Divide 64.5 by 0.015, leaving the answer in standard form

Can someone please help me with this, btw don’t mind the white-out

inysia [295]

Answer:

x< 10

Step-by-step explanation:

To start the absolute difference between 20 and 5 is 15. Next, write out an equation for the problem:

x * 1.5 < 15,

Then divide by 1.5 on both sides...

Boom! x < 10.

7 0

3 years ago

Please help me with this !!!!!!!!!!!!!!!!

sergij07 [2.7K]

Answer:

option D. 29

Step-by-step explanation:

since the selling price per mow on weekdays is 45 so divide the total earned with the price

1305 / 45 = <em>29</em>

<em>this is the answer</em>

4 0

3 years ago

Miss smith’s pay is directly proportional to the number of hours she work at the airport. Her pay for 22 hours is $297.How much

inna [77]

Answer:

$540

Step-by-step explanation:

You need first to calculate the hour rate that is payment divided to total number of hours done.

297/22 = 13.5

Now we know that she is earning $13.5/1 hour

We just need to multiply by 22 hours and we have 13.5*22 = $540

3 0

3 years ago

Read 2 more answers

A garden has 8 white roses and 11 red roses. One rose is plucked from the garden and then one more rose is plucked without repla

liberstina [14]

P(r/w) is the probability of picking a red rose at first picking and a white rose at second picking.
P(w/r) is the probability of picking a white rose at first picking and a red rose at second picking.

P(r/w) = $\frac{11}{19}$ × $\frac{8}{18}$ = $\frac{88}{342}$
P(w/r) = $\frac{8}{19}$ × $\frac{11}{18}$ = $\frac{88}{342}$

Notice that the second fraction is out of 18 because the second picking of rose will be out of 18 since the first rose is not replaced.

P(r/w) equals to P(w/r)

3 0

4 years ago

Life Expectancies In a study of the life expectancy of people in a certain geographic region, the mean age at death was years an

Sphinxa [80]

Answer:

The probability that the mean life expectancy of the sample is less than X years is the p-value of $Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$ , in which $\mu$ is the mean life expectancy, $\sigma$ is the standard deviation and n is the size of the sample.

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 normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

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

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

We have:

Mean $\mu$ , standard deviation $\sigma$ .

Sample of size n:

This means that the z-score is now, by the Central Limit Theorem:

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

Find the probability that the mean life expectancy will be less than years.

The probability that the mean life expectancy of the sample is less than X years is the p-value of $Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$ , in which $\mu$ is the mean life expectancy, $\sigma$ is the standard deviation and n is the size of the sample.

8 0

3 years ago