O GRAPHS, FUNCTIONS, AND SEQUENCES

6x7x6x9 this one is for you

Papessa [141]

Answer:HOPE THIS HELPS! :D

2268

Step-by-step explanation:

6x7=42

42x6=252

252x9=2268

so, 6x7x6x9=2268

8 0

3 years ago

Help!!<br><br> State two ways that you could use a slope of -5/3 to graph a line.

kondor19780726 [428]

Answer:

rise/run

Step-by-step explanation:

-5 how high/ low you go on y axis

3 how far you move left on the graph

7 0

2 years ago

Thomas is throwing a graduation party for his son Michael. He went to the store and bought a total of 34 apples. The number of g

kiruha [24]

Answer:

budd

Step-by-step explanation:

xuamz xjxs piozj lloakx xxjx zzmdsjs

8 0

2 years ago

A square wrestling mat has a side length of (4x - 1) feet.

son4ous [18]

Answer:

$(4x-1)^2$

Step-by-step explanation:

Area of a square = $l^2$

If l = $(4x - 1)$ ,

Then area = $(4x - 1)^2$

HOPE THIS HELPS

7 0

2 years ago

Suppose we want to see if American children have higher levels of cholesterol than the average child (i.e., in the entire world

lina2011 [118]

Answer:

$t=\frac{201-190}{\frac{10}{\sqrt{25}}}=5.5$

$p_v =P(t_{(24)}>5.5)=0.00000589$

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 true mean is higher than 190

Step-by-step explanation:

Data given and notation

$\bar X=201$ represent the mean

$s=10$ represent the sample standard deviation for the sample

$n=25$ sample size

$\mu_o =190$ represent the value that we want to test

$\alpha=0.05$ 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 is higher than 190, the system of hypothesis would be:

Null hypothesis: $\mu \leq 190$

Alternative hypothesis: $\mu > 190$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

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

$t=\frac{201-190}{\frac{10}{\sqrt{25}}}=5.5$

P-value

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

$df=n-1=25-1=24$

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

$p_v =P(t_{(24)}>5.5)=0.00000589$

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 true mean is higher than 190

3 0

3 years ago