Express sin F as a fraction in simplest terms. D 17 10 E F

Solve for L. P = 2L + 2W L = 2/(P - W) L = 2(W - 2P) L = (P - 2W)/2

yanalaym [24]

Answer:

L = (P - 2W)/2

Step-by-step explanation:

This is the formula for perimeter

P = 2L + 2W

Sustract 2W on both sides:

P-2W= 2L

Divide both sides by 2:

(P-2W)/2 = L

L= (P-2W)/2

6 0

2 years ago

T-6=8-t how do solve? pls include steps

umka2103 [35]

First, add t to each side:
2t - 6 = 8
Add 6 to each side:
2t = 14
Divide both sides by 2:
t = 7

Hope this helps!! :)

8 0

2 years ago

Among 400 randomly selected drivers in the 16 minus 18 age bracket, 293 were in a car crash in the last year. If a driver in th

professor190 [17]

Answer:

The approximate probability is 0.7325.

Is not unlikely that a driver in that age bracket is involved in a car crash during a year, so there are reasons to be concerned.

Step-by-step explanation:

The approximate probability can be estimated using the sample proportion:

$p=\dfrac{X}{n}=\dfrac{293}{400}=0.7325$

A probability of 0.7325 is high, so it is not unlikely that a driver in that age bracket is involved in a car crash during a year.

As this value is high, there is reasons to be concerned about it.

4 0

2 years ago

X intercepts for the equation y=-3(x-2)^2+1

nignag [31]

Answer:

Step-by-step explanation:

5 0

2 years ago

When grading an exam, 90% of a professor's 50 students passed. If the professor randomly selected 10 exams, what is the probabil

Damm [24]

Using the binomial distribution, it is found that there is a:

a) 0.9298 = 92.98% probability that at least 8 of them passed.

b) 0.0001 = 0.01% probability that fewer than 5 passed.

For each student, there are only two possible outcomes, either they passed, or they did not pass. The probability of a student passing is independent of any other student, hence, the binomial distribution is used to solve this question.

<h3>What is the binomial probability distribution formula?</h3>

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

In this problem:

90% of the students passed, hence $p = 0.9$ .

The professor randomly selected 10 exams, hence $n = 10$ .

Item a:

The probability is:

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)$

In which:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{10,8}.(0.9)^{8}.(0.1)^{2} = 0.1937$

$P(X = 9) = C_{10,9}.(0.9)^{9}.(0.1)^{1} = 0.3874$

$P(X = 10) = C_{10,10}.(0.9)^{10}.(0.1)^{0} = 0.3487$

Then:

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) = 0.1937 + 0.3874 + 0.3487 = 0.9298$

0.9298 = 92.98% probability that at least 8 of them passed.

Item b:

The probability is:

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

Using the binomial formula, as in item a, to find each probability, then adding them, it is found that:

$P(X < 5) = 0.0001$

Hence:

0.0001 = 0.01% probability that fewer than 5 passed.

You can learn more about the the binomial distribution at brainly.com/question/24863377

3 0

1 year ago