Volume8cm what is side lengths

Find sin theta if cos theta is 4/5

Dima020 [189]

Answer:

Step-by-step explanation:

hello :

cos²θ +sin²θ = 1 and cosθ =4/5

(4/5)² + sin²θ = 1 so : sin²θ = 1 - 16/25

sin²θ = 9/25 = (3/5)²

sinθ = 3/5 or sinθ = - 3/5 because : (3/5)²= (- 3/5)²= 9/25

5 0

2 years ago

Richard has just been given an l0-question multiple-choice quiz in his history class. Each question has five answers, of which o

myrzilka [38]

Answer:

a) 0.0000001024 probability that he will answer all questions correctly.

b) 0.1074 = 10.74% probability that he will answer all questions incorrectly

c) 0.8926 = 89.26% probability that he will answer at least one of the questions correctly.

d) 0.0328 = 3.28% probability that Richard will answer at least half the questions correctly

Step-by-step explanation:

For each question, there are only two possible outcomes. Either he answers it correctly, or he does not. The probability of answering a question correctly is independent of any other question. This means that we use the binomial probability distribution 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.

Each question has five answers, of which only one is correct

This means that the probability of correctly answering a question guessing is $p = \frac{1}{5} = 0.2$

10 questions.

This means that $n = 10$

A) What is the probability that he will answer all questions correctly?

This is $P(X = 10)$

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

$P(X = 10) = C_{10,10}.(0.2)^{10}.(0.8)^{0} = 0.0000001024$

0.0000001024 probability that he will answer all questions correctly.

B) What is the probability that he will answer all questions incorrectly?

None correctly, so $P(X = 0)$

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

$P(X = 0) = C_{10,0}.(0.2)^{0}.(0.8)^{10} = 0.1074$

0.1074 = 10.74% probability that he will answer all questions incorrectly

C) What is the probability that he will answer at least one of the questions correctly?

This is

$P(X \geq 1) = 1 - P(X = 0)$

Since $P(X = 0) = 0.1074$ , from item b.

$P(X \geq 1) = 1 - 0.1074 = 0.8926$

0.8926 = 89.26% probability that he will answer at least one of the questions correctly.

D) What is the probability that Richard will answer at least half the questions correctly?

This is

$P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

In which

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

$P(X = 5) = C_{10,5}.(0.2)^{5}.(0.8)^{5} = 0.0264$

$P(X = 6) = C_{10,6}.(0.2)^{6}.(0.8)^{4} = 0.0055$

$P(X = 7) = C_{10,7}.(0.2)^{7}.(0.8)^{3} = 0.0008$

$P(X = 8) = C_{10,8}.(0.2)^{8}.(0.8)^{2} = 0.0001$

$P(X = 9) = C_{10,9}.(0.2)^{9}.(0.8)^{1} \approx 0$

$P(X = 10) = C_{10,10}.(0.2)^{10}.(0.8)^{0} \approx 0$

So

$P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.0264 + 0.0055 + 0.0008 + 0.0001 + 0 + 0 = 0.0328$

0.0328 = 3.28% probability that Richard will answer at least half the questions correctly

8 0

3 years ago

W^2-8w-20 factor the polynomial.

Mademuasel [1]

Your answer would be (w+2)(w−10). If you want me to explain please don't hesitate to ask.

6 0

3 years ago

Read 2 more answers

Consider and encode the following problem. The salt has been stolen! Well, it was found that the culprit was either the Caterpil

GenaCL600 [577]

The caterpillar ate the salt

3 0

3 years ago

Breyers is a major producer of ice cream and would like to test if the average American consumes more than 17 ounces of ice crea

asambeis [7]

Answer:

The conclusion for this hypothesis test would be that the average American consumes less than or equal to 17 ounces of ice cream per month.

Step-by-step explanation:

We are given that Breyers is a major producer of ice cream and would like to test if the average American consumes more than 17 ounces of ice cream per month.

A random sample of 25 Americans was found to consume an average of 19 ounces of ice cream last month. The standard deviation for this sample was 5 ounces.

Let $\mu$ = average ounces of ice cream consumed by American per month

So, Null Hypothesis, $H_0$ : $\mu \leq$ 17 ounces {means that the average American consumes less than or equal to 17 ounces of ice cream per month}

Alternate Hypothesis, $H_A$ : $\mu$ > 17 ounces {means that the average American consumes more than 17 ounces of ice cream per month}

The test statistics that will be used here is One-sample t test statistics as we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample average = 19 ounces

s = sample standard deviation = 5 ounces

n = sample of Americans = 25

So, test statistics = $\frac{19-17}{\frac{5}{\sqrt{25} } }$ ~ $t_2_4$

= 2

The value of the test statistics is 2.

Now at 0.025 significance level, the t table gives critical value of 2.06 at 24 degree of freedom for right-tailed test. Since our test statistics is less than the critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the the average American consumes less than or equal to 17 ounces of ice cream per month.

8 0

3 years ago