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stira [4]
2 years ago
9

Sergei buys a rectangular rug for his living room. He measures the diagonal of the rug to be 18 feet. The length of the rug is 3

feet longer than the width.
What are the approximate dimensions of the rug? Round each dimension to the nearest tenth of a foot.

10.4 feet by 7.4 feet
11.1 feet by 8.1 feet
13.4 feet by 10.4 feet
14.1 feet by 11.1 feet
Mathematics
2 answers:
Novosadov [1.4K]2 years ago
7 0

Answer:

D. 14.1 feet by 11.1 feet.

Step-by-step explanation:

Let w be the width of the rug.

We have been given that the length of the rug is 3 feet longer than the width. So the length of the rug would be w+3.

We have been given that the measure of the diagonal of the rug to be 18 feet. To find the width of rug we will use Pythagoras theorem as rug is rectangular.

w^2+(w+3)^2=18^2

w^2+w^2+6w+9=324

2w^2+6w+9=324

2w^2+6w+9-324=0

2w^2+6w-315=0

We will use quadratic formula to solve for w.

w=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

w=\frac{-6\pm \sqrt{6^2-4*2*-315}}{2*2}

w=\frac{-6\pm \sqrt{36+2520}}{4}

w=\frac{-6\pm \sqrt{2556}}{4}

w=\frac{-6}{4}\pm \frac{\sqrt{2556}}{4}

w=\frac{-6}{4}\pm \frac{50.55689863}{4}

Since the width cannot be negative, so width of the rug would be:

w=\frac{-6}{4}+\frac{50.55689863}{4}

w=-1.5+12.639224659

w=11.139224\approx 11.1

Therefore, the width of rug would be 11.1 feet.

Since length of rug is 3 feet longer than width, so width of the rug would be 11.1+3=14.1,

Therefore, the option D is the correct choice.

ivanzaharov [21]2 years ago
3 0

14.1 feet by 11.1 feet is the correct one

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Two random samples are taken from private and public universities
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Answer:

Step-by-step explanation:

For private Institutions,

n = 20

Mean, x1 = (43120 + 28190 + 34490 + 20893 + 42984 + 34750 + 44897 + 32198 + 18432 + 33981 + 29498 + 31980 + 22764 + 54190 + 37756 + 30129 + 33980 + 47909 + 32200 + 38120)/20 = 34623.05

Standard deviation = √(summation(x - mean)²/n

Summation(x - mean)² = (43120 - 34623.05)^2+ (28190 - 34623.05)^2 + (34490 - 34623.05)^2 + (20893 - 34623.05)^2 + (42984 - 34623.05)^2 + (34750 - 34623.05)^2 + (44897 - 34623.05)^2 + (32198 - 34623.05)^2 + (18432 - 34623.05)^2 + (33981 - 34623.05)^2 + (29498 - 34623.05)^2 + (31980 - 34623.05)^2 + (22764 - 34623.05)^2 + (54190 - 34623.05)^2 + (37756 - 34623.05)^2 + (30129 - 34623.05)^2 + (33980 - 34623.05)^2 + (47909 - 34623.05)^2 + (32200 - 34623.05)^2 + (38120 - 34623.05)^2 = 1527829234.95

Standard deviation = √(1527829234.95/20

s1 = 8740.22

For public Institutions,

n = 20

Mean, x2 = (25469 + 19450 + 18347 + 28560 + 32592 + 21871 + 24120 + 27450 + 29100 + 21870 + 22650 + 29143 + 25379 + 23450 + 23871 + 28745 + 30120 + 21190 + 21540 + 26346)/20 = 25063.15

Summation(x - mean)² = (25469 - 25063.15)^2+ (19450 - 25063.15)^2 + (18347 - 25063.15)^2 + (28560 - 25063.15)^2 + (32592 - 25063.15)^2 + (21871 - 25063.15)^2 + (24120 - 25063.15)^2 + (27450 - 25063.15)^2 + (29100 - 25063.15)^2 + (21870 - 25063.15)^2 + (22650 - 25063.15)^2 + (29143 - 25063.15)^2 + (25379 - 25063.15)^2 + (23450 - 25063.15)^2 + (23871 - 25063.15)^2 + (28745 - 25063.15)^2 + (30120 - 25063.15)^2 + (21190 - 25063.15)^2 + (21540 - 25063.15)^2 + (26346 - 25063.15)^2 = 1527829234.95

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This is a test of 2 independent groups. Let μ1 be the mean out-of-state tuition for private institutions and μ2 be the mean out-of-state tuition for public institutions.

The random variable is μ1 - μ2 = difference in the mean out-of-state tuition for private institutions and the mean out-of-state tuition for public institutions.

We would set up the hypothesis. The correct option is

-B. H0: μ1 = μ2 ; H1: μ1 > μ2

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(x1 - x2)/√(s1²/n1 + s2²/n2)

t = (34623.05 - 25063.15)/√(8740.22²/20 + 3766.55²/20)

t = 9559.9/2128.12528473889

t = 4.49

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [8740.22²/20 + 3766.55²/20]²/[(1/20 - 1)(8740.22²/20)² + (1/20 - 1)(3766.55²/20)²] = 20511091253953.727/794331719568.7114

df = 26

We would determine the probability value from the t test calculator. It becomes

p value = 0.000065

Since alpha, 0.01 > than the p value, 0.000065, then we would reject the null hypothesis. Therefore, at 1% significance level, the mean out-of-state tuition for private institutions is statistically significantly higher than public institutions.

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Answer:

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