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Dimas [21]
3 years ago
10

Plz help.... 2|x-3|-5=7

Mathematics
2 answers:
maksim [4K]3 years ago
8 0

Answer:

x=9            x=-3

Step-by-step explanation:

2|x-3|-5=7

Add 5 to each side

2|x-3|-5+5=7+5

2|x-3|=12

Divide by 2

2/2|x-3|=12/2

|x-3|=6

There are two solutions to an absolute value equation, one positive and one negative

x-3 =6           x-3 = -6

Add 3 to each side

x-3+3 = 6+3      x-3+3 = -6+3

x=9                     x = -3

blagie [28]3 years ago
6 0

Answer:

x = -3 and x = 9.

Step-by-step explanation:

2|x - 3| - 5 = 7

2|x - 3| = 12

|x - 3| = 6

x - 3 = 6

x = 9

-(x - 3) = 6

-x + 3 = 6

-x = 3

x = -3

Hope this helps!

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A tobacco company claims that the amount of nicotene in its cigarettes is a random variable with mean 2.2 and standard deviation
Aleksandr-060686 [28]

Answer:

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

Step-by-step explanation:

To solve this question, 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}}

In this problem, we have that:

\mu = 2.2, \sigma = 0.3, n = 100, s = \frac{0.3}{\sqrt{100}} = 0.03

What is the approximate probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true?

This is 1 subtracted by the pvalue of Z when X = 3.1. So

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

By the Central Limit Theorem

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

Z = \frac{3.1 - 2.2}{0.03}

Z = 30

Z = 30 has a pvalue of 1.

1 - 1 = 0

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

4 0
4 years ago
What is the answer to this question? (100 points pls answer!)
oksian1 [2.3K]

Answer : 14 Meters

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4 0
3 years ago
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A standard deck of cards contains 52 cards. One card is selected from the deck. ​(a) Compute the probability of randomly selecti
Masja [62]

Answer:

a) 1/2 = 50%

b) 3/4 = 75%

c)  1 / 52 or 1,9%

Step-by-step explanation:

In a standard deck of cards, there are 52 cards in total:

13 are hearts, 13 are diamonds, 13 are clubs and 13 are spades.

​(a) Compute the probability of randomly selecting a club or spade

How many cards are a club or a spade?

C = 13 clubs + 13 spades = 26 cards

Out of the 52 total, that means that:

P (club or spade) = 26/52 = 1/2 = 50%

​(b) Compute the probability of randomly selecting a club or spade or heart. ​

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C = 13 clubs + 13 spades  + 13 hearts = 39 cards

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P (club or spade or heart) = 39/52 = 3/4 = 75%

(c) Compute the probability of randomly selecting a two or diamond.

There's only ONE two of diamond in  regular deck of cards, so...

P(2 of diamond) = 1 / 52 or 1,9%

7 0
3 years ago
Find the equation of the line that goes through the points (-2, 5) and (3, 1).
sveta [45]

Answer:

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Step-by-step explanation:

4 0
3 years ago
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irina1246 [14]

Answer:

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Step-by-step explanation:

You have to Multiply Your diameter by 3.14. Then you multiply that bye Your height which is 7. Your answer will then be 175.84

7 0
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