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3 years ago

-3(x + 2)2 + 4 = -8 what’s the answer

2 answers:

8 0

X=0

-3(0+2)2+4 distribute -3 to the parentheses-PEMDAS
(0-6)2+4 multiply 2 by the parentheses
(0-12)+4 solve the equation
= -8

Gnesinka [82]3 years ago

7 0

Answer:

x = 0

Step-by-step explanation:

First, multiply the -3 by two (-3(x + 2)2 + 4 = -8) -- -6(x+2)+4 = -8.

Then, subtract 4 from both sides (-6(x+2)+4 = -8) -- -6(x+2) = -12

After that, divide both sides by -6 (-6(x+2) = -8) -- $\frac{-6\left(x+2\right)}{-6}=\frac{-12}{-6}$

Simplify -- x+2=2

Subtract 2 from both sides (x+2=2) -- x = 0

Hope this helped! Good luck with future math problems :)

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3 years ago

A line intersects the points (3,-4) and

MrRa [10]

Answer:

y = - 3x + 5

Step-by-step explanation:

The equation of a line in slope- intercept form is

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m = (y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) = (3, - 4) and (x₂, y₂ ) = (1, 2)

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4 years ago

In a given year, the average annual salary of a NFL football player was $189,000 with a standard deviation of $20,500. If a samp

nika2105 [10]

Answer:

15.15% probability that the sample mean will be $192,000 or more.

Step-by-step explanation:

To solve this problem, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$