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3 years ago

5

Simplify the following expression: 9^−53 ⋅ 9^37 Please also explain your answer. Thanks!

1 answer:

Alexus [3.1K]3 years ago

8 0

Answer: 9^-16

Step-by-step explanation:

a^b*a^c= a^b+c

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Combine like terms. 12m2+6q-9m2

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3m2+6q
m2 is a like term so take the 9m2 from the 12m2

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3 years ago

One polygon has a side of length 3 feet.A similar polygon has a corresponding side of length 9 feet.the ratio of the perimeter o

Olegator [25]

The ratio of the smaller to the larger is 3:9. That can also be written as a fraction, $\frac{3}{9}$ , which of course can be reduced to $\frac{1}{3}$ or 1:3. That's your ratio.

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3 years ago

Helppp me <br> Idk the AWNSER

LekaFEV [45]

The answer to number one is b

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3 years ago

According to the Knot, 22% of couples meet online. Assume the sampling distribution of p follows a normal distribution and answe

Ann [662]

Using the <em>normal distribution and the central limit theorem</em>, we have that:

a) The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

b) There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

c) There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

It 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, which is the percentile of X.

By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1 - p)}{n}}$ , as long as $np \geq 10$ and $n(1 - p) \geq 10$ .

In this problem:

22% of couples meet online, hence p = 0.22.

A sample of 150 couples is taken, hence n = 150.

Item a:

The mean and the standard error are given by:

$\mu = p = 0.22$

$s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.22(0.78)}{150}} = 0.0338$

The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

Item b:

The probability is <u>one subtracted by the p-value of Z when X = 0.25</u>, hence:

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

By the Central Limit Theorem:

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

$Z = \frac{0.25 - 0.22}{0.0338}$

Z = 0.89

Z = 0.89 has a p-value of 0.8133.

1 - 0.8133 = 0.1867.

There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

Item c:

The probability is the <u>p-value of Z when X = 0.2 subtracted by the p-value of Z when X = 0.15</u>, hence:

X = 0.2:

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

$Z = \frac{0.2 - 0.22}{0.0338}$

Z = -0.59

Z = -0.59 has a p-value of 0.2776.

X = 0.15:

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

$Z = \frac{0.15 - 0.22}{0.0338}$

Z = -2.07

Z = -2.07 has a p-value of 0.0192.

0.2776 - 0.0192 = 0.2584.

There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

To learn more about the <em>normal distribution and the central limit theorem</em>, you can check brainly.com/question/24663213

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2 years ago

A box contains one yellow, two red, and three green balls. Two balls are randomly chosen without replacement. Define the followi

Vinil7 [7]

Answer:

Step-by-step explanation:

Given that a box contains one yellow, two red, and three green balls.

Two balls are randomly chosen without replacement.

A) one ball is yellow:

This means one is yellow and other can be any other colour

i.e. drawing one yellow and one from 5 remaining.

This can be as (y,r) (y,g) (r,y) or (g,y)

where y for yellow, r for red and g for green.

B) { At least one ball is red }

This means the two balls should not be different from red.

The favorable outcomes are two red, (one red, one yellow) (one green, one red). But should not be (green,yellow)

C) :{ Both balls are green }

Here the balls drawn should be only green, green and not any other combination.

D) { Both balls are of the same col

Since yellow is one, there cannot be two yellows

Hence only possibilities are (red, red) or (green, green)

7 0

3 years ago