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3 years ago

Rewrite 1/5and1/2 as a unit rate

Mathematics

1 answer:

denpristay [2]3 years ago

4 0

Answer:

1/2 = 0.5

1/5 = 0.2

Step-by-step explanation:

Hope this helps!!!

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True or false: the standard deviation of a set of data values is never a negative number

Anestetic [448]

True I think !! Not to sure

8 0

3 years ago

A recipe for salad dressing calls for 3 tablespoons of oil for every 2 tablespoons of vinegar. The line represents the relations

pochemuha

Answer:

3:2, 3 to 2

Step-by-step explanation:

4 0

3 years ago

Can you find 2/3 of 120

polet [3.4K]

The only thing you should do is to multiply 2/3 by 120 so you will have 2*120/3 = 80 :)))
i hope this is helpful
have a nice day

6 0

3 years ago

3(2 + 4) = (3•2)+(3•4) is an<br> example of which property?

melomori [17]

Answer:

Distributive Property

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

The lengths of pregnancies are normally distributed with a mean of days and a standard deviation of days. a. Find the probabilit

Alik [6]

Answer:

a) The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of $Z = \frac{X - \mu}{\sigma}$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

b) We have to find X when Z has a p-value of $\frac{a}{100}$ , and X is given by: $X = \mu - Z\sigma$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

Step-by-step explanation:

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.

In this question:

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

a. Find the probability of a pregnancy lasting X days or longer.

The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of $Z = \frac{X - \mu}{\sigma}$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

b. If the length of pregnancy is in the lowest a%, then the baby is premature. Find the length that separates premature babies from those who are not premature.

We have to find X when Z has a p-value of $\frac{a}{100}$ , and X is given by: $X = \mu - Z\sigma$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

8 0

3 years ago