Round to the nearest hundredth

FIRST PERSON TO ANSWER THIS CORRECTLY GETS BRAINLEIST

cupoosta [38]

Answer:

Nine less than twice a number 2x-9

Seven divided by a number 7-x

Ninety-nine plus the difference of eleven and five 99+(11-5)

Four-fifths more than three times. Number 3x+4/5

Step-by-step explanation:

Please mark brainliest

5 0

2 years ago

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A board game that normally costs $30 is on sale for 25 percent off. What is the sale price of the game?

Paha777 [63]

We can use two different ways to solve this equation:

Because the board is 25 off, we can multiply the price and subtract the result:

30 - 0.25(30) = 30 - 7.5 = $22.50

We can also solve by multiplying the total price by 0.75

(1 - 0.25)(30) = 0.75(30) = $22.50

The sales price is $22.50

4 0

3 years ago

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Find the GCF of the following literal terms.<br><br> m^7n^4p^3 and mn^12p^5

Sidana [21]

I hope this helps you

3 0

2 years ago

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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 $\mu = 522, \sigma = 106$