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3 years ago

13

Hannah and Francine have $120. Hannah and Peter have $230. Peter has 6times as much money as Francine. How much money does Hanna

h have?

1 answer:

lara31 [8.8K]3 years ago

4 0

I believe Hannah has 98$

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HELP PLEASE! URGENT!

finlep [7]

Answer:

$\boxed{B. \: 5 {x}^{6} + 5 {x}^{5} + 8 {x}^{2} - x + 2}$

Step-by-step explanation:

$= > (2 {x}^{6} + 8 {x}^{2} - x + 1) + (3 {x}^{6} + 5 {x}^{5} + 1) \\ \\ = > 2 {x}^{6} + 8 {x}^{2} - x + 1 + 3 {x}^{6} + 5 {x}^{5} + 1 \\ \\ = > 2 {x}^{6} + 3 {x}^{6} + 5 {x}^{5} + 8 {x}^{2} - x + 1 + 1 \\ \\ = > 5 {x}^{6} + 5 {x}^{5} + 8 {x}^{2} - x + 2$

8 0

3 years ago

Please help, I really need help

7nadin3 [17]

So the teacher bought 1/4 of the brownie pan. And it came out to $4.
I hope this helps! :)

5 0

3 years ago

Integrala x la a treia ori ln la a doua dx va rog

Studentka2010 [4]

I don't speak Romanian, but the closest translation for this suggests you're trying to compute

$\displaystyle \int x^3 \ln(x)^2 \, dx$

Integrate by parts:

$\displaystyle \int x^3 \ln(x)^2 \, dx = uv - \int v \, du$

where

u = ln(x)² ⇒ du = 2 ln(x)/x dx

dv = x³ dx ⇒ v = 1/4 x⁴

$\implies \displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac12 \int x^3 \ln(x) \, dx$

Integrate by parts again:

$\displaystyle \int x^3 \ln(x) \, dx = u'v' - \int v' du'$

where

u' = ln(x) ⇒ du' = dx/x

dv' = x³ dx ⇒ v' = 1/4 x⁴

$\implies \displaystyle \int x^3 \ln(x) \, dx = \frac14 x^4 \ln(x) - \frac14 \int x^3 \, dx$

So, we have

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac12 \left(\frac14 x^4 \ln(x) - \frac14 \int x^3 \, dx \right)$

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac18 \int x^3 \, dx$

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac18 \left(\frac14 x^4\right) + C$

$\displaystyle \int x^3 \ln(x)^2 \, dx = \frac14 x^4 \ln(x)^2 - \frac18 x^4 \ln(x) + \frac1{32} x^4 + C$

$\boxed{\displaystyle \int x^3 \ln(x)^2 \, dx = \frac1{32} x^4 \left(8\ln(x)^2 - 4\ln(x) + 1\right) + C}$

3 0

2 years ago

What function can be used to model data pairs that have a common ratio?

liubo4ka [24]

Correct answer is C.

For example 2, 4, 8, 16, 32...

$2=\frac{4}{2} = \frac{8}{4} = \frac{16}{8} = \frac{32}{16} =...$

$y=2^x$

4 0

3 years ago

Read 2 more answers

A random survey of 1000 students nationwide showed a mean ACT score of 21.1. Ohio was not used. A survey of 500 randomly selecte

vodka [1.7K]

Answer:

We conclude that the mean Ohio score is below the national average.

Step-by-step explanation:

We are given that a random survey of 1000 students nationwide showed a mean ACT score of 21.1. Ohio was not used.

A survey of 500 randomly selected Ohio scores showed a mean of 20.8. The population standard deviation is 3.

<u><em>Let </em></u> $\mu$ <u><em> = mean Ohio scores.</em></u>

So, Null Hypothesis, $H_0$ : $\mu$ $\geq$ 21.1 {means that the mean Ohio score is above or equal the national average}

Alternate Hypothesis, $H_A$ : $\mu$ < 21.1 {means that the mean Ohio score is below the national average}

The test statistics that would be used here <u>One-sample z test statistics</u> as we know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n}}}$ ~ N(0,1)

where, $\bar X$ = sample mean Ohio score = 20.8

$\sigma$ = population standard deviation = 3

n = sample of Ohio = 500

So, <u><em>test statistics</em></u> = $\frac{20.8-21.1}{\frac{3}{\sqrt{500}}}$

= -2.24

The value of z test statistics is -2.24.

<em>Now, at 0.1 significance level the z table gives critical value of -1.2816 for left-tailed test.</em><em> Since our test statistics is less than the critical values of z as -2.24 < 1.2816, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><em><u>we reject our null hypothesis</u></em><em>.</em>

Therefore, we conclude that the mean Ohio score is below the national average.

4 0

3 years ago