Ask question

Ask question

All categories

2 years ago

9

Please help me I really need it please I will friend request you I will give you points, even though I don't have some right now

I will still give you a lot. I will make you brainliest whatever you want... Please

1 answer:

guapka [62]2 years ago

5 0

Answer:

9.53939 = 9.53 or rounded to 9.54

Step-by-step explanation:

You might be interested in

Kyle is a basketball player. His bottle was full at the beginning of the game. At the end of the first quarter, he drank 5/7 of

gizmo_the_mogwai [7]

Answer:

1 4/35

Step-by-step explanation:

All I did was do 5/7 + 2/5.

5 0

3 years ago

The blue dot is at what value on the number line? 5

Eduardwww [97]

Answer:

value of blue dot is -6

Step-by-step explanation:

-6 is visible if you go 3 units to the left of -3. -3 - 3 = -6.

4 0

2 years ago

Calculating the degrees of freedom, the sample variance, and the estimated standard error for evaluations using the t statistic

Yuliya22 [10]

Answer:

a) For the first part we have a sample of n =10 and we want to find the degrees of freedom, and we can use the following formula:

$df = n-1= 10-1=9$

d.9

b) $s^2 = \frac{SS}{n-1}= \frac{600}{41-1}= 15$

a.15

c) For this case we have the sample size n = 25 and the sample variance is $s^2 =400$ , the standard error can founded with this formula:

$SE = \frac{s^2}{\sqrt{n}}= \frac{400}{\sqrt{25}}= 80$

Step-by-step explanation:

Part a

For the first part we have a sample of n =10 and we want to find the degrees of freedom, and we can use the following formula:

$df = n-1= 10-1=9$

d.9

Part b

From a sample we know that n=41 and SS= 600, where SS represent the sum of quares given by:

$SS = \sum_{i=1}^n (X_i -\bar X)^2$

And the sample variance for this case can be calculated from this formula:

$s^2 = \frac{SS}{n-1}= \frac{600}{41-1}= 15$

a.15

Part c

For this case we have the sample size n = 25 and the sample variance is $s^2 =400$ , the standard error can founded with this formula:

$SE = \frac{s^2}{\sqrt{n}}= \frac{400}{\sqrt{25}}= 80$

8 0

3 years ago

Connor believed that he did not do well in his ABC physics test. All ABC physics test scores are normally distributed with a mea

jenyasd209 [6]

Answer:

Yes! he passed

Step-by-step explanation:

8 0

3 years ago

Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$

Viktor [21]

When $n=0$ , we have

$5^{3^0} + 1 = 5^1 + 1 = 6$

$3^{0 + 1} = 3^1 = 3$

and of course 3 | 6. ("3 divides 6", in case the notation is unfamiliar.)

Suppose this is true for $n=k$ , that

$3^{k + 1} \mid 5^{3^k} + 1$

Now for $n=k+1$ , we have

$5^{3^{k+1}} + 1 = 5^{3^k \times 3} + 1 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k}\right)^3 + 1^3 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k} + 1\right) \left(\left(5^{3^k}\right)^2 - 5^{3^k} + 1\right)$

so we know the left side is at least divisible by $3^{k+1}$ by our assumption.

It remains to show that

$3 \mid \left(5^{3^k}\right)^2 - 5^{3^k} + 1$

which is easily done with Fermat's little theorem. It says

$a^p \equiv a \pmod p$

where $p$ is prime and $a$ is any integer. Then for any positive integer $x$ ,

$5^3 \equiv 5 \pmod 3 \implies (5^3)^x \equiv 5^x \pmod 3$

Furthermore,

$5^{3^k} \equiv 5^{3\times3^{k-1}} \equiv \left(5^{3^{k-1}}\right)^3 \equiv 5^{3^{k-1}} \pmod 3$

which goes all the way down to

$5^{3^k} \equiv 5 \pmod 3$

So, we find that

$\left(5^{3^k}\right)^2 - 5^{3^k} + 1 \equiv 5^2 - 5 + 1 \equiv 21 \equiv 0 \pmod3$

QED

5 0

1 year ago