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3 years ago

What is the first operation used in solving this equation 2x + 5 = 10

Mathematics

2 answers:

vovikov84 [41]3 years ago

8 0

Subtraction. you subtract 5 from both sides.

Zielflug [23.3K]3 years ago

8 0

Answer is subtraction.

First you would subtract 5 on both sides
2x= 5
Then divide by 2
X= 5/2

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8 + 8( 8 - n ) = 40 + 8n

Ierofanga [76]

n=2

8+8(8-n)=40+8n (distribute 8 through the parenthesis)

8+64-8n=40-8n (add the numbers)

72-8n=40+8n (move the variable to the left side and change its sign)

72-8n+8n=40

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-16n = -32 (divide both sides by -16)

n=2

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$30

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Step-by-step explanation:

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2 years ago

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2 years ago

Leila is going for a run. She runs at a speed of 5 miles per hour for 12.5

Ksivusya [100]

She ran for 2 and a half hours
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2 years ago

2. The quality assurance department inspects its production line. The product either fails or passes the inspection. Past experi

Oksana_A [137]

Answer:

(a) $E(X) = 950$

(b) $COV = 0.007255$

Step-by-step explanation:

The given problem can be solved using binomial distribution since the product either fails or passes, the probability of failure or success is fixed and there are n repeated trials.

probability of failure = q = 0.05

probability of success = p = 1 - 0.05 = 0.95

number of trials = n = 1000

(a) What is the expected number of non-defective units?

The expected number of non-defective units is given by

$E(X) = n \times p \\\\E(X) = 1000 \times 0.95 \\\\E(X) = 950$

(b) what is the COV of the number of non-defective units?

The coefficient of variance is given by

$COV = \frac{\sigma}{E(X)}$

Where the standard deviation is given by

$\sigma = \sqrt{n \times p\times q} \\\\\sigma = \sqrt{1000 \times 0.95\times 0.05} \\\\\sigma = 6.892$

So the coefficient of variance is

$COV = \frac{6.892}{950}$

$COV = 0.007255$

(c) What is the probability of having more than 980 non-defective units?

We can use the Normal distribution as an approximation to the Binomial distribution since n is quite large and so is p.

$P(X > 980) = 1 - P(X < 980)\\\\P(X > 980) = 1 - P(Z < \frac{x - \mu}{\sigma} )\\\\$

We need to consider the continuity correction factor whenever we use continuous probability distribution (Normal distribution) to approximate discrete probability distribution (Binomial distribution).

$P(X > 980) = 1 - P(Z < \frac{979.5 - 950}{6.892} )\\\\P(X > 980) = 1 - P(Z < \frac{29.5}{6.892} )\\\\P(X > 980) = 1 - P(Z < 4.28)\\\\$

The z-score corresponding to 4.28 is 0.99999

$P(X > 980) = 1 - 0.99999\\\\P(X > 980) = 0.00001\\\\$

So it means that it is very unlikely that there will be more than 980 non-defective units.

8 0

2 years ago