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3 years ago

Solve this problem.

Mathematics

1 answer:

Katyanochek1 [597]3 years ago

3 0

Answer:

Let x = the width of the smaller rectangle.

The length of the smaller rectangle is 2x - 1.

Area is A = lw

So the area of the smaller rectangle is A = (x)(2x - 1) = 2x^2 - x

Step-by-step explanation:

hope it helped

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Rounding and math strategies

Otrada [13]

304 would be the correct answer

5 0

3 years ago

The sum of the squares of three consecutive integers is 509. Determine the integers.

GalinKa [24]

Answer:

12, 13, 14

Step-by-step explanation:

Denote the integers as:

x+1

x+2

The sum of their squares, so that would be;

(x^(2)) + (( x + 1 )^(2)) + (( x + 2 )^(2)) = 509

write out the squares

x^2 + x^2 + 2x + 1 + x^2 + 4x + 4 = 509

combine like terms

3x^2 + 6x + 5 = 509

inverse operations

3x^2 + 6x + 5 = 509

-5 -5

3x^2 + 6x = 504

factor

3x^2 + 6x = 504

3 ( x^2 + 2x ) =504

Inverse operations

3 ( x^2 + 2x ) = 504

/3 /3

x^2 + 2x = 168

Factor again

x ( x + 2 ) = 168

At this point, it should be obvious that x is 12 (because 12 * 14 = 168)

So now substitute back into the consecutive numbers

x = 12

x + 1 = 13

x + 2 = 14

6 0

3 years ago

There are 2,000 eligible voters in a precinct. A total of 500 voters are randomly selected and asked whether they plan to vote f

Ann [662]

Answer:

$0.7 - 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.647$

$0.7 + 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.753$

And the 99% confidence interval would be given (0.647;0.753).

So the correct answer would be:

a. 0.647 and 0.753

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

The estimated population proportion for this case is:

$\hat p = \frac{350}{500}=0.7$

The confidence interval would be given by this formula

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

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.