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Romashka-Z-Leto [24]
2 years ago
12

A construction worker is building a wall. The wall is 17.5 yds long. He needs to place a stud every 15 inches. The first stud ha

s already been placed at one end of the wall. How many additional studs are needed, if the last stud will be placed at the other end of the wall? First fill in the blanks on the left side using the ratios shown. Then write your answer.

Mathematics
1 answer:
evablogger [386]2 years ago
5 0

The correct answers are:

Box-1: \frac{3\thinspace ft}{1\thinspace yd}

Box-2: \frac{12\thinspace \thinspace in}{1\thinspace ft}

Box-3: \frac{1\thinspace stud}{15\thinspace in}

Box-4: <em>42</em> studs

Explanation:

Given:

Wall is 17.5 yds long.

<em>In 1 yard, there are 3 feet.</em>

Therefore,

\frac{17.5\thinspace yd}{1} * \frac{3\thinspace ft}{1\thinspace yd}

In Box-1, you should put \frac{3\thinspace ft}{1\thinspace yd}.

<em>In 1 foot, there are 12 inches.</em>

Therefore,

\frac{17.5\thinspace yd}{1} * \frac{3\thinspace ft}{1\thinspace yd}*\frac{12\thinspace in}{1\thinspace ft}

In Box-2, you should put \frac{12\thinspace \thinspace in}{1\thinspace ft}.

<em>He needs to place 1 stud every 15 inches.</em>

Therefore,

\frac{17.5\thinspace yd}{1} * \frac{3\thinspace ft}{1\thinspace yd}*\frac{12\thinspace in}{1\thinspace ft}*\frac{1\thinspace stud}{15\thinspace in}

In Box-3, you should put \frac{1\thinspace stud}{15\thinspace in}.

<em>Now multiple all the numbers, you will get the number of studs.</em>

<em>\frac{17.5\thinspace yd}{1} * \frac{3\thinspace ft}{1\thinspace yd}*\frac{12\thinspace in}{1\thinspace ft}*\frac{1\thinspace stud}{15\thinspace in} = 42\thinspace studs</em>

In Box-4, write 42 studs.

-i

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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 with step-by-step explanation:

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Initial condition

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We have to show that Fibonacci numbers satisfies the recurrence relation.

The recurrence relation of Fibonacci numbers

f_n=f_{n-1}+f_{n-2},f_0=0,f_1=1

Apply this

f_n=(f_{n-2}+f_{n-3})+f_{n-2}=2f_{n-2}+f_{n-3}

f_n=2(f_{n-3}+f_{n-4})+f_{n-3}=3f_{n-3}+2f_{n-4}

f_n=3(f_{n-4}+f_{n-5})+2f_{n-4}=5f_{n-4}+3f_{n-5}

Substitute n=2

f_2=f_1+f_0=1+0=1

f_3=f_2+f_1=1+1=2

f_4=f_3+f_2=2+1=3

Hence, the Fibonacci numbers satisfied the given recurrence relation .

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Now replace n by 5n

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Let f_{5k}=5q

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It is multiple of 5 .Therefore, it is divisible by 5.

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