Need help with math points and brainlest

4.6(x - 3) = -0.4x + 16.2

kolbaska11 [484]

Hey!

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Steps To Solve:

4.6(x - 3) = -0.4x + 16.2

~Distributive property

4.6x - 13.8 = -0.4x + 16.2

~Add 0.4 to both sides

4.6x - 13.8 + 0.4= -0.4x + 16.2 + 0.4

~Simplify

5x - 13.8 = 16.2

~Add 13.8 to both sides

5x - 13.8 + 13.8 = 16.2 + 13.8

~Simplify

5x = 30

Divide 5 to both sides

5x/5 = 30/5

~Simplify

x = 6

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Answer:

$\large\boxed{x~=~6}$

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Hope This Helped! Good Luck!

8 0

3 years ago

Read 2 more answers

The manager of a supermarket would like to determine the amount of time that customers wait in a check-out line. He randomly sel

garri49 [273]

Complete question:

The manager of a supermarket would like to determine the amount of time that customers wait in a check-out line. He randomly selects 45 customers and records the amount of time from the moment they stand in the back of a line until the moment the cashier scans their first item. He calculates the mean and standard deviation of this sample to be barx = 4.2 minutes and s = 2.0 minutes. If appropriate, find a 90% confidence interval for the true mean time (in minutes) that customers at this supermarket wait in a check-out line

Answer:

(3.699, 4.701)

Step-by-step explanation:

Given:

Sample size, n = 45

Sample mean, x' = 4.2

Standard deviation $\sigma$ = 2.0

Required:

Find a 90% CI for true mean time

First find standard error using the formula:

$S.E = \frac{\sigma}{\sqrt{n}}$

$= \frac{2}{\sqrt{45}}$

$= \frac{2}{6.7082}$

$SE = 0.298$

Standard error = 0.298

Degrees of freedom, df = n - 1 = 45 - 1 = 44

To find t at 90% CI,df = 44:

Level of Significance α= 100% - 90% = 10% = 0.10

$t_\alpha_/_2_, _d_f = t_0_._0_5_, _d_f_=_4_4 = 1.6802$

Find margin of error using the formula:

M.E = S.E * t

M.E = 0.298 * 1.6802

M.E = 0.500938 ≈ 0.5009

Margin of error = 0.5009

Thus, 90% CI = sample mean ± Margin of error

Lower limit = 4.2 - 0.5009 = 3.699

Upper limit = 4.2 + 0.5009 = 4.7009 ≈ 4.701

Confidence Interval = (3.699, 4.701)

5 0

3 years ago

A line passes through point A (14,21). A second point on the line has an x-value that is 125% of the x-value of point A and a y-

seropon [69]

Answer:

The equation of the line in point-slope form is $y-21 = - \frac{3}{2}\cdot (x-14)$ .

Step-by-step explanation:

According to the statement, let $A(x,y) = (14,21)$ and $B(x,y) = (1.25\cdot x_{A},0.75\cdot y_{A})$ . The equation of the line in point-slope form is defined by the following formula: