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Paladinen [302]
3 years ago
11

Simplify: (6x2 + x - 7)(4x - 6) HELPP

Mathematics
1 answer:
Mariulka [41]3 years ago
3 0

Answer:

24x^3−32x^2-34x+42

Step-by-step explanation:

Distribute 4x(6x^2+x-7) and -6(6x^2+x-7)

You get 24x^3-32x^2-34x+42-32x^2+42

Then simplify the answer by adding/subtracting each variable, since you can only simplify by like terms. And you get your answer

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Answer:

8

Step-by-step explanation:

Formula for height of triangle

A÷BX2

In this case, 48÷12=4×2=8

Hope you understand ☺️

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Answer:

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Step-by-step explanation:

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A study of 4,430 U.S. adults found that 2,913 were obese or overweight. The 95% confidence interval for the proportion of adults
IrinaK [193]

Answer:

0.658 - 1.96\sqrt{\frac{0.658(1-0.658)}{4430}}=0.644

0.658 + 1.96\sqrt{\frac{0.658(1-0.658)}{4430}}=0.672

We are 95% confident that the true proportion of adults overweight is between 0.644 and 0.672

Step-by-step explanation:

We can begin founding the estimated proportion of people overweight:

\hat p =\frac{2913}{4430}= 0.658

We need to find a critical value for the confidence level using the normal standard distribution. We know that 95% is the confidence level, then the significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96

The confidence interval for the true population proportion of interest is given by the following formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

Replacing the data provided we got:

0.658 - 1.96\sqrt{\frac{0.658(1-0.658)}{4430}}=0.644

0.658 + 1.96\sqrt{\frac{0.658(1-0.658)}{4430}}=0.672

We are 95% confident that the true proportion of adults overweight is between 0.644 and 0.672

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3 years ago
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Step-by-step explanation:

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