BOBTHEBUILDER can do it

Expand (4x-3y)^4 using pascal's triangle ...?

jenyasd209 [6]

Using row 4:

coefficients are: 1, 4, 6, 4, 1 

a^4 + a^3b + a^2b^2 + ab^3 + b^4 

Now adding the coefficients: 

1a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + 1b^4 

Substitute a and b: 

a = 4x 

b = -3y 

1(4x)^4 + 4(4x)^3(-3y) + 6(4x)^2(-3y)^2 + 4(4x)(-3y)^3 + 1(-3y)^4 

Now simplify the above: 

256x^4 - 768x^3y + 864x^2y^2 - 432xy^3 + 81y^4

8 0

3 years ago

Can somebody help me wit this ? Cause Ion understand this .

Triss [41]

Answer:

wow that's nice of your

4 0

3 years ago

Read 2 more answers

What is (2ab-a+2) - (-ab-2)

Greeley [361]

Step-by-step explanation:

iknow but its hard to expalin

8 0

2 years ago

Find the area of the regular polygon

Vsevolod [243]

Answer:

433.4

Step-by-step explanation:

Height will be (14/2)/tan36 which is 9.63, then area for one triangle is 1/2*hb which is .5*9.63*7 is 43.34, times 10 you get 433.4

7 0

3 years ago

Life Expectancies In a study of the life expectancy of people in a certain geographic region, the mean age at death was years an

Sphinxa [80]

Answer:

The probability that the mean life expectancy of the sample is less than X years is the p-value of $Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$ , in which $\mu$ is the mean life expectancy, $\sigma$ is the standard deviation and n is the size of the sample.

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 normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

We have:

Mean $\mu$ , standard deviation $\sigma$ .

Sample of size n:

This means that the z-score is now, by the Central Limit Theorem:

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

Find the probability that the mean life expectancy will be less than years.

The probability that the mean life expectancy of the sample is less than X years is the p-value of $Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$ , in which $\mu$ is the mean life expectancy, $\sigma$ is the standard deviation and n is the size of the sample.

8 0

2 years ago