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4 years ago

54x2y - 12x2y2 distributive property

1 answer:

3 0

<span>Use the distributive property to factor the expression below.
54x2y - 12x2y2
-----
GCF = 6x^2y
---

Factor:
= 6x^2y(9 - 2y)</span>

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Describe the solutions of (2 − 5)^2 ≤ t in words.

natita [175]

Answer:

two minus five raised to the power of two is greater or equal to t

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3 years ago

Carla’s team won 3 of its 5 games played. Elena’s team won games at the same rate. How many games were played if Elena’s team wo

garik1379 [7]

3/5=12/x

reciprical of both sides

5/3=x/12

times twelve on both sides

x=20

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2 years ago

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What is the answer 2c+(3c-4)=a-4

gavmur [86]

Answer:idk

Step-by-step explanation:

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3 years ago

2x-4/5=4<br> plsss answer quick

nikdorinn [45]

Answer:

Step-by-step explanation:

$\frac{2x - 4}{5} =4$

First of all multiply through by 5 to eliminate the fraction

That's

$5 \times \frac{2x - 4}{5} = 4 \times 5 \\ 2x - 4 = 20$

Next using the addition property add 4 to both sides of the equation

We have

$2x - 4 + 4 = 20 + 4 \\ 2x = 24$

<u>Divide both sides by 2</u>

$\frac{2x}{2} = \frac{24}{2}$

We have the final answer as

Hope this helps you

6 0

3 years ago

Read 2 more answers

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