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3 years ago

Factor the expression below. x2 - 4x + 4 A. 4(x2 - x + 1) B. (x - 2)(x + 2) C. (x + 2)(x + 2) D. (x - 2)(x - 2)

Mathematics

1 answer:

Tanzania [10]3 years ago

4 0

Answer:

D. (x - 2)(x - 2)

Step-by-step explanation:

x2 - 4x + 4 = x² - 2x - 2x +4

= x(x-2) - 2(x - 2)

= (x-2)(x-2)

HINT: - 2x +4 = -2 * x + (-2) * (-2) = -2 ( x -2)

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Answer:

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Step-by-step explanation:

Given the data:

Payout ($) (x) : 0 2 4 8 10

Probability p(x) : 0.35 0.2 0.1 0.2 0.15

The expected winning ; E(X) = Σ(x * p(x))

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3 years ago

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Answer:

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3 years ago

A number multiplied by -8, subtracted from the sum of 14 and five times the number

Charra [1.4K]

Let us suppose the number is x. Now we try to make an algebraic expression going step by step.

Step 1:

Five times the number means 5 multiplied by x.

So it becomes 5x.

Step 2:

Sum of 14 and five times the number means 5x+14.

Step 3:

Number multiplied by -8 means -8x.

Step 4:

Subtract the expression in step (3) from expression in step (2)

5x+14-(-8x)

Now we can simplify it further,

5x+14+8x

Combining like terms,

13x+14

Answer: The final expression becomes 13x+14.

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3 years ago

Read 2 more answers

This is the answer to Jo question

Dahasolnce [82]

Answer:

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Step-by-step explanation:

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2 years ago

A ferry will safely accommodate 82 tons of passenger cars. Assume that the meanweight of a passenger car is 2 tons with standard

Shalnov [3]

Answer:

The probability that the maximum safe-weight will be exceeded is <u>0.0455 or 4.55%</u>.

Step-by-step explanation:

Given:

Maximum safe-weight of 37 cars = 82 tons

∴ Maximum safe-weight of 1 car (x) = 82 ÷ 37 = 2.22 tons (Unitary method)

Mean weight of 1 car (μ) = 2 tons

Standard deviation of 37 cars = 0.8 tons

So, standard deviation of 1 car is given as:

$\sigma=\frac{0.8}{\sqrt{37}}=0.13$

Probability that maximum safe-weight is exceeded, P(x > 2.22) = ?

The sample is normally distributed (Assume)

Now, let us determine the z-score of the mean weight.

The z-score is given as:

$z=\frac{x-\mu}{\sigma}\\\\z=\frac{2.22-2}{0.13}\\\\z=\frac{0.22}{0.13}=1.69$

Now, finding P(x > 2.22) is same as finding P(z > 1.69).

From the z-score table of normal distribution curve, the value of area under the curve for z < 1.69 is 0.9545.

But we need the area under the curve for z > 1.69.

So, we subtract from the total area. Total area is 1 or 100%.

So, $P(z > 1.69) = 1 - P(z < 1.69)$

$P(z>1.69)=1-0.9545=0.0455\ or\ 4.55\%$

Therefore, the probability that the maximum safe-weight will be exceeded is 0.0455 or 4.55%.

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3 years ago