Please help, I've tried learning this multiple times but just can't get it down

Find the island height of a pyramid with a volume of 432 ft^3 if the base is a square with a side length of 12 ft. Round to the

dmitriy555 [2]

So the picture below shows what the slant height is. It also shows that the pyramid can form a right triangle, in which we can use the pythagorean theorem, which is $leg^2+leg^2=hypotenuse^2$ .

But first, we have to find the height of this pyramid. The volume formula for a square pyramid is $A=\frac{lwh}{3}$ . Since we know its a square pyramid, the length and width are going to be the same (12 ft), and we know that the volume is 432 ft^3. Using this info, we can solve for the height:

$432=\frac{12*12*h}{3}\\ 1296=12*12*h\\ 1296=144*h\\ 9=h$

So now we can use the pythagorean theorem (remember to split the 12 by 2, since the right triangle has only half the length of the base). Our equation will be solved as such:

$6^2+9^2=s^2\\ 36+81=s^2\\ 117=s^2\\ 10.8=s$

In short, the slant height is 10.8 ft.

3 0

3 years ago

What is the midpoint of (-2,3) and (10,3) *simple answer please

Allushta [10]

The "midpoint formula" is the same as the "average formula:"

x-coordinate of the midpoint = average of -2 and 10:

= (-2 + 10) / 2 = 4

y-coordinate of the midpoint = average of 3 and 3:

= (3+3) / 2 = 3

Thus, the midpoint of the given line segment is (4,3).

5 0

3 years ago

It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute

Molodets [167]

Answer:

10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Normal probability distribution

Problems of normally distributed samples can be 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.

Mildly obese

Normally distributed with mean 375 minutes and standard deviation 68 minutes. So $\mu = 375, \sigma = 68$

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?

So $n = 6, s = \frac{68}{\sqrt{6}} = 27.76$

This probability is 1 subtracted by the pvalue of Z when X = 410.

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

$Z = \frac{410 - 375}{27.76}$

$Z = 1.26$

$Z = 1.26$ has a pvalue of 0.8962.

So there is a 1-0.8962 = 0.1038 = 10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

Lean

Normally distributed with mean 522 minutes and standard deviation 106 minutes. So $\mu = 522, \sigma = 106$