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stich3 [128]
3 years ago
13

3x-2=2(x-5)find the value of x ​

Mathematics
2 answers:
kipiarov [429]3 years ago
5 0

Now we have to,

find the required value of x.

Let's begin,

→ 3x-2 = 2(x-5)

→ 3x-2 = 2x-10

→ 3x-2x = -10+2

→ x = -8

Hence, value of x is -8.

kherson [118]3 years ago
3 0

Answer:

x = -8

Step-by-step explanation:

3x - 2 = 2 ( x + 5

Solve for x.

Let's solve,

3x - 2 = <u>2 ( x + 5 )</u>

Step 1:- Distribute 2.

3x - 2 = 2 × x + 2 × 5

3x <u>- 2</u> = 2x - 10

Step 2 :- Move constant to the right-hand and change their sign.

3x = 2x <u>- 10 + 2</u>

Step 3:- Add -10 and 2.

3x = <u>2x</u> - 8

Step 4 :- Move variable to the left-hand side and change their sign.

<u>3x - 2x</u> = -8

Step 5 :- Subtract 2x from 2x.

x = -8

Hence, value of x = -8.

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Find the value of (-4)(6)(-7)<br> -178<br> 168<br> 178<br> -168
kakasveta [241]

<u><em>Answer:</em></u>

168

<u><em>Explanation:</em></u>

<u>Before we begin, remember the following:</u>

+ve * +ve = +ve                     -ve * -ve = +ve

+ve * -ve = -ve                       -ve * +ve = +ve

<u>Now, for the given problem we have:</u>

(-4) * (6) * (-7)

<u>Let's take the first two terms:</u>

(-4) * (6)

Based on the above rules, the product will be negative

<u>Therefore, </u>

(-4) * (6) = -24

<u>Now, the expression became:</u>

(-24) * (-7)

Again, based on the above rule, the product here will be positive

<u>Therefore,</u>

(-24) * (-7) = 168

Hope this helps :)

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Please help with this math question
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Domain: {-3}
Range: {8, 7, 6, 5}
This is not a function.
5 0
3 years ago
Assume that foot lengths of women are normally distributed with a mean of 9.6 in and a standard deviation of 0.5 in.a. Find the
Makovka662 [10]

Answer:

a) 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.

b) 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

c) 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.

Step-by-step explanation:

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 \frac{\sigma}{\sqrt{n}}.

Normal probability distribution

Problems of normally distributed samples can be 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.

In this problem, we have that:

\mu = 9.6, \sigma = 0.5.

a. Find the probability that a randomly selected woman has a foot length less than 10.0 in

This probability is the pvalue of Z when X = 10.

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

Z = \frac{10 - 9.6}{0.5}

Z = 0.8

Z = 0.8 has a pvalue of 0.7881.

So there is a 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.

b. Find the probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

This is the pvalue of Z when X = 10 subtracted by the pvalue of Z when X = 8.

When X = 10, Z has a pvalue of 0.7881.

For X = 8:

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

Z = \frac{8 - 9.6}{0.5}

Z = -3.2

Z = -3.2 has a pvalue of 0.0007.

So there is a 0.7881 - 0.0007 = 0.7874 = 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

c. Find the probability that 25 women have foot lengths with a mean greater than 9.8 in.

Now we have n = 25, s = \frac{0.5}{\sqrt{25}} = 0.1.

This probability is 1 subtracted by the pvalue of Z when X = 9.8. So:

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

Z = \frac{9.8 - 9.6}{0.1}

Z = 2

Z = 2 has a pvalue of 0.9772.

There is a 1-0.9772 = 0.0228 = 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.

5 0
3 years ago
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