Round of 94136 to the nearest whole number

What is 5 Left-bracket cosine (StartFraction pi Over 4 EndFraction) + I sine (StartFraction pi Over 4 EndFraction) right-bracket

Liono4ka [1.6K]

Answer:

C

Step-by-step explanation:

I took it on edg 2020

7 0

3 years ago

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Pls help I give brainliest to the best!!!!!!!!!!

Neporo4naja [7]

Answer:

m

2

−2m+3

Step-by-step explanation:

6 0

3 years ago

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-18x-17=-53

nydimaria [60]

-18x-17=-53 /add 17 to both sides

-18x-17+17=-53+17

-18x=-36

x=(-36)/(-18)

x=2

5 0

3 years ago

Find the appropriate rejection regions for the large-sample test statistic z in these cases. (Round your answers to two decimal

Usimov [2.4K]

Answer:

a) We have that the significance is given by $\alpha =0.01$ and we know that we have a right tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 1% of the area on the right and 99% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.01,0,1)"

And we got for this case $z_{crit}=2.33$

So then the rejection region would be $z>2.33$

b) We have that the significance is given by $\alpha =0.05$ , $\alpha/2 =0.025$ and we know that we have a two tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 2.5% of the area on the right and 97.5% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.025,0,1)"

And we got for this case $z_{crit}=\pm 1.96$

So then the rejection region would be $z>1.96 \cup z$

Step-by-step explanation:

Part a

We have that the significance is given by $\alpha =0.01$ and we know that we have a right tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 1% of the area on the right and 99% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.01,0,1)"

And we got for this case $z_{crit}=2.33$

So then the rejection region would be $z>2.33$

Part b

We have that the significance is given by $\alpha =0.05$ , $\alpha/2 =0.025$ and we know that we have a two tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 2.5% of the area on the right and 97.5% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.025,0,1)"

And we got for this case $z_{crit}=\pm 1.96$

So then the rejection region would be $z>1.96 \cup z$

7 0

3 years ago

The binomial x-2 is a factor of the polynomial below. f(x)=x3+x2+nx+10 What is the value of n?

Nutka1998 [239]

<h3>Answer: n = -11</h3>

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Explanation:

Since x-2 is a factor of f(x), this means f(2) = 0.

More generally, if x-k is a factor of p(x), then p(k) = 0. This is a special case of the remainder theorem.

So if we plugged x = 2 into f(x), we'd get

f(x) = x^3+x^2+nx+10

f(2) = 2^3+2^2+n(2)+10

f(2) = 8+4+2n+10

f(2) = 2n+22

Set this equal to 0 and solve for n

2n+22 = 0

2n = -22

n = -22/2

n = -11 is the answer

Therefore, x-2 is a factor of f(x) = x^3+x^2-11x+10

Plug x = 2 into that updated f(x) function to find....

f(x) = x^3+x^2-11x+10

f(2) = 2^3+2^2-11(2)+10

f(2) = 8+4-22+10

f(2) = 0

Which confirms our answer.

3 0

3 years ago