All categories

3 years ago

Cscθ - sinθ = _____

1 answer:

galina1969 [7]3 years ago

4 0

So you maybe know that $\csc x = \frac{1}{\sin x}$ . Now let us try to simplify the expression

$\csc x - \sin x = \frac{1}{\sin x} - \sin x = \frac{1}{\sin x} - \frac{\sin^2 x }{\sin x}=\frac{1-\sin^2 x}{\sin x} =$

$= \frac{\cos^2 x}{\sin x} = \cot x \cdot \cos x$

You might be interested in

For his yearly vacation, Colton plans to spend $350 on train tickets, $200 on food, $150 on accommodations, and $50 on miscellan

scoundrel [369]

Answer: He would spend 750$ on the entire vacation.

Step-by-step explanation: Simple addition, 350+150 (Always do the two odd ones first) which equals 500, 500+200 equals 700, and 700+50 equals 750, which gives your answer of 750$'s for the whole vacation.

5 0

3 years ago

Read 2 more answers

In a southern statek, it was revealed that 5% of all automobiles in the state did not pass inspection. of the next ten automobil

lbvjy [14]

Well it’s simple just do this

4 0

3 years ago

Help !!!! my math teacher wants me to answer this, and I have to give the answer in 10 minutes

mariarad [96]

45 degrees for x.
Do 180-160 = 20 (Since 180 is a straight line)
The 115, x, and 20 form another straight line, so take 180-20-115. You get 45.

7 0

3 years ago

The average heart pumps about 80ml of Blood each contraction. if your heart rate was 70 beats per minute, calculate the volume o

Anuta_ua [19.1K]

The heart beat rate is 70beats/minute with 80ml of blood pumped every beat. Then, the volume of blood your heart pumps in one minute would be:

volume of blood per minute= volume per beat * beat per minute
volume of blood per minute= 80ml/beat * 70beat/minute= 560ml/ minute

volume of blood per hour would be: 560ml/minute * 60 minute/hour= 33.600 ml/hour

5 0

3 years ago

41% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the

lys-0071 [83]

Answer:

a) 0.2087 = 20.82% probability that the number of U.S. adults who have very little confidence in newspapers is exactly five.

b) 0.1834 = 18.34% probability that the number of U.S. adults who have very little confidence in newspapers is at least six.

c) 0.3575 = 35.75% probability that the number of U.S. adults who have very little confidence in newspapers is less than four.

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they have very little confidence in newspapers, or they do not. The answers of each adult are independent, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

41% of U.S. adults have very little confidence in newspapers.

This means that $p = 0.41$

You randomly select 10 U.S. adults.

This means that $n = 10$

(a) exactly five

This is $P(X = 5)$ . So

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

$P(X = 5) = C_{10,5}.(0.41)^{5}.(0.59)^{5} = 0.2087$

0.2087 = 20.82% probability that the number of U.S. adults who have very little confidence in newspapers is exactly five.

(b) at least six

This is:

$P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

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

$P(X = 6) = C_{10,6}.(0.41)^{6}.(0.59)^{4} = 0.1209$

$P(X = 7) = C_{10,7}.(0.41)^{7}.(0.59)^{3} = 0.0480$

$P(X = 8) = C_{10,8}.(0.41)^{8}.(0.59)^{2} = 0.0125$

$P(X = 9) = C_{10,9}.(0.41)^{9}.(0.59)^{1} = 0.0019$

$P(X = 10) = C_{10,10}.(0.41)^{10}.(0.59)^{0} = 0.0001$

Then

$P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.1209 + 0.0480 + 0.0125 + 0.0019 + 0.0001 = 0.1834$

0.1834 = 18.34% probability that the number of U.S. adults who have very little confidence in newspapers is at least six.

(c) less than four.

This is:

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

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

$P(X = 0) = C_{10,0}.(0.41)^{0}.(0.59)^{10} = 0.0051$

$P(X = 1) = C_{10,1}.(0.41)^{1}.(0.59)^{9} = 0.0355$

$P(X = 2) = C_{10,2}.(0.41)^{2}.(0.59)^{8} = 0.1111$

$P(X = 3) = C_{10,3}.(0.41)^{3}.(0.59)^{7} = 0.2058$

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0051 + 0.0355 + 0.1111 + 0.2058 = 0.3575$

0.3575 = 35.75% probability that the number of U.S. adults who have very little confidence in newspapers is less than four.

5 0

3 years ago