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1 year ago

Find m∠DEA if m∠DEA= x + 30, m∠AEF= x + 132, and m∠DEF= 146 degrees

1 answer:

6 0

The numerical sum of the degree measures of m ∠DEA and m ∠AEF and m ∠DEF is 360°; The numerical measures of the angles is,

m ∠DEA = 56°

m ∠AEF = 158°

m ∠DEF = 146°

Based on the given data,

m ∠DEA= x + 30,

m ∠AEF= x + 132, and

m ∠DEF= 146 degrees

If the sum of two linear angles is 360° then, they are known as supplementary angles.

∠A + ∠B + ∠C = 360°, (∠A and ∠B and ∠C are linear angles.)

So,

We can write,

m ∠AEF + m ∠DEA + m ∠DEF = 360°

( x + 132) + (x + 30) + 146 = 360°

x + 30 + x + 132 + 146 = 360°

2x + 308 = 360°

2x = 360° - 308

x = 52/2

x =26

Now, we will substitute the value of x = 26° in the ∠DEA and ∠AEF, hence we get:

m ∠DEA = x + 30

m ∠DEA = 26 + 30

m ∠DEA = 56 degrees

Also,

m ∠AEF = x + 132

m ∠AEF = 26 + 132

m ∠AEF = 158

Hence,

m ∠DEA + m ∠AEF + m ∠DEF = 360°

56 + 158 + 146 = 360°

360° = 360°

Therefore,

Therefore, the numerical sum of the degree measures of m ∠DEA and m ∠AEF and m ∠DEF is 360°; The numerical measures of the angles is,

m ∠DEA = 56°

m ∠AEF = 158°

m ∠DEF = 146°

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Ten times the square of a non zero number is equal to ninety times the number. what is the number?

ANTONII [103]

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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

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