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3 years ago

8

Consider the following statement. Five less than 4 times a number equals 2 more than 3 times the number. a. Write an equation to

represent the statement. Use n for the unknown number. b. Solve the equation. c. Explain how you solved the equation.

2 answers:

stich3 [128]3 years ago

7 0

4n-5 = 3n+2 thats taking the word problem and writing it out
Now solve

4n-5=3n+2
n = 7

prisoha [69]3 years ago

5 0

n = unknown number

5 less than 4 times a number = 4n-5

2 more than 3 times a number = 3n+2

so you have 4n-5 = 3n +2

add 5 to both sides:

4n=3n+7

subtract 3n from both sides:

n = 7

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Hi there!

First you simplify.

Simplified = 9x+(−5)+(−8)+x

Then you combine the like terms.

<span><span><span><span><span>9x</span>+(<span>−5)</span></span>+(<span>−8)</span></span>+x (*-5 and -8 are like terms) (9x and x are like terms)

</span></span><span><span><span>Combined it's~ (<span><span>9x</span>+x</span>)</span>+<span>(<span><span>−5</span>+<span>−8</span></span>)

</span></span></span><span><span><span>Solve and that comes to 10x</span>+<span>−<span>13 which is your answer. :)

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Step-by-step explanation:

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Step 1.

Calculate measure of angle α:

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Step 2.

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$\dfrac{9^o}{360^o}=\dfrac{1}{40}$

Step 3.

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$C=2\pi r\to C=2\pi\cdot4000=8000\pi\ mi$

Step 4.

The length of arc is equal 1/40 of the circumference:

$\dfrac{1}{40}C=\dfrac{1}{40}\cdot8000\pi=200\pi\ mi$

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3 years ago

Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s

stellarik [79]

Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.

Then we have $n_{1} = 25$ , $\bar{x}_{1} = 2.04$ , $\sigma_{1} = 0.010$ and $n_{2} = 20$ , $\bar{x}_{2} = 2.07$ , $\sigma_{2} = 0.015$ . The pooled estimate is given by

$\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552$

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative).

The test statistic is $T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$ and the observed value is $t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257$ . T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value $t_{0}$ falls inside RR, we reject the null hypothesis at the significance level of 0.05

b. The p-value for this test is given by $2P(T$ 0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.

c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

$(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}$ , i.e.,

$-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}$

where $t_{0.025}$ is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So

$-0.03\pm(2.0167)(0.012459)(0.3)$ , i.e.,

(-0.0225, -0.0375)

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3 years ago