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3 years ago

The letter b represents a nonzero constant. Solve 2bx-bx= -8 for x?

Mathematics

1 answer:

zhuklara [117]3 years ago

3 0

Answer:

x = -8/b

Step-by-step explanation:

2bx-bx= -8

bx = -8

Divide by b since b is nonzero

x = -8/b

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Six yards of rope costs $5.50 while four yards costs $4.75

Amiraneli [1.4K]

The best option is the first one (Six yards of rope costs $5.50). It is the lowest unit cost.

<u>Step-by-step explanation:</u>

For the given question case, first we should compare the unit costs of the choices provided and evaluate the best option.

Choice 1:

Six yards of rope costs $ 5.50

$\frac{5.50 \text { dollar}}{6 \text { yard }}=0.9167\ dollar/yard$

Choice 2:

Four yards costs $ 4.75

$\frac{4.75 \text { dollar}}{4 \text { yard }}=1.1875\ dollar/yard$

Thus, the best choice is the first one. Because it is the lowest unit cost.

3 0

3 years ago

Find the product. Write your answer in scientific notation. (3 x 10^2)×(4×10^5)

suter [353]

Answer:

12 * 10 ^ 7

Step-by-step explanation:

3 * 10 ^ 2 * 4 * 10 ^5

= 12 * 10^ 2 + 5

= 12 * 10 ^ 7

Hope it will help :)

7 0

3 years ago

A boy pushes on box of mass 43 kg and it moves at an acceleration of 8.6m/s

madam [21]

If you want to work out the force then F=MA, therefore 43*8.6=369.8 Newtons

7 0

3 years ago

Read 2 more answers

Simplify |3 − 11| − (15 ÷ 3 + 2) 2

Vikentia [17]

|-8| - (5 + 2)2
8-(7)2
8-14= -6

Follow order of operations

6 0

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)

8 0

3 years ago