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3 years ago

(Divide the following polynomials (x5 + y5) ÷ (x + y)

1 answer:

7 0

$\text{Use}\ a^5+b^5=(a + b) (a^4 - a^3 b + a^2 b^2 - a b^3 + b^4).\\\\(x^5+y^5):(x+y)=\dfrac{x^5+y^5}{x+y}=\dfrac{(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)}{x+y}\\\\=\boxed{x^4-x^3y+x^2y^2-xy^3+y^4}$

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A bag had 7 orange rocks , 6 green rock , 1 white rock ,7 black rocks , and 2 yellow you randomly pull a rock from the bag ,keep

daser333 [38]

Answer:

$Probability = \frac{7}{253}$

Step-by-step explanation:

Given

$Orange = 7$

$Green = 6$

$White = 1$

$Black = 7$

$Yellow = 2$

Required

Probability of selecting a yellow then orange rock

From the question, we understand that the probability is that of without replacement.

Since the yellow, is first picked.

We need to determine the probability of picking a yellow rock, first.

$P(Yellow) = \frac{n(Yellow)}{Total}$

$P(Yellow) = \frac{2}{7 + 6 + 1 + 7 + 2}$

$P(Yellow) = \frac{2}{23}$

The rock has reduced by 1;

Next is to determine the probability of picking an orange rock

$P(Orange) = \frac{n(Orange)}{Total - 1}$

$P(Orange) = \frac{7}{7 + 6 + 1 + 7 + 2- 1}$

$P(Orange) = \frac{7}{22}$

The required probability is calculated as thus:

$Probability = P(Yellow) * P(Orange)$

$Probability = \frac{2}{23} * \frac{7}{22}$

$Probability = \frac{14}{506}$

$Probability = \frac{7}{253}$

4 0

3 years ago

Find 415 of 619 in simplest form,

Finger [1]

Answer:

415/619

Step-by-step explanation:

619 is a prime number, so the fraction is in simplest form already.

8 0

3 years ago

For which points is the y coordinate less that -1.5

Masja [62]

The bottom 2.

I can’t see the point names cause its blurry but those bottom 2 points should be the answers.

Y coordinates are how high or low
X coordinates are how far left or right

Brainliest if this helps?

3 0

2 years ago

a recipe for one pumpkin pie calls for 1 1/4 cups of sugar Elena only has 1/2 cup of sugar and she needs to make for pumpkin pie

seraphim [82]

Answer: she would need four and a half cups more of sugar.

Step-by-step explanation:

You have to add together all of them sugar she would have needed so (1 1/4 times 4) and then subtract what she already had and that gives you 4 and 1/2 cups she would need. Hope this helps

7 0

3 years ago

A personnel manager is concerned about absenteeism. She decides to sample employee records to determine if absenteeism is distri

nlexa [21]

Answer:

$df=categor-1=6-1=5$

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got $\chi^2_{critc}= 11.0705$

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Solution to the problem

For this case we want to test:

H0: Absenteeism is distributed evenly throughout the week

H1: Absenteeism is NOT distributed evenly throughout the week

We have the following data:

Monday Tuesday Wednesday Thursday Friday Saturday Total

12 9 11 10 9 9 60

The level of significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{60}{6}= 10$ and the expected value is the same for all the days since that's what we want to test.

now we can calculate the statistic:

$\chi^2 = \frac{(12-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(11-10)^2}{10}+\frac{(10-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(9-10)^2}{10}=0.8$

Now we can calculate the degrees of freedom (We know that we have 6 categories since we have information for 6 different days) for the statistic given by:

$df=categor-1=6-1=5$

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got $\chi^2_{critc}= 11.0705$

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

4 0

2 years ago