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3 years ago

What is the solution to the linear equation?

2 answers:

8 0

First, you want to make all your fractions have a common denominator. Your common denominator would be 6, so multiplying each /3 fraction by 2 and each /2 fraction by 3, you get 4/6x-3/6=2/6+5/6x. You can now combine your fractions getting (4x-3)/6=(2+5x)/6. Now, reduce out the fractions to get 4x-3=2+5x, subtract 4x from both sides, -3=2+x, subtract 2 from both sides, x=-5, or answer choice A.

solniwko [45]3 years ago

6 0

Answer:

A) -5

Step-by-step explanation:

You might be interested in

Evaluate 3x for x = -2, x = 1, and x = 3.

jok3333 [9.3K]

Answer:

A) 1/9, 3, 27

Step-by-step explanation:

i think you mean 3^x (and not 3x).

because otherwise none of the answer options withdraw makes any sense.

3^-2 = 1/(3²) = 1/9

3¹ = 3

3³ = 27

3 0

2 years ago

Consider the population of all 1-gallon cans of dusty rose paint manufactured by a particular paint company. Suppose that a norm

Artemon [7]

Answer:

a) 0.5.

b) 0.8413

c) 0.8413

d) 0.6826

e) 0.9332

f) 1

Step-by-step explanation:

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.

In this problem, we have that:

$\mu = 6, \sigma = 0.2$

(a) P(x > 6) =

This is 1 subtracted by the pvalue of Z when X = 6. So

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

$Z = \frac{6-6}{0.2}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5.

1 - 0.5 = 0.5.

(b) P(x < 6.2)=

This is the pvalue of Z when X = 6.2. So

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

$Z = \frac{6.2-6}{0.2}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

(c) P(x ≤ 6.2) =

In the normal distribution, the probability of an exact value, for example, P(X = 6.2), is always zero, which means that P(x ≤ 6.2) = P(x < 6.2) = 0.8413.

(d) P(5.8 < x < 6.2) =

This is the pvalue of Z when X = 6.2 subtracted by the pvalue of Z when X 5.8.

X = 6.2

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

$Z = \frac{6.2-6}{0.2}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

X = 5.8

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

$Z = \frac{5.8-6}{0.2}$