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4 years ago

Factorise fully 4x^2+28x^3

Mathematics

2 answers:

olga_2 [115]4 years ago

8 0

Answer:

4x²×(1+7x)

Step-by-step explanation:

Novosadov [1.4K]4 years ago

4 0

Answer:

$4x^2+28x^3 : 4x^2 (7x + 1)$

Step-by-step explanation:

$4x^2+28x^3$

Apply exponent rule: $\:a^{b+c}=a^ba^c$

$x^3=xx^2$

$=28xx^2+4x^2$

Rewrite $28$ as $4\cdot \:7$ :

$=4\cdot \:7xx^2+4x^2$

Factor out common term $4x^2$ :

$=4x^2\left(7x+1\right)$

Hope I helped, if so may I get brainliest and a thanks?

Thank you, have a good day! =)

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In 1992, the population of flamingos in Brazil was 15,500. Today, there are 32,860 flamingos in Brazil. What is the percent chan

Andrej [43]

232 bc 15500 divided by 100 is 155 then 155 times 232 is 32860

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2 years ago

At the movie theater all snacks and all movie tickets are priced the same. Four snacks and three movie tickets cost $40. Two sna

Nataliya [291]

We know that four snacks and three movie tickets cost $40 dollars, and when you take away 2 snacks but still keep the tickets, its $32.

2 snacks= $8

What i did to find the cost of each ticket is I divided 8 by 2 (8/2) and ended getting $4

1 snack= $4

Now to find the cost of each ticket, i divided 2 snacks ($8) and got this equation:

32 - 8= 24

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Now to find the amount for each ticket, i divided 24 by 3 (24/3) and got the answer:

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Therefore, snacks are $4 and movie tickets are $8

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<em>Thank you! Btw can i please get brainly :3</em>

7 0

3 years ago

Experian would like to test the hypothesis that the average credit score for an adult in Virginia is different from the average

aliya0001 [1]

Answer:

a. We fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

b. The 95% confidence interval for the true difference of means is -2.2468 and 36.2468. There is a probability of 95% that the true difference of means $\mu_{1}-\mu_{2}$ is between -2.2468 and 36.2468. This confidence interval contains the number 0, this is consistent with the results we got in a. Because we fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

Step-by-step explanation:

Let $\mu_{1}-\mu_{2}$ be the true difference between the average credit score for an adult in Virginia and the average credit score for an adult in North Carolina. We have the large sample sizes $n_{1} = 40$ and $n_{2} = 35$ , the unbiased point estimate for $\mu_{1}-\mu_{2}$ is $\bar{x}_{1} - \bar{x}_{2}$ , i.e., 699-682 = 17.

The standard error is given by $\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$ , i.e.,

$\sqrt{\frac{(44)^{2}}{40}+\frac{(41)^{2}}{35}}$ = 9.8198.

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 rejection region is given by RR = {z | z < -1.96 or z > 1.96} where -1.96 and 1.96 are the 2.5th and 97.5th quantiles of the standard normal distribution respectively. The test statistic is $Z = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}}$ and the observed value is $z_{0} = \frac{17}{9.8198} = 1.7312$ . Because 1.7312 does not fall inside RR, we fail to reject the null hypothesis.

b. The endpoints for a 95% confidence interval for $\mu_{1}-\mu_{2}$ is given by $17\pm (z_{0.05/2})9.8198$ , i.e., $17\pm (z_{0.025})9.8198$ where $z_{0.025}$ is the 2.5th quantile of the standard normal distribution, i.e., -1.96, so, we have 17-(1.96)(9.8198) and 17+(1.96)(9.8198), i.e., -2.2468 and 36.2468. There is a probability of 95% that the true difference of means $\mu_{1}-\mu_{2}$ is between -2.2468 and 36.2468. This confidence interval contains the number 0, this is consistent with the results we got in a. Because we fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.