The correct answer is actually around 21%. Are you sure the question wasn't asking what the approximate percent decrease from 85 to 70 was? Because in that case it would be around 17%.

Step-by-step explanation:

6 0

2 years ago

Helppp plzzzz N+5n=

ankoles [38]

I’m pretty sure it’s 6n

3 0

3 years ago

The ages of students in a school are normally distributed with a mean of 15 years and a standard deviation of 2 years. Approxima

worty [1.4K]

We have been given that the ages of students in a school are normally distributed with a mean of 15 years and a standard deviation of 2 years.

We are asked to find the percentage of students that are between 14 and 18 years old.

First of all, we will find z-score corresponding to 14 and 18 using z-score formula.

$z=\frac{x-\mu}{\sigma}$

$z=\frac{14-15}{2}$

$z=\frac{-1}{2}$

$z=-0.5$

Similarly, we will find the z-score corresponding to 18.

$z=\frac{18-15}{2}$

$z=\frac{3}{2}$

$z=1.5$

Now we will find the probability of getting a z-score between $-0.5$ and $1.5$ that is $P(-0.5$ .

$P(-0.5$

Using normal distribution table, we will get:

$P(-0.5$

Let us convert $0.62465$ into percentage.

$0.62465\times 100\%=62.465\%\approx 62.5\%$

Therefore, approximately $62.5\%$ of the students are between 14 and 18 years old.

8 0

3 years ago

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No pic/link just type answer pls.

Genrish500 [490]

Answer:

This problem requires you to use the equation for the volume of a cone. The equation is $\pi r^{2}\frac{h}{3}$ . Plug in the corresponding numbers to solve the equation. The equation should be $\pi (3\frac{3}{16})^{2} \frac{9\frac{3}{4} }{3}$ . Simplify the equation to get 103.74 $in^{3}$ .

7 0

3 years ago

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