Find sin θ, cot θ, and csc θ, where θ is the angle shown in the figure.

Mathematics

1 answer:

Natalka [10]3 years ago

8 0

Step-by-step explanation:

everything can be found in the picture

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The question is in the picture above. Please help

kow [346]

Answer:

it is 1

Step-by-step explanation:

5 0

3 years ago

-2x-7≥41 in algaebra for

dexar [7]

Answer:

They are equal. x= -24

Step-by-step explanation:

-2x - 7 ≥ 41

First you isolate the variable

-2x -7 ≥ 41

+7 +7

-2x≥48

then divide each side by -2

x≥-24

Then go back and plug in

-2(-24)-7=

48-7=41

Therefore

41=41

4 0

3 years ago

Discrete R.V. Assume a student shows up for EGR 280 completely unprepared and the instructor gives a pop quiz. Assume there are

Tom [10]

Answer:

a) p=0.2

b) probability of passing is 0.01696

c) The expected value of correct questions is 1.2

Step-by-step explanation:

a) Since each question has 5 options, all of them equally likely, and only one correct answer, then the probability of having a correct answer is 1/5 = 0.2.

b) Let X be the number of correct answers. We will model this situation by considering X as a binomial random variable with a success probability of p=0.2 and having n=6 samples. We have the following for k=0,1,2,3,4,5,6

$P(X=k) = \binom{n}{k}p^{k}(1-p)^{n-k} = \binom{6}{k}0.2^{k}(0.8)^{6-k}$ .

Recall that $\binom{n}{k}= \frac{n!}{k!(n-k)!}$ In this case, the student passes if X is at least four correct questions, then

$P(X\geq 4) = P(X=4)+P(X=5)+P(X=6)=\binom{6}{4}0.2^{4}(0.8)^{6-4}+\binom{6}{5}0.2^{5}(0.8)^{6-5}+\binom{6}{6}0.2^{6}(0.8)^{6-6}= 0.01696$

c)The expected value of a binomial random variable with parameters n and p is $E[X] = np$ . IN our case, n=6 and p =0.2. Then the expected value of correct answers is $6\cdot 0.2 = 1.2$