How do you solve it and what's the answer because this has been stumping me for a while

which sequences are arithmetic? check all that apply. –8.6, –5.0, –1.4, 2.2, 5.8, … 2, –2.2, 2.42, –2.662, 2.9282, … 5, 1, –3, –

sergiy2304 [10]

In an arithmetic sequence, the next term is found by adding a constant term to each number to arrive at the next number. The common difference can be found by subtracting the first term from the second term.

-8.6, -5.0, -1.4, 2.2, 5.8....the common difference here is 3.6
-8.6 + 3.6 = -5.0
-5.0 + 3.6 = -1.4
1.4 + 3.6 = 2.2
2.2 + 3.6 = 5.8
so this IS an arithmetic sequence.

2,-2.2, 2.42, -2.662, 2.9282...there is no common difference..so this is not an arithmetic sequence

5,1,-3,-7,-11....common difference is -4
5 + (-4) = 1
1 + (-4) = -3
-3 + (-4) = -7
-7 + (-4) = -11
this IS an arithmetic sequence

-3,3,9,15,21...common difference is 6
-3 + 6 = 3
3 + 6 = 9
9 + 6 = 15
15 + 6 = 21
this IS an arithmetic sequence

-6.2, -3.1, -1.55, -0.775, -0.3875...this is not an arithmetic sequence

4 0

3 years ago

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A state randomly selects 30 high school and collects data on the average 2 points

Vesnalui [34]

I would be a nice person and answer this, but I’m dumb so sorry 4-4x

5 0

3 years ago

A random sample of 16 students selected from the student body of a large university had an average age of 25 years and a populat

eimsori [14]

Answer:

$t=\frac{25-24}{\frac{2}{\sqrt{16}}}=2$

$p_v =P(t_{15}>2)=0.0320$

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the true mean is significant higher than 24 years.

Step-by-step explanation:

1) Data given and notation

$\bar X=25$ represent the sample mean

$s=2$ represent the standard deviation for the sample

$n=16$ sample size

$\mu_o =24$ represent the value that we want to test

$\alpha$ 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)

Confidence =0.95 or 95%

$\alpha=0.05$

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the mean is higher than 24, the system of hypothesis would be:

Null hypothesis: $\mu \leq 24$

Alternative hypothesis: $\mu > 24$

We don't know the population deviation, so for this case 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:

$t=\frac{25-24}{\frac{2}{\sqrt{16}}}=2$

Calculate the P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=16-1=15$

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

$p_v =P(t_{15}>2)=0.0320$

Conclusion

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the true mean is significant higher than 24 years.

3 0

3 years ago

I gvee brainlilsttttt

Lady_Fox [76]

Answer:

g or h whichever u think is best

Step-by-step explanation:

7 0

3 years ago

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Hii please help i’ll give brainliest!!

cupoosta [38]