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3 years ago

Math problem easy question! Giving brainlist

Mathematics

2 answers:

Oksanka [162]3 years ago

8 0

Answer:

It would be A

Step-by-step explanation:

Ket [755]3 years ago

4 0

A: multiplication

Hope this helped, if not, don’t hate.

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3 years ago

Many utility companies promote energy conservation by offering discount rates to consumers who keep their energy usage below cer

aleksley [76]

Answer:

a. 0.1681 = 16.81% probability that all five qualify for the favorable rate.

b. 0.5283 = 52.83% probability that at least four qualify for the favorable rates

Step-by-step explanation:

For each Puerto Rico resident, there are only two possible outcomes. Either they qualify for discounted rates, or they do not. The probability of a person in the sample qualifying for discounted rates is independent of any other person in the sample. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

70% of the island residents of Puerto Rico have reduced their electricity usage sufficiently to qualify for discounted rates.

This means that $p = 0.7$

Five residential subscribers are randomly selected from San Juan, Puerto Rico

This means that $n = 5$

a. All five qualify for the favorable rate

This is P(X = 5). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{5,5}.(0.7)^{5}.(0.3)^{0} = 0.1681$

0.1681 = 16.81% probability that all five qualify for the favorable rate.

b. At least four qualify for the favorable rates

This is

$P(X \geq 4) = P(X = 4) + P(X = 5)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 4) = C_{5,4}.(0.7)^{4}.(0.3)^{1} = 0.3602$

$P(X = 5) = C_{5,5}.(0.7)^{5}.(0.3)^{0} = 0.1681$

$P(X \geq 4) = P(X = 4) + P(X = 5) = 0.3602 + 0.1681 = 0.5283$

0.5283 = 52.83% probability that at least four qualify for the favorable rates

5 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

3 years ago