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Alex
3 years ago
15

3(x+4)=5(x-2) I really need help i have no idea how to do it

Mathematics
2 answers:
Oksana_A [137]3 years ago
6 0

<u>answer:</u>  x = 11

<u>work:</u>

3(x+4)=5(x-2)  | distribute 3 and 5 into the parentheses.

3x + 12 = 5x - 10  | subtract 5x and move it over to 3x, and solve.

-2x + 12 = -10  | subtract 12 and move it over to -10, and solve.

-2x = -22  | divide by -2.

x = 11  | final answer.

hope this helps! ❤ from peachimin

Alexxandr [17]3 years ago
3 0

Gotcha!

3(x+4)=5(x-2)

Distributive Property

3x+12=5x-10

Subtract 5x from 3x & 12 from -10

So far, you would have: -2x=-22

Divide both sides by -2!

Answer: x=11

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5 0
2 years ago
Assume the readings on thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. Find the pro
lisov135 [29]

Answer:

0.0326 = 3.26% probability that a randomly selected thermometer reads between −2.23 and −1.69.

The sketch is drawn at the end.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean of 0°C and a standard deviation of 1.00°C.

This means that \mu = 0, \sigma = 1

Find the probability that a randomly selected thermometer reads between −2.23 and −1.69

This is the p-value of Z when X = -1.69 subtracted by the p-value of Z when X = -2.23.

X = -1.69

Z = \frac{X - \mu}{\sigma}

Z = \frac{-1.69 - 0}{1}

Z = -1.69

Z = -1.69 has a p-value of 0.0455

X = -2.23

Z = \frac{X - \mu}{\sigma}

Z = \frac{-2.23 - 0}{1}

Z = -2.23

Z = -2.23 has a p-value of 0.0129

0.0455 - 0.0129 = 0.0326

0.0326 = 3.26% probability that a randomly selected thermometer reads between −2.23 and −1.69.

Sketch:

3 0
3 years ago
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