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3 years ago

Type the correct answer in each box.

Mathematics

1 answer:

MariettaO [177]3 years ago

4 0

Answer: G

Step-by-step explanation:

Data: Point which has the coordinates of (-2.5, 5.5)=x

Only step, find those two coordinates on the table

Explanation: Since the question is asking you to find which point has those two coordinates, all you have to do is instead of finding all the coordinates of the points, just search for the coordinates provided and see what point it lands on.

For this problem, it lands on Point G so the answer is Point G(or just G)

I hope this helps!

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Find the general solution of <img src="https://tex.z-dn.net/?f=2cos%282x%20%2B%20%5Cfrac%7B%5Cpi%20%7D%7B4%7D%20%29%20%3D%20%5Cs

Aleks04 [339]

Answer:

$\large\boxed{x=-\dfrac{\pi}{4}+k\pi\ \text{or}\ x=k\pi\ \text{for}\ k\in\mathbb{Z}}$

Step-by-step explanation:

$2\cos\left(2x+\dfrac{\pi}{4}\right)=\sqrt2\qquad\text{divide both sides by 2}\\\\\cos\left(2x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt2}{2}\to2x+\dfrac{\pi}{4}=\pm\dfrac{\pi}{4}+2k\pi\\\\2x+\dfrac{\pi}{4}=-\dfrac{\pi}{4}+2k\pi\ \vee\ 2x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+2k\pi\qquad\text{subtract}\ \dfrac{\pi}{4}\ \text{from all sides}\\\\2x=-\dfrac{2\pi}{4}+2k\pi\ \vee\ 2x=2k\pi\qquad\text{divide all sides by 2}\\\\x=-\dfrac{\pi}{4}+k\pi\ \vee\ x=k\pi\ \text{for}\ k\in\mathbb{Z}$

5 0

3 years ago

Leakage from underground gasoline tanks at service stations can damage the environment. It is estimated that 25% of these tanks

Gnesinka [82]

Answer:

a) 3.75

b) 23.61% probability that fewer than 3 tanks will be found to be leaking

c) 0% the probability that at least 600 of these tanks are leaking

Step-by-step explanation:

For each tank there are only two possible outcomes. EIther they leak, or they do not. The probability of a tank leaking is independent of other tanks. So 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.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

To solve question c), i am going to approximate the binomial distribution to the normal.

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

It is estimated that 25% of these tanks leak.

This means that $p = 0.25$

15 tanks chosen at random

This means that $n = 15$

a.What is the expected number of leaking tanks in such samples of 15?

$E(X) = np = 15*0.25 = 3.75$

b.What is the probability that fewer than 3 tanks will be found to be leaking?

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

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

$P(X = 0) = C_{15,0}.(0.25)^{0}.(0.75)^{15} = 0.0134$

$P(X = 1) = C_{15,1}.(0.25)^{1}.(0.75)^{14} = 0.0668$

$P(X = 2) = C_{15,2}.(0.25)^{2}.(0.75)^{13} = 0.1559$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0134 + 0.0668 + 0.1559 = 0.2361$

23.61% probability that fewer than 3 tanks will be found to be leaking

c.Now you do a larger study, examining a random sample of 2000 tanks nationally. What is the probability that at least 600 of these tanks are leaking?

Now we have n = 2000. So

$\mu = E(X) = np = 2000*0.25 = 500$