Part A: If (26)x = 1, what is the value of x? Explain your answer. (5 points)

What is the volume of the cone below?

lara [203]

Answer:

the world may never know

Step-by-step explanation:

7 0

3 years ago

2.1/1.488 <br> Divide. Round to the tenths place

muminat

Answer:

1.4

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

A tower stand on a horizontal plane. P is the top and Q is the bottom of the tower. A and B are two points on this plane such th

pochemuha

Answer:

32√5

Step-by-step explanation:

We have the right triangles PQA and PQB as well as the given right triangle QAB.

cot(PAQ) = 2/5 = QA/PQ

cot(PBQ) = 3/5 = QB/PQ

cot(PAQ) / cot(PBQ) = (2/5) / (3/5) = 2/3

cot(PAQ) / cot(PBQ) = (QA/PQ) / (QB/PQ) = QA / QB

QA / QB = 2/3

QA = (2/3) QB

QB = (3/2) QA

By the Pythagorean Theorem we have:

(QA)² + 32² = (QB)²

(QA)² + 32² = (3/2 QA)²

(QA)² + 1024 = (9/4) (QA)²

(5/4) (QA)² = 1024

(QA)² = (4/5)1024 = 4096/5

QA = 64/√5

Solve for PQ.

cot(PAQ) = QA/PQ

PQ = QA / cot(PAQ)

PQ = (64/√5) / (2/5) = 32√5

The height of the tower is 32√5.

7 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

Danielle runs 1/5 of a mile in 4.5 minutes. if she continues at the same speed, how long will it take her to run two miles?

motikmotik

(1/5) / 4.5 = 2 / m....1/5 mile to 4.5 minutes = 2 miles to m minutes
cross multiply
(1/5)(m) = (2)(4.5)
1/5m = 9
m = 9 * 5/1
m = 45 minutes <===

4 0

3 years ago