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mixer [17]
3 years ago
12

10(x-2) = -20 Solve

Mathematics
2 answers:
ale4655 [162]3 years ago
7 0

Answer:

x=0

Step-by-step explanation:

10(x-2)=-20

10x-20=-20

add 20 on both sides.

10x=0

10 times 0 =0

x=0


AveGali [126]3 years ago
6 0

10(x - 2) = -20      <em>divide both sides by 10</em>

x - 2 = -2      <em>add 2 to both sides</em>

<h3>x = 0</h3>

othrer method

10(x - 2) = -20       <em>use distributive property a(b - c) = ab - ac</em>

10x - (10)(2) = -20

10x - 20 = -20      <em>add 20 to both sides</em>

10x = 0      <em>divide both sides by 10</em>

<h3>x = 0</h3>
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Answer:

(a) The probability distribution of the random variable <em>X</em> is Binomial.

(b) The mean and standard deviation of the 25-question test are 5 and 2 respectively.

(c) The value of <em>x</em> is 0.34.

Step-by-step explanation:

The random variable <em>X </em>is defined as the number of correct answers given by a person who is guessing each answer on a 25-question exam.

There are five possible answer for every question.

This implies that the probability of getting a correct answer is:

<em>P</em> (X) = 0.20.

There are a total of <em>n</em> = 25 questions.

Every answer is independent of the others.

(a)

The random variable <em>X</em> has  finite number of independent trials (i.e. 25 questions). There are only two outcomes for each trial, i.e. Success = correct answer and Failure = wrong answer. Each trial has the same probability of success (, i.e. P (X) = 0.25).

Thus, the probability distribution of the random variable <em>X</em> is Binomial with parameters <em>n</em> = 25 and <em>p</em> = 0.20.

(b)

Compute the mean of the random variable <em>X</em> as follows:

E(X)=np\\=25\times 0.20\\=5

Compute the standard deviation of the random variable <em>X</em> as follows:

SD(X)=\sqrt{np(1-p)}\\=\sqrt{25\times 0.20\times (1-0.20)}\\=2

Thus, the mean and standard deviation of the 25-question test are 5 and 2 respectively.

(c)

The sample is large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

1. np ≥ 5

2. n(1 - p) ≥ 5

Check the conditions as follows:

np=25\times 0.20=5=5\\\\n(1-p)=25\times (1-0.20)=20>5

Thus, a Normal approximation to binomial can be applied.

So,  X\sim N(5, 2^{2})

It is provided that the minimum passing score foe the test is such that only 1% of students who are guessing will pass the test.

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⇒ P (Z < z) = 0.01

The value of <em>z</em> is -2.33.

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z=\frac{x-\mu}{\sigma}\\\\-2.33=\frac{x-5}{2}\\\\x=5-(2.33\times 2)\\\\x=0.34

Thus, the value of <em>x</em> is 0.34.

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