Xy^2/xy + x^2y^2<br> ANSWER ASAP WILL GIVE BRAINLY

Write out the form of the partial fraction decomposition of the fraction. Do not determine the numerical values of the coefficie

anyanavicka [17]

Factorizing the denominator gives

$\dfrac{x^4}{(x^3+x)(x^2-x+7)} = \dfrac{x^4}{x(x^2+1)(x^2-x+7)} = \dfrac{x^3}{(x^2+1)(x^2-x+7)}$

Then the partial fraction decomposition would take the form

$\dfrac{x^4}{(x^3+x)(x^2-x+7)} = \dfrac{a_0x + b_0}{x^2+1} + \dfrac{a_1x + b_1}{x^2-x+7}$

7 0

2 years ago

One of the more famous anecdotes in the field of statistics is known as the "Lady Tasting Tea" and involves the renowned statist

mote1985 [20]

Answer:

$n=\frac{0.3(1-0.3)}{(\frac{0.01}{1.96})^2}=8067.36$

And rounded up we have that n=8068

Step-by-step explanation:

The margin of error for the proportion interval is given by this:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

We want for this case a 95% of confidence desired, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The margin of error for this case is $ME =\pm 0.01$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

The estimated proportion for this case is $\hat p=0.3$ . And replacing into equation (b) the values from part a we got:

$n=\frac{0.3(1-0.3)}{(\frac{0.01}{1.96})^2}=8067.36$

And rounded up we have that n=8068

7 0

3 years ago

Please don’t troll include work

zheka24 [161]

4 1/8-2 3/4= 4 1/8 - 2 6/8= 1 3/8

2 3/5 + 1 5/6 = 2 18/30 + 1 25/30= 3 43/30 = 4 13/30

8 9/12 + 2 3/8= 8 18/24 + 2 9/24= 10 27/24 = 11 3/24= 11 1/8

3 1/3 - 1 1/2= 3 2/6 - 1 3/6 = 2 8/6 - 1 3/6= 1 5/6

4 0

2 years ago

A grocery store clerk put only cans of tomato soup and cans of chicken soup on a shelf. The ratio of the number of cans of chick

larisa [96]

Answer:

9 cans of tomato soup

Step-by-step explanation:

We are told that only cans of tomato soup and cans of chicken soup are put on the shelf.

Now, we are told that the ratio of the number of cans of chicken noodle soup to the total number of cans on the shelf was 8 to 17.

In fraction, this is 8/17

Now, this implies that there are 8 cans of chicken noodle soup while there are a total of 17 cans on the shelf.

Thus,

Since only cans of chicken noodles and cans of tomato soup are on the shelf, it means;

Cans of tomato soup = 17 - 8 = 9

8 0

3 years ago

Consider the following hypothesis test.H0:μ1−μ2=0 Ha:μ1−μ2≠0The following results are for two independent samples taken from the

Julli [10]

Answer:

a) $z =\frac{104-106}{\sqrt{\frac{8.4^2}{80} +\frac{7.6^2}{70}}}= -1.53$

b) $p_v =2*P(z$

c) Since the p value is higher than the significance level provided we have enogh evidence to FAIL to reject the null hypothesis and we can't conclude that the true means are different at 5% of significance

Step-by-step explanation:

Information given

$\bar X_{1}= 104$ represent the mean for 1

$\bar X_{2}= 106$ represent the mean for 2

$\sigma_{1}= 8.4$ represent the population standard deviation for 1

$\sigma_{2}= 7.6$ represent the population standard deviation for 2

$n_{1}=80$ sample size for the group 1

$n_{2}=70$ sample size for the group 2

z would represent the statistic

Hypothesis to test

We want to check if the two means for this case are equal or not, the system of hypothesis would be:

H0: $\mu_{1}=\mu_{2}$

H1: $\mu_{1} \neq \mu_{2}$

The statistic would be given by:

$z =\frac{\bar X_1-\bar X_2}{\sqrt{\frac{\sigma^2_1^2}{n_1} +\frac{\sigma^2_2^2}{n_2}}}=$ (1)

Part a

Replacing we got:

$z =\frac{104-106}{\sqrt{\frac{8.4^2}{80} +\frac{7.6^2}{70}}}= -1.53$

Part b

The p value would be given by this probability:

$p_v =2*P(z$

Part c

Since the p value is higher than the significance level provided we have enogh evidence to FAIL to reject the null hypothesis and we can't conclude that the true means are different at 5% of significance

6 0

2 years ago

Xy^2/xy + x^2y^2 ANSWER ASAP WILL GIVE BRAINLY