All categories

4 years ago

Help i'll make you the brainliest

Mathematics

1 answer:

9966 [12]4 years ago

7 0

Answer:

D) b = $\frac{S-2ac}{2a + 2c}$

Step-by-step explanation:

First, move all the terms that do not include b to the left side with S:

S = 2ab + 2bc + 2ac

S - 2ac = 2ab + 2bc

Now, factor out 2b from the right side:

S - 2ac = 2b(a + c)

Divide both sides by 2(a + c):

b = $\frac{S-2ac}{2(a + c)}$

Finally, multiply out the denominator:

b = $\frac{S-2ac}{2a + 2c}$

You might be interested in

Suppose data made available through a health system tracker showed health expenditures were $10,348 per person in the United Sta

nignag [31]

Answer:

a) 30.08% probability the sample mean will be within $100 of the population mean.

b) 0% probability the sample mean will be greater than $12,600

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the Central Limit Theorem.

Normal probability distribution

When the distribution is normal, we use 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.

In this question, we have that:

$\mu = 10348, \sigma = 2500, n = 100, s = \frac{2500}{\sqrt{100}} = 250$

a. What is the probability the sample mean will be within Â±$100 of the population mean?

This is the pvalue of Z when X = 100 divided by s subtracted by the pvalue of Z when X = -100 divided by s. So

$Z = \frac{100}{250} = 0.4$

$Z = -\frac{100}{250} = -0.4$

Z = 0.4 has a pvalue of 0.6554, Z = -0.4 has a pvalue of 0.3556

0.6554 - 0.3546 = 0.3008

30.08% probability the sample mean will be within $100 of the population mean.

b. What is the probability the sample mean will be greater than $12,600?

This is 1 subtracted by the pvalue of Z when X = 12600. So

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

By the Central Limit Theorem

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

$Z = \frac{12600 - 10348}{250}$

$Z = 9$

$Z = 9$ has a pvalue of 1.

1 - 1 = 0

0% probability the sample mean will be greater than $12,600

5 0

3 years ago

Who can help me d e f thanks

12345 [234]

$y = (2ax^2 + c)^2 (bx^2 - cx)^{-1}$

Product rule:

$y' = \bigg((2ax^2+c)^2\bigg)' (bx^2-cx)^{-1} + (2ax^2+c)^2 \bigg((bx^2-cx)^{-1}\bigg)'$

Chain and power rules:

$y' = 2(2ax^2+c)\bigg(2ax^2+c\bigg)' (bx^2-cx)^{-1} - (2ax^2+c)^2 (bx^2-cx)^{-2} \bigg(bx^2-cx\bigg)'$

Power rule:

$y' = 2(2ax^2+c)(4ax) (bx^2-cx)^{-1} - (2ax^2+c)^2 (bx^2-cx)^{-2} (2bx - c)$

Now simplify.

$y' = \dfrac{8ax (2ax^2+c)}{bx^2 - cx} - \dfrac{(2ax^2+c)^2 (2bx-c)}{(bx^2-cx)^2}$

$y' = \dfrac{8ax (2ax^2+c) (bx^2 - cx) - (2ax^2+c)^2 (2bx-c)}{(bx^2-cx)^2}$

$y = \dfrac{3bx + ac}{\sqrt{ax}}$

Quotient rule:

$y' = \dfrac{\bigg(3bx+ac\bigg)' \sqrt{ax} - (3bx+ac) \bigg(\sqrt{ax}\bigg)'}{\left(\sqrt{ax}\right)^2}$

$y'= \dfrac{\bigg(3bx+ac\bigg)' \sqrt{ax} - (3bx+ac) \bigg(\sqrt{ax}\bigg)'}{ax}$

Power rule:

$y' = \dfrac{3b \sqrt{ax} - (3bx+ac) \left(-\frac12 \sqrt a \, x^{-1/2}\right)}{ax}$

Now simplify.

$y' = \dfrac{3b \sqrt a \, x^{1/2} + \frac{\sqrt a}2 (3bx+ac) x^{-1/2}}{ax}$

$y' = \dfrac{6bx + 3bx+ac}{2\sqrt a\, x^{3/2}}$

$y' = \dfrac{9bx+ac}{2\sqrt a\, x^{3/2}}$

$y = \sin^2(ax+b)$

Chain rule:

$y' = 2 \sin(ax+b) \bigg(\sin(ax+b)\bigg)'$

$y' = 2 \sin(ax+b) \cos(ax+b) \bigg(ax+b\bigg)'$

$y' = 2a \sin(ax+b) \cos(ax+b)$

We can further simplify this to

$y' = a \sin(2(ax+b))$

using the double angle identity for sine.

7 0

2 years ago

Solve for x.<br> 2(3x+8) = 40<br> Simplify your answer as much as possible.

maxonik [38]

6x+16=40
40-16=24
24=6x
X=4

4 0

3 years ago

Read 2 more answers

What are the solutions to the equation 3(x – 4)(x + 5) = 0?

iris [78.8K]

Answer:

x=4 and x= - 5

Step-by-step explanation:

In order to solve this we should re-arrange the equation:

$(x-4)(x+5)=\frac{0}{3}$

This is equal to:

$(x-4)(x+5)=0$

Then we can seperate this into:

$(x-4)=0\\(x+5)=0$

So solving for x in both cases we get

$x-4=0 \\x=4\\And\\x+5=0\\x=-5\\$

4 0

3 years ago

James wants to enlarge a photograph that is 6 inches wide and 4 inches tall so that it fits into the frame shown above. How tall

Andrej [43]

You would set up a proportion 6/4 = 30/x you then can do what i call the fish and multiply 30 by 4 and the divide by 6 which will give you x which your answer is 20

4 0

4 years ago