Look at the photo, it has my question but ignore where it says '1. 5 Min' please

Ten kids can go to the water park for $95. How many can go if there is $200?

posledela

Answer:

only 2 kid's could go since the price of both is 95 for one and times it by two it's 190

5 0

3 years ago

What is the determinant of k= 6 8 0 3

zmey [24]

Answer:

18

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

A betta fish tank is in the shape of a cube. The volume of the fish tank is 216

GarryVolchara [31]

To find this answer, we need to understand how to get the Volume.

We get the volume by the formula: L x W x H

L = Length

W = Width

H = Height

So, we can see that there are 3 units required to know for getting the volume, if that's true, then, to do the inverse operation of finding the volume. That can be done by using thie formula:

$V=a^3$

a= edge

Solving for a

a=V⅓=216⅓=6

<h2>So hence, each of the Cube shaped Fish Tank is 6in³</h2>

now, let's look at how to find whether our answer is right or wrong

6x6=36

36x6=216

So hence, our answer is right!

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