Which statement is correct?

What is the area of the polygon

irakobra [83]

The area or answer is 45

5 0

3 years ago

Need help on this problem it helps if you zoom in what kind of triangle is it

sladkih [1.3K]

D. an equilateral triangle

Two of $\triangle BDC$ 's angles are 60 degrees, so that means its last angle will also be 60 degrees.

60 + 60 = 120, 180 - 120 = 60

Equiangular triangles are always equilateral.

5 0

4 years ago

Read 2 more answers

Given cos x = 12/13 and sin x = 5/13 What is ratio for tan x?

arlik [135]

Answer:

tan x = 5/12

Step-by-step explanation:

given:

- sin x = 5/13

- cos x = 12/13

to find

- tan x

solving

tan x = sin x / cos x

tan x = (5 / 13) / (12 / 13)

tan x = 5 / 12

6 0

3 years ago

Read 2 more answers

Suppose the number of corn kernels on an ear of corn is normally distributed. For a random sample of ears of corn, the confidenc

Dafna11 [192]

Answer:

The error bound is of 13.5 corn kernels.

Step-by-step explanation:

Any confidence interval has a lower bound and an upper bound. The error bound is the difference between the values of those two bounds, divided by 2.

In this problem, we have that:

Lower bound: 858.50

Upper bound: 885.50

Error bound: (885.50-858.50)/2 = 13.5

The error bound is of 13.5 corn kernels.

6 0

3 years ago

Suppose that the number of calls per hour to an answering service follows a Poisson process with rate 4. Suppose that 3/4's of t

Paraphin [41]

Answer:

(a) The probability that in one hour exactly two men and three women will call the answering service is 0.01374.

(b) The probability that three men will make phone calls before three women do is 0.00687.

Step-by-step explanation:

Let X = number of calls per hour to an answering service.

The average number of calls per hour is, λ = 4.

The random variable X follows a Poisson distribution with parameter λ = 4.

The probability mass function of X is:

$P(X=x)=\frac{e^{-4}4^{x}}{x!};\ x=0,1,2,3...$

Let Y = number of calls made by a man.

The probability that a call is made by a man is, $p=\frac{3}{4}$ .

A randomly selected call is made by a man independently of the other calls.

Te random variable Y follows a Binomial distribution with parameters n and p.

The probability mass function of Y is:

$P(Y=y)={n\choose y}\times(\frac{3}{4})^{y}\times (\frac{1}{4})^{n-y};\ y=0,1,2,3...$

The random variable X are independent of each other.

(a)

Compute the probability that in one hour exactly two men and three women will call the answering service as follows:

P(X = 5, Y = 2) = P (X = 5) × P (Y = 2)

$=\frac{e^{-4}4^{5}}{5!}\times [{5\choose 2}\times(\frac{3}{4})^{2}\times (\frac{1}{4})^{5-2}]\\=0.1563\times [10\times 0.5625\times 0.015625]\\=0.01374$

Thus, the probability that in one hour exactly two men and three women will call the answering service is 0.01374.

(b)

The random variable Z can be defined as the number of calls made by women.

The random variable Z is defined as the number of failures before a fixed number of successes (z) . That is, the number of calls made by men before a specific number of women made the calls.

The random variable Z follows a Negative Binomial distribution.

The probability mass function of Z is:

$P(Z=z)={n-1\choose z-1}\times (\frac{1}{4})^{z}\times (\frac{3}{4})^{n-z}$

Compute the probability that three men will make phone calls before three women do as follows:

P (X = 6, Z = 3) = P (X = 6) × P (Z = 3)

$=\frac{e^{-4}4^{6}}{6!}\times [{6-1\choose 3-1}\times (\frac{1}{4})^{3}\times (\frac{3}{4})^{6-3}]\\=0.1042\times [10\times 0.015625\times 0.421875]\\=0.00687$

Thus, the probability that three men will make phone calls before three women do is 0.00687.

8 0

3 years ago