In Ivy's neighborhood, 40% of the homes have a swimming pool. If there are 80 homes , how many have a pool

Which graph corresponds to the following inequality? A. B. C. D.

zlopas [31]

I can see the question but if you add a picture maybe i’ll understand it a little more

6 0

2 years ago

54.7912 Rounded to the nearest hundredth

kupik [55]

Answer:

54.79 because if it's 4 and lower, you let them stay the same but if its five and above you let them go higher.

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 $\mu = 522, \sigma = 106$

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes?

So $n = 6, s = \frac{106}{\sqrt{6}} = 43.27$

This probability is 1 subtracted by the pvalue of Z when X = 410.

$Z = \frac{X - \mu}{s}$

$Z = \frac{410 - 523}{43.27}$

$Z = -2.61$

$Z = -2.61$ has a pvalue of 0.0045.

So there is a 1-0.0045 = 0.9955 = 99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes

6 0

3 years ago

What is the product a2/5 (10)?

melisa1 [442]

The correct answer is 2a^2

8 0

3 years ago

If b is the midpoint of ac and if c is the midpoint of bd, then what percent of cd is ac

Sergeu [11.5K]

Answer:

ac is 200% of cd

Step-by-step explanation:

To answer this question, we shall be making a visual representation.

Let’s take it one at a time.

Given;

b is the midpoint of ac

a b c

What this means is that a-b represents 50%, while bc represents another 50%

okay, we move on:

c is the midpoint of bd

b c d

What this means is that bc is 50%, while cd is another 50%.

So let’s combine the representations;

a 50% b 50% c 50% d

Now what does the question says again?

What percent of cd is ac?

From a to c, we can see two 50% which means 100%

while cd is just the regular 50%

So we can see that ac is actually twice cd

What we are saying here is , if cd is x, then ac is 2x

So we can restructure our question to mean, what percent of x is 2x?

That is simply 2x/x * 100 and that is 200%

7 0

3 years ago

Read 2 more answers