If 5 woodchucks could chuck 50 cords of

18 less than the product of 42 and x write as an algebraic expression

Afina-wow [57]

Less than = subtraction = -
Product = multiplication

(42 + x) - 18

4 0

3 years ago

Find the greatest common factor of 36 and 54

8090 [49]

Hey there!

To find the greatest common factor you will need to find all the factors of 36 and 54. (A factor is a number that can be divided into another number).

54 - 1, 2, 3, 6, 9, 18, 27.

36 - 1, 2, 3, 4, 6, 9, 12, 18.

As you can see here, the factor that is the greatest is 18. Therefore, that is your answer.

Hope this helps! :)

8 0

3 years ago

Read 2 more answers

A researcher claims that the mean of the salaries of elementary school teachers is greater than the mean of the salaries of seco

Brut [27]

Answer:

$t=\frac{(48250-45630)}{\sqrt{\frac{3900^2}{26}+\frac{5530^2}{24}}}}=1.921$

$df=n_{A}+n_{B}-2=26+24-2=48$

Since is a one sided test the p value would be:

$p_v =P(t_{(48)}>1.921)=0.0303$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the mean for elementary school teachers is significantly higher than the mean for secondary teachers at 5% of significance

Step-by-step explanation:

Data given and notation

$\bar X_{A}=48250$ represent the mean of elementary teachers

$\bar X_{B}=45630$ represent the mean for secondary teachers

$s_{A}=3900$ represent the sample standard deviation for elementary teacher

$s_{B}=5530$ represent the sample standard deviation for secondary teachers

$n_{A}=26$ sample size selected

$n_{B}=24$ sample size selected

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean of the salaries of elementary school teachers is greater than the mean of the salaries of secondary school teachers, the system of hypothesis would be:

Null hypothesis: $\mu_{A}-\mu_{B}\leq 0$

Alternative hypothesis: $\mu_{A}-\mu_{B}>0$

We don't know the population deviations, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{(\bar X_{A}-\bar X_{B})}{\sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{B}}{n_{B}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{(48250-45630)}{\sqrt{\frac{3900^2}{26}+\frac{5530^2}{24}}}}=1.921$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n_{A}+n_{B}-2=26+24-2=48$

Since is a one sided test the p value would be:

$p_v =P(t_{(48)}>1.921)=0.0303$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the mean for elementary school teachers is significantly higher than the mean for secondary teachers at 5% of significance

5 0

3 years ago

If Josie wants to buy cupcakes and cookies for a party with a budget of

kolezko [41]

Answer:

50 cupcakes, 150 cookies

Step-by-step explanation:

3 0

3 years ago

The following statement are true about complex fraction,except?

Paraphin [41]

You didn’t even put the choices...