Whats -3/5×-10/3×-2×9

Which graph shows a linear function.<br> A) <br> B) <br> C) <br> D)

Hitman42 [59]

The First one (A) because it’s a straight line

3 0

3 years ago

Grain is falling from a chute onto the ground, forming a conical pile whose diameter is always three times its height. how high

bezimeni [28]

Given that grain is falling from a chute onto the ground, forming a conical pile whose diameter is always three times its height.

So if D is the diameter and h is the height of the conical pile then we can write:

D=3h

We know that diameter = 2r, where r is the radius

2r=3h

$r=\frac{3h}{2}$

Volume of conical pile is given by formula

$V=\frac{1}{3}\pi r^2h$

Given that volume is 1110 cubic feet.

Now plug the values of Volume and r into equation of volume

$1110=\frac{1}{3}\pi (\frac{3h}{2})^2h$

$1110=\frac{1}{3}\pi\cdot\frac{9h^2}{4}\cdot h$

$1110=\pi\cdot\frac{3h^3}{4}$

$1110\cdot\frac{4}{3\pi}=h^3$

$471.098631552=h^3$

take cube root of both sides

7.78103342467=h

Hence height is approx 7.78 feet.

7 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 $\mu = 522, \sigma = 106$