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3 years ago

6

Rectangle A is similar to rectangle B. Rectangle B has sides that are 4 times longer than rectangle A. what is the relationship

between the areas of retangles A and B?

1 answer:

Irina18 [472]3 years ago

4 0

Rectangle B is 16 times the size of Rectangle A. Hope this helps

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If angle 1 is 125 and angle 3 is 3x -1 what is x

Amiraneli [1.4K]

Answer: X means you have to times

Step-by-step explanation: So you will have to do 3x3x3

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2 years ago

To decrease your household water usage by 10% a month means that a household using 4000 gallons of water a month would cut down

a_sh-v [17]

Answer:

3600

Step-by-step explanation:

0.1*4000=400

4000-400=3600 gallons

8 0

2 years ago

Help.........................<br><br>

garri49 [273]

<h3>Answer: Choice A</h3>

$x^2\left(\sqrt[4]{x^2}\right)$

=====================================================

Explanation:

The fourth root of x is the same as x^(1/4)

I.e,

$\sqrt[4]{x} = x^{1/4}$

The same applies to x^10 as well

$\sqrt[4]{x^{10}} = \left(x^{10}\right)^{1/4}$

Multiply the exponents 10 and 1/4 to get 10/4

$\sqrt[4]{x^{10}} = \left(x^{10}\right)^{1/4} = x^{10*1/4} = x^{10/4}$

$\sqrt[4]{x^{10}} = x^{10/4}$

-----------------------

If we have an expression in the form x^(m/n), with m > n, then we can simplify it into an equivalent form as shown below

$x^{m/n} = x^a\sqrt[n]{x^b}$

The 'a' and 'b' are found through dividing m/n

m/n = a remainder b

'a' is the quotient, b is the remainder

-----------------------

The general formula can easily be confusing, so let's replace m and n with the proper numbers. In this case, m = 10 and n = 4

m/n = 10/4 = 2 remainder 2

We have a = 2 and b = 2

So

$x^{m/n} = x^a\sqrt[n]{x^b}$

turns into

$x^{10/4} = x^2\sqrt[4]{x^2}$

which means

$\sqrt[4]{x^{10}} = {x^2} \sqrt[4]{x^2}$

7 0

2 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

1.What is the mean of the given distribution, and which type of skew does it exhibit? {4.5, 3, 1, 2, 4, 3, 6, 4.5, 4, 5, 2, 1, 3

marysya [2.9K]

Mean = (4.5 + 3 + 1 + 2 + 4 + 3 + 6 + 4.5 + 4 + 5 + 2 + 1 + 3 + 4 + 3 + 2)/16 = 50/16 = 3.125

The data set arranged in order is 1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4.5, 4.5, 5, 6

The mean is relatively in the middle, therefore if exhibits center skewness.

4 0

2 years ago