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3 years ago

OK right image and easier to read lol

Mathematics

1 answer:

Brums [2.3K]3 years ago

6 0

Answer:

you want us to do all that

Step-by-step explanation:

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The sugar content of the syrup is canned peaches is normally distributed. Assumethe can is designed to have standard deviation 5

andreyandreev [35.5K]

Answer: 0.50477

Step-by-step explanation:

Given : The sugar content of the syrup is canned peaches is normally distributed.

We assume the can is designed to have standard deviation $\sigma=5$ milligrams.

The sampling distribution of the sample variance is chi-square distribution.

Also,The data yields a sample standard deviation of $s=4.8$ milligrams.

Sample size : n= 10

Test statistic for chi-square : $\chi^2=\dfrac{s^2(n-1)}{\sigma^2}$

i.e. $\chi^2=\dfrac{(4.8)^2(10-1)}{(5)^2}=8.2944$

Now, P-value = $P(\chi^2>8.2944)=0.50477$ [By using the chi-square distribution table for p-values.]

Hence, the chance of observing the sample standard deviation greater than 4.8 milligrams = 0.50477

8 0

3 years ago

PLEASE HELP URGENT WILL GIVE BRAINLIEST

Vadim26 [7]

The answer is 2.3716 have a great day y’all

4 0

2 years ago

Factor out this expression: 4x² - 9y2

Marta_Voda [28]

Answer:

(2x+3y)(2x-3y)

Step-by-step explanation:

a² - b² = (a + b)(a - b)

4x² - 9y²

= (2x)² - (3y)²

= (2x + 3y)(2x - 3y)

5 0

3 years ago

What is the inverse of the function

Ilya [14]

Then the inverse is y = (x + 2) / 3.

6 0

3 years ago

This is a question on my partial fractions homework, but no matter what I try I can't figure it out..

Ierofanga [76]

$\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{a_1x+a_0}{(x+1)^2}+\dfrac b{x+2}$
$\implies\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{(a_1x+a_0)(x+2)+b(x+1)^2}{(x+1)^2(x+2)}$
$\implies x^2+x+1=(a_1+b)x^2+(2a_1+a_0+2b)x+(2a_0+b)$
$\implies\begin{cases}a_1+b=1\\2a_1+a_0+2b=1\\2a_0+b=1\end{cases}\implies a_1=-2,a_0=-1,b=3$

So you have

$\displaystyle\int_0^2\frac{x^2+x+1}{(x+1)^2(x+2)}\,\mathrm dx=-2\int_0^2\frac x{(x+1)^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$
$=\displaystyle-2\int_1^3\dfrac{x-1}{x^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$

where in the first integral we substitute $x\mapsto x+1$ .

$=\displaystyle-2\int_1^3\left(\frac1x-\frac1{x^2}\right)\,\mathrm dx-\frac1{1+x}\bigg|_{x=0}^{x=2}+3\ln|x+2|\bigg|_{x=0}^{x=2}$
$=-2\left(\ln|x|+\dfrac1x\right)\bigg|_{x=1}^{x=3}-\dfrac23+3(\ln4-\ln2)$
$=-2\left(\ln3+\dfrac13-1\right)-\dfrac23+3\ln2$
$=\dfrac23+\ln\dfrac89$

4 0

3 years ago