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nlexa [21]
2 years ago
6

The owner of a construction company would like to know if his current work teams can build room additions quicker than the time

allotted for by the contract. A random sample of 16 room additions completed recently revealed an average completion time of 0.32 days (with s = 0.2 days) faster than contracted. Is this strong evidence that the teams can complete room additions in less than the contract times? The appropria
Mathematics
1 answer:
frez [133]2 years ago
8 0

Answer:

The claim that the current work teams can build room additions quicker than the time allotted for by the contract has strong statistical evidence.

Step-by-step explanation:

We have to test the hypothesis to prove the claim that the work team can build room additions quicker than the time allotted for by the contract.

The null hypothesis is that the real time used is equal to the contract time. The alternative hypothesis is that the real time is less thant the allotted for by the contract.

H_0: \mu=0\\\\H_1: \mu

The significance level, as a storng evidence is needed, is α=0.01.

The estimated standard deviation is:

s_M=\frac{s}{\sqrt{n}}=\frac{0.2}{\sqrt{16}} =0.2/4=0.05

As the standard deviation is estimated, we use the t-statistic with (n-1)=15 degrees of freedom.

For a significance level of 0.01, right-tailed test, the critical value of t is t=2.603.

Then, we calculate the t-value for this sample:

t=\frac{x-\mu}{s_M} =\frac{-0.32-0}{0.05}= -6.4

As the t-statistic lies in the rejection region, the null hypothesis is rejected. The claim that the current work teams can build room additions quicker than the time allotted for by the contract has strong statistical evidence.

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Which of the following are true statements about David Hilbert?
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he invented the point

he was a German mathematician

he  Hilbert's improvements to geometry are still used in textbooks today

he developed Hilbert's axioms

Step-by-step explanation:

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2 years ago
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Aristotle was born 384 B.C . Ron Howard was born in 1952 AD . How many years apart were they born ?
rodikova [14]

Answer: 2336 years

Step-by-step explanation:

It's simple, you just add them together :)

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2 years ago
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Jorge is scheduled to work 36 hours this week. He has already worked 27.25 hours. How much more hours does gorge have to work th
lesantik [10]

Answer:

Step-by-step explanation:

Beth earns $54 per day and $10 for each extra hour she works. Ray earns $60 per day and $8 for each extra hour he puts in. They both work five days a week. The equations show their weekly earnings with respect to how many extra hours they work.

Beth: y = 270 + 10x

Ray: y = 300 + 8x

This system is graphed below.

6 0
2 years ago
Jake scored 17 out of 25 points on a math quiz. He then scored 16 out of 20 points on a spelling quiz. On which quiz did Jake sc
Nikolay [14]

Answer:

Jake scored more on the spelling quiz. 80% is more than 68% so therefore he scored more on the quiz. To find 80% and 68%, you can use a percent of proportion or divide and multiply by 100.

Step-by-step explanation:

To find the percent of a score, you divide.

17/25 (division) = 0.68

0.68 is a decimal so to make it a percent you multiply by 100 = 68%

Or you can do it with fractions:

17/25 = ?/100: 68

16/20 = 0.8

Again, 0.8 times 100 = 80%

Or you can do it with fractions:

16/20 = ?/100: 80

He scored more on the spelling quiz, I hope this helps (justification on the answer).

7 0
1 year ago
three positive numbers are in Arithmetic Progression (A.P). the sum of the squares of the three numbers is 155. while the sum of
Alecsey [184]
Given:
Three numbers in an AP, all positive.
Sum is 21.
Sum of squares is 155.
Common difference is positive.

We do not know what x and y stand for.  Will just solve for the three numbers in the AP.
Let m=middle number, then since sum=21, m=21/3=7
Let d=common difference.
Sum of squares
(7-d)^2+7^2+(7+d)^2=155
Expand left-hand side
3*7^2-2d^2=155
d^2=(155-147)/2=4
d=+2 or -2
=+2  (common difference is positive)

Therefore the three numbers of the AP are
{7-2,7,7+2}, or
{5,7,9}


6 0
2 years ago
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