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1 year ago

ip a standard sheet of paper measures 8 1/2 by 11 inches. find the area of one such sheet of paper in

Mathematics

1 answer:

SOVA2 [1]1 year ago

7 0

The area of the standard sheet of paper is 93 1/2 square inches.

The standard sheet of paper is a rectangle and to find its area, multiply the length by the width.

The area of the rectangle is the region enclosed by the perimeter of the rectangle and is equal to product of length and width.

If the standard sheet of paper has a length of 11 inches and a width of 8 1/2 inches, then:

A = l * w

A = 11 x 8 1/2

A = 93 1/2 square inches

Hence, the area of the standard sheet of paper is 93 1/2 square inches.

To learn more about area of rectangle: brainly.com/question/25292087

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The manager of a store that specializes in selling tea decides to experiment with a new blend. She will mix some Earl Grey tea t

marishachu [46]

Let us say that:

x = mass from Earl Grey Tea

y = mass from Orange Pekoe Tea

So that:

x + y = 600

and that:

6 x + 4 y = 4.5 * 600

6 (600 – y) + 4 y = 2700

3600 – 6y + 4y = 2700

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x = 600 – 450 = 150 pounds

Answer:

<span> Earl Grey Tea = 150 pounds</span>

<span>Orange Pekoe Tea = 450 pounds</span>

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3 years ago

When you solve the equation x + 1 = 2x for x, the solution is 1. 1) Square both sides of the equation, and verify that your solu

tatiyna

Answer:

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3 years ago

Please answer as fast as possible thanks

svp [43]

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3 years ago

Many high school students take the AP tests in different subject areas. In 2007, of the 144,796 students who took the biology ex

Nataly [62]

Answer:

$(0.582-0.485) - 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.0942$

$(0.582-0.485) + 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.09978$

And the 90% confidence interval would be given (0.0942;0.09978).

We are confident at 90% that the difference between the two proportions is between $0.0942 \leq p_A -p_B \leq 0.09978$

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".

$p_A$ represent the real population proportion female for Biology

$\hat p_A =\frac{84199}{144796}=0.582$ represent the estimated proportion female for biology

$n_A=144796$ is the sample size for A

$p_B$ represent the real population proportion female for calculus AB

$\hat p_B =\frac{102598}{211693}=0.485$ represent the estimated proportion female for Calculus AB

$n_B=211693$ is the sample size required for B

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

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

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 90% confidence interval the value of $\alpha=1-0.90=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution.