Without dividing, select the number that is divisible by 3.

What is the <img src="https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B7%7D" id="TexFormula1" title="\sqrt[3]{7}" alt="\sqrt[3]{7}" alig

zvonat [6]

The answer you want will be 1.91293\dots

4 0

3 years ago

Write an equation then solve

frutty [35]

$21.50 - $4 = $17.50 If you start with $21.50, subtract the $4 entrance fee, you have $17.50 left to spend on rides. To figure out how many rides you can go on, divide $17.50 by $2.50. You can go on 7 rides. Here's the equation:

(21.5-4)/2.5=amount of rides

(21.5-4)/2.5= 7 rides

7 0

3 years ago

Read 2 more answers

Maine has a cold climate in the winter. Which statement about the probability of temperatures falling below 32°F in Maine during

Marianna [84]

Answer:

the probability is 100

this is the most suitable ans

4 0

3 years ago

Identify the function shown in this graph

alexandr402 [8]

It’s c since the line goes down 5 and the y intercept is 3 so it would be y=-5x+3

4 0

3 years ago

A data set with a mean of 34 and a standard deviation of 2.5 is normally distributed

tresset_1 [31]

Answer:

a) $z= \frac{34-34}{2.5}= 0$

$z= \frac{39-34}{2.5}= 2$

And we want the probability from 0 to two deviations above the mean and we got 95/2 = 47.5 %

b) $P(X$

$z= \frac{31.5-34}{2.5}= -1$

So one deviation below the mean we have: (100-68)/2 = 16%

c) $z= \frac{29-34}{2.5}= -2$

$z= \frac{36.5-34}{2.5}= 1$

For this case below 2 deviation from the mean we have 2.5% and above 1 deviation from the mean we got 16% and then the percentage between -2 and 1 deviation above the mean we got: (100-16-2.5)% = 81.5%

Step-by-step explanation:

For this case we have a random variable with the following parameters:

$X \sim N(\mu = 34, \sigma=2.5)$

From the empirical rule we know that within one deviation from the mean we have 68% of the values, within two deviations we have 95% and within 3 deviations we have 99.7% of the data.

We want to find the following probability:

$P(34 < X$

We can find the number of deviation from the mean with the z score formula:

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

And replacing we got

$z= \frac{34-34}{2.5}= 0$

$z= \frac{39-34}{2.5}= 2$

And we want the probability from 0 to two deviations above the mean and we got 95/2 = 47.5 %

For the second case:

$P(X$

$z= \frac{31.5-34}{2.5}= -1$

So one deviation below the mean we have: (100-68)/2 = 16%

For the third case:

$P(29 < X$

And replacing we got:

$z= \frac{29-34}{2.5}= -2$

$z= \frac{36.5-34}{2.5}= 1$

For this case below 2 deviation from the mean we have 2.5% and above 1 deviation from the mean we got 16% and then the percentage between -2 and 1 deviation above the mean we got: (100-16-2.5)% = 81.5%

7 0

3 years ago