Can y’all help me i’m taking a test

Given the numbers 36 60 and 72 write the numbers as a product of their prime factors

PSYCHO15rus [73]

Answer:

36:2 times 2 times 3 times 3.

60:2 times 2 times 3 times 5

72:2 times 2 times 2 times 3 times 3

Step-by-step explanation:

just keep in mind that when it says times it means multiplication sign.

hope this helps

6 0

3 years ago

What is equivalent to 16/81

Lady_Fox [76]

There are a lot of things that are equivalent to 16/81.

16/81 = 0.197530864

Or we can keep multiplying 2 to the numerator and denominator:

16/81 = 32/162 = 64/324 = 128/648

4 0

3 years ago

Read 2 more answers

Suppose a continuous probability distribution has an average of μ=35 and a standard deviation of σ=16. Draw 100 times at random

yulyashka [42]

Answer:

To use a Normal distribution to approximate the chance the sum total will be between 3000 and 4000 (inclusive), we use the area from a lower bound of 3000 to an upper bound of 4000 under a Normal curve with its center (average) at 3500 and a spread (standard deviation) of 160 . The estimated probability is 99.82%.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For sums, we have that the mean is $\mu*n$ and the standard deviation is $s = \sigma \sqrt{n}$

In this problem, we have that:

$\mu = 100*35 = 3500, \sigma = \sqrt{100}*16 = 160$

This probability is the pvalue of Z when X = 4000 subtracted by the pvalue of Z when X = 3000.

X = 4000

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

By the Central Limit Theorem

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

$Z = \frac{4000 - 3500}{160}$

$Z = 3.13$

$Z = 3.13$ has a pvalue of 0.9991

X = 3000

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

$Z = \frac{3000 - 3500}{160}$

$Z = -3.13$

$Z = -3.13$ has a pvalue of 0.0009

0.9991 - 0.0009 = 0.9982

So the correct answer is:

To use a Normal distribution to approximate the chance the sum total will be between 3000 and 4000 (inclusive), we use the area from a lower bound of 3000 to an upper bound of 4000 under a Normal curve with its center (average) at 3500 and a spread (standard deviation) of 160 . The estimated probability is 99.82%.

5 0

3 years ago

If f(x) = 4x ^ 2 and g(x) = x + 1 , find (fg)(x) ).

Irina-Kira [14]

Answer:

$\boxed{\sf \ C \ is \ the \ correct \ answer \ ! \ }$