$

Jlenok [28]3 years ago

6 0

Answer:

The volume of the triangular prism is V=72\ cm^3

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The area of a rectangle is 14 m2 , and the length of the rectangle is 3 m less than double the width. find the dimensions of the

muminat

Width =3.5, length = 4

Let's make a few equations to summarize what we know.:
a = 14
l = 2w -3
a = wl

Now let's substitute the expression for l into the expression for area and solve.
a = w(2w - 3)
a = 2w^2 - 3w
14 = 2w^2 - 3w

0 = 2w^2 - 3w - 14

which factors into
(2w - 7) (w + 2)

Solving for w
2w - 7 = 0
2w = 7
w = 3.5

w+2 = 0
w = -2

So w can be either 3.5 or -2. Since a negative width doesn't make sense, w = 3.5

Let's verify
w = 3.5
l = 2w - 3
l = 2*3.5 - 3
l = 7 - 3
l = 4
a = wl
a = 3.5 * 4
a = 14

So the width is 3.5 and the length is 4.

3 0

3 years ago

The annual precipitation amounts in a certain mountain range are normally distributed with a mean of 104 inches, and a standard

dsp73

Answer:

91.92% probability that the mean annual precipitation during 49 randomly picked years will be less than 106.8 inches

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$