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3 years ago

What is 2.31 divided by 3?

Mathematics

2 answers:

zimovet [89]3 years ago

8 0

2.31 / 3, the answer is 0.77

Kobotan [32]3 years ago

5 0

.77 is the answer. I hope this helps!

You might be interested in

Which expression is equivalent to 36^-1/2

inn [45]

ANSWER

$( {36})^{ (- \frac{1}{2}) } = \frac{1}{6}$

EXPLANATION

The given expression is

${36}^{( - \frac{1}{2} )}$
This is having a negative index. We must first of all change to a positive index.

Recall that,

${a}^{ - m} = \frac{1}{ {a}^{m} }$

We apply this law of exponents to get,

${36}^{( - \frac{1}{2} )} = \frac{1}{ {36}^{( \frac{1}{2} )} }$

We cab rewrite the given expression to obtain;

${36}^{( - \frac{1}{2} )} = \frac{1}{ \sqrt{36} }$

This will simplify to give us,

${36}^{( - \frac{1}{2} )} = \frac{1}{ 6 }$

6 0

3 years ago

Read 2 more answers

Help!!!!!!!!!!!!!!!!!!!!!

Stells [14]

your answer is B. I believe im not 100% sure though.

7 0

3 years ago

45% of what distance is 135 miles

horrorfan [7]

300 miles. Hope it helps

8 0

3 years ago

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a number has a 4 in the thousands place the number also has a digit whose value is 10 times the value of the 4 on the thousands

maks197457 [2]

Well I can still answer that question without the following and it will be 44,000

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