All categories

3 years ago

Which best describes the relationship among the volumes of hemisphere X, cylinder Y, and cone Z?

Mathematics

1 answer:

kogti [31]3 years ago

6 0

Answer: I saw on another post on here with the same question that someone said it was

B. The sum of the volumes of X and Z equals the volume of Y.

You might be interested in

What segment is congruent to AC? A. BD B. BE C. CE D. none

Phoenix [80]

You can find the segment congruent to AC by finding another segment with the same length. So first, you need to find the length of AC.

C - A = AC
0 - (-6) = AC Cancel out the double negative
0 + 6 = AC
6 = AC

Now, find another segment that also has a length of 6.

D - B = BD
2 - (-2) = BD Cancel out the double negative
2 + 2 = BD
4 = BD
4 ≠ 6

E - B = BE
4 - (-2) = BE Cancel out the double negative
4 + 2 = BE
6 = BE
6 = 6

So, the segment congruent to AC is B. BE .

8 0

4 years ago

An instructor who taught two sections of engineering statistics last term, the first with 25 students and the second with 35, de

Amiraneli [1.4K]

Answer:

a) P=0.1721

b) P=0.3528

c) P=0.3981

Step-by-step explanation:

This sampling can be modeled by a binominal distribution where p is the probability of a project to belong to the first section and q the probability of belonging to the second section.

a) In this case we have a sample size of n=15.

The value of p is p=25/(25+35)=0.4167 and q=1-0.4167=0.5833.

The probability of having exactly 10 projects for the second section is equal to having exactly 5 projects of the first section.

This probability can be calculated as:

$P=\frac{n!}{(n-k)!k!}p^kq^{n-k}= \frac{15!}{(10)!5!}\cdot 0.4167^5\cdot0.5833^{10}=0.1721$

b) To have at least 10 projects from the 2nd section, means we have at most 5 projects for the first section. In this case, we have to calculate the probability for k=0 (every project belongs to the 2nd section), k=1, k=2, k=3, k=4 and k=5.

We apply the same formula but as a sum:

$P(k\leq5)=\sum_{k=0}^{5}\frac{n!}{(n-k)!k!}p^kq^{n-k}$

Then we have:

$P(k=0)=0.0003\\P(k=1)=0.0033\\P(k=2)=0.0165\\P(k=3)=0.0511\\P(k=4)=0.1095\\P(k=5)=0.1721\\\\P(k\leq5)=0.0003+0.0033+0.0165+0.0511+0.1095+0.1721=0.3528$

c) In this case, we have the sum of the probability that k is equal or less than 5, and the probability tha k is 10 or more (10 or more projects belonging to the 1st section).

The first (k less or equal to 5) is already calculated.

We have to calculate for k equal to 10 or more.

$P(k\geq10)=\sum_{k=10}^{15}\frac{n!}{(n-k)!k!}p^kq^{n-k}$

Then we have

$P(k=10)=0.0320\\P(k=11)=0.0104\\P(k=12)=0.0025\\P(k=13)=0.0004\\P(k=14)=0.0000\\P(k=15)=0.0000\\\\P(k\geq10)=0.032+0.0104+0.0025+0.0004+0+0=0.0453$

The sum of the probabilities is

$P(k\leq5)+P(k\geq10)=0.3528+0.0453=0.3981$

8 0

3 years ago

Sally did some counting look at her work explain why you think sally counted this way 177,178,179,180,190,200,220,211,212,213,21

dimulka [17.4K]

The first and last four numbers each have a difference of one between them. The fifth number has a difference of 10 between it and the previous number. The middle number has a difference of 20 between itself and the two numbers that surround it. Counting in this way could have been a result of a lot of things to count, and spot checking along the way (when the numbers have a difference of 1).

8 0

4 years ago

Calculate the bearing of Y from X

nlexa [21]

Answer:

bearing of Y from X is 074°

Step-by-step explanation:

the bearing of Y from X is the clockwise measure of the angle from the northline N to Y , that is

180° - 106° = 74°

the 3- figure bearing of Y from X is 074°

7 0

2 years ago

V=u+at u=2 a= -5 t=1/2 work out the value of v.

Usimov [2.4K]

Answer:

(-1)/2

Step-by-step explanation:

Evaluate u + a t where u = 2, a = -5 and t = 1/2:

u + a t = 2 - 5/2

Put 2 - 5/2 over the common denominator 2. 2 - 5/2 = (2×2)/2 - 5/2:

(2×2)/2 - 5/2

2×2 = 4:

4/2 - 5/2

4/2 - 5/2 = (4 - 5)/2:

(4 - 5)/2

4 - 5 = -1:

Answer: (-1)/2

6 0

3 years ago