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3 years ago

Evaluate sin(Arcos(-9/√145)).

Mathematics

2 answers:

tino4ka555 [31]3 years ago

8 0

Use the fact that $\sin x = \sqrt{1-\cos^2 x}$ to solve it.

$\sin(\arccos(-\frac{9}{\sqrt{145}}))=\sqrt{1-\cos^2(\arccos(-\frac{9}{\sqrt{145}}))} =\\= \sqrt{1-(-\frac{9}{\sqrt{145}})^2}=\frac{8}{\sqrt{145}}$

makkiz [27]3 years ago

7 0

Answer:

Step-by-step explanation:

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Ackie got 3 wrong out of 20 questions on her social studies quiz.What percent did she answer correctly on the quiz

taurus [48]

Answer:

85%

Step-by-step explanation:

If she got 3 wrong then it means she got 20-3 right, so she got 17 questions right.

From there we get the fraction: 17/20

To make the a percentage we times both the top and bottom by 5 (because 20 x 5 = 100)

By doing that we get 17x5/20x5

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2 years ago

How is the graph of y=2(3)^x+1 translated from the graph of y = 2(3)^x

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y=2(3)x+1

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3 years ago

A recipe calls for 10 cups of oatmeal. Benjamin wants to make 1/5 fewer servings than the recipe makes.

miskamm [114]

The answer is c: .2 cups of oatmeal

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3 years ago

Read 2 more answers

Mark earned a 64 on his last science quiz. If Mark’s z-score was -2.8 and the standard deviation was 10, find the mean grade on

docker41 [41]

Answer:

he score is 2.15 standard deviations above the mean.

Step-by-step explanation:

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3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Anettt [7]

Answer:

Choice C: approximately 121 green beans will be 13 centimeters or shorter.

Step-by-step explanation:

What's the probability that a green bean from this sale is shorter than 13 centimeters?

Let the length of a green bean be $X$ centimeters.

$X$ follows a normal distribution with

mean $\mu = 11.2$ and
standard deviation $\sigma = 2.1$ .

In other words,

$X\sim \text{N}(11.2, 2.1^{2})$ ,

and the probability in question is $X \le 13$ .

Z-score table approach:

Find the z-score of this measurement:

$\displaystyle z= \frac{x-\mu}{\sigma} = \frac{13-11.2}{2.1} = 0.857143$ . Closest to 0.86.

Look up the z-score in a table. Keep in mind that entries on a typical z-score table gives the probability of the left tail, which is the chance that $Z$ will be less than or equal to the z-score in question. (In case the question is asking for the probability that $Z$ is greater than the z-score, subtract the value from table from 1.)

$P(X\le 13) = P(Z \le 0.857143) \approx 0.8051$ .

"Technology" Approach

Depending on the manufacturer, the steps generally include:

Locate the cumulative probability function (cdf) for normal distributions.
Enter the lower and upper bound. The lower bound shall be a very negative number such as $-10^{9}$ . For the upper bound, enter $13$
Enter the mean and standard deviation (or variance if required).
Evaluate.

For example, on a Texas Instruments TI-84, evaluating $\text{normalcdf})(-1\text{E}99,\;13,\;11.2,\;2.1 )$ gives $0.804317$ .

As a result,

$P(X\le 13) = 0.804317$ .

Number of green beans that are shorter than 13 centimeters:

Assume that the length of green beans for sale are independent of each other. The probability that each green bean is shorter than 13 centimeters is constant. As a result, the number of green beans out of 150 that are shorter than 13 centimeters follow a binomial distribution.

Number of trials $n$ : 150.
Probability of success $p$ : 0.804317.

Let $Y$ be the number of green beans out of this 150 that are shorter than 13 centimeters. $Y\sim\text{B}(150,0.804317)$ .

The expected value of a binomial random variable is the product of the number of trials and the probability of success on each trial. In other words,

$E(Y) = n\cdot p = 150 \times 0.804317 = 120.648\approx 121$

The expected number of green beans out of this 150 that are shorter than 13 centimeters will thus be approximately 121.

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3 years ago