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3 years ago

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one linear expression is subtracted from a second linear expression and the difference is x-5. When is the difference when the s

econd linear expression is subtracted from the fist?

1 answer:

sineoko [7]3 years ago

6 0

The answer is 5× if is wrong sorry might help you

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Can someone please help me with question 1, I will mark you as a brainest and I'll thank you

Leya [2.2K]

I believe it is 86.4

8 0

3 years ago

The yearbook club had a meeting. The meeting had 21 people, which is one-third of the club. How many people are in the club?

VladimirAG [237]

Answer:

63 because if 1/3 of the club is gone then we need to multiply 21 times 21 to find the total. Which is 63.

3 0

4 years ago

If a=-1,b=-6 and c=2 then what is the value of b2-4ac?

grandymaker [24]

Answer:

-4

Step-by-step explanation:

(-6)(2)-4(-1)(2)

(-12)-(-8)

-12+8=-4

4 0

4 years ago

Stella wants to buy colored pencils that cost $3.50. She has 1 dollar, 5 quarters, 4 dimes, and 7 pennies. Does she have enough

Andrews [41]

Answer:

no

Step-by-step explanation:

Lets she how much money she has

1.00 dollars

5 * .25 = 1.25 quarters

4 * .10 = .40 dimes

7 * .01 = .07 pennies

Add it all together

1 + 1.25+ .40 +.07 =2.72

3.50 - 2.72 =.78

She is 78 cents short

6 0

3 years ago

How many students must be randomly selected to estimate the mean weekly earnings of students at one college? We want 95% confide

kari74 [83]

Answer:

The sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:

$CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

The margin of error of a (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:

$MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

The information provided is:

<em>σ</em> = $60

<em>MOE</em> = $2

The critical value of <em>z</em> for 95% confidence level is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Compute the sample size as follows:

$MOE=z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

$n=[\frac{z_{\alpha/2}\times \sigma }{MOE}]^{2}$

$=[\frac{1.96\times 60}{2}]^{2}$

$=3457.44\\\approx 3458$

Thus, the sample of students required to estimate the mean weekly earnings of students at one college is of size, 3458.

8 0

3 years ago