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3 years ago

5

When 0.3(4x - 8) - 0.5(2.4x + 4) is simplified what is the resulting expression?

1 answer:

LUCKY_DIMON [66]3 years ago

4 0

0.3(4x-8)-0.5(2.4x+4)
1.2x-2.4-1.2x+2
x-0.4

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It rained 0.53 inches on Monday.

jok3333 [9.3K]

Answer:

0.33

Step-by-step explanation:

0.53

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0.33

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3 years ago

Read 2 more answers

Air has a density of 0.001226 g/mL. What volume of air would have a mass of 1.0 lb? A. 2.7 mL B. 815.6 mL C. 37 mL D. 3.7 × 102

Andre45 [30]

Answer:

$Volume = 3.7\ * 10^2\ L$

Step-by-step explanation:

Given

$Density = 0.001226 g/mL$

$Mass = 1.0\ lb$

Required

Determine the volume of air

Density is calculated using:

$Density = \frac{Mass}{Volume}$

Substitute values for Density and Mass

$0.001226 g/mL = \frac{1.0\ lb}{Volume}$

Convert lb to grams

$0.001226 g/mL = \frac{1.0 * 453.592\ g}{Volume}$

$0.001226 g/mL = \frac{453.592\ g}{Volume}$

Solve for Volume

$Volume = \frac{453.592\ g}{0.001226 g/mL}$

$Volume = 369977.161501\ mL$

Convert mL to L

$Volume = 369977.161501\ L * 0.001$

$Volume = 369.977161501\ L$

Represent using scientific notation

$Volume = 3.69977161501\ * 10^2\ L$

Approximate

$Volume = 3.7\ * 10^2\ L$

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3 years ago

Write the equation of the line with slope = -3 and passes through (2,7)

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Answer:

Step-by-step explanation:

y - 7 = -3(x - 2)

y - 7 = -3x + 6

y = -3x + 13

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3 years ago

Find equation to this line

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Answer:

the equation of this line is y= 1x+6

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3 years ago

At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population pro

Gnom [1K]

Answer:

A sample of 1068 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population proportion?

We need a sample of n.

n is found when M = 0.03.

We have no prior estimate of $\pi$ , so we use the worst case scenario, which is $\pi = 0.5$

Then

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.03 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.03\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.03}$

$(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.03})^{2}$

$n = 1067.11$

Rounding up

A sample of 1068 is needed.

8 0

2 years ago