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ki77a [65]
2 years ago
9

In order to determine the average price of hotel rooms in Atlanta, a sample of 63 hotels were selected. It was determined that t

he test statistic (z) was $-1.53. We would like to test whether or not the average room price is significantly different from $110. Population standard deviation is known to us. Compute the p-value.
Mathematics
1 answer:
olchik [2.2K]2 years ago
5 0

Answer:

Pvalue = 0.063008

Step-by-step explanation:

Given :

Sample size, n = 63

The test statistic = - 1.53

To test if average room price is significantly different from $110 ; μ = $110

H0 : μ = 110

H1 : μ < 110

From the test statistic value obtained, we could see that the test is left tailed

The Pvalue :

P(Z < - 1.53) = 0.063008

If the claim is to be tested at 5% level of confidence then

Pvalue > 0.05 ; Hence, we fail to reject the. NULL. Hence, average price is not significantly different from $110

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GarryVolchara [31]
X/8 - 10 = x/3

bring the values with the variable on one side

x/3 - x/8 = 10

find a common denominator and subtract these fractions

8x/24 - 3x/24 = -10

now subtract fractions

8x - 3x = 5x

5x/24 = -10

multiply both sides by 24

5x = -240

divide both sides by 5

x = -48

the number is -48
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Whats the slope of the line help pls
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Step-by-step explanation:

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2 years ago
1-What is the sum of the series? ​∑j=152j​ Enter your answer in the box.
tangare [24]

Answer:

Please see the Step-by-step explanation for the answers

Step-by-step explanation:

1)

∑\left \ {{5} \atop {j=1}} \right. 2j

The sum of series from j=1 to j=5 is:

∑ = 2(1) + 2(2) + 2(3) + 2(4) + 2(5)

  =  2 + 4 + 6 + 8 + 10

∑ = 30

2)

This question is not given clearly so i assume the following series that will give you an idea how to solve this:

∑\left \ {{4} \atop {k=1}} \right. 2k²

The sum of series from k=1 to j=4 is:

∑ = 2(1)² + 2(2)² + 2(3)² + 2(4)²

  = 2(1) + 2(4) + 2(9) + 2(16)

  =  2 + 8 + 18 + 32

∑ = 60

∑\left \ {{4} \atop {k=1}} \right. (2k)²

∑ = (2*1)² + (2*2)² + (2*3)² + (2*4)²

  = (2)² + (4)² + (6)² + (8)²

  = 4 + 16 + 36 + 64

∑ = 120

∑\left \ {{4} \atop {k=1}} \right. (2k)²- 4

∑ = (2*1)²-4 + (2*2)²-4 + (2*3)²-4 + (2*4)²-4

  = (2)²-4 + (4)²-4 + (6)²-4 + (8)²-4

  = (4-4) + (16-4) + (36-4) + (64-4)

  = 0 + 12 + 32 + 60

∑ = 104

∑\left \ {{4} \atop {k=1}} \right. 2k²- 4

∑ = 2(1)²-4 + 2(2)²-4 + 2(3)²-4 + 2(4)²-4

  = 2(1)-4 + 2(4)-4 + 2(9)-4 + 2(16)-4

  = (2-4) + (8-4) + (18-4) + (32-4)

  = -2 + 4 + 14 + 28

∑ = 44

3)

∑\left \ {{6} \atop {k=3}} \right. (2k-10)

∑ = (2×3−10) + (2×4−10) + (2×5−10) + (2×6−10)  

  = (6-10) + (8-10) + (10-10) + (12-10)

  = -4 + -2 + 0 + 2  

∑ = -4

4)

1+1/2+1/4+1/8+1/16+1/32+1/64

This is a geometric sequence where first term is 1 and the common ratio is 1/2 So

a = 1

This can be derived as

1/2/1 = 1/2 * 1 = 1/2

1/4/1/2 = 1/4 * 2/1 = 1/2

1/8/1/4 = 1/8 * 4/1  = 1/2

1/16/1/8 = 1/16 * 8/1  = 1/2

1/32/1/16 = 1/32 * 16/1  = 1/2

1/64/1/32 = 1/64 * 32/1  = 1/2

Hence the common ratio is r = 1/2

So n-th term is:

ar^{n-1} = 1(\frac{1}{2})^{n-1}

So the answer that represents the series in sigma notation is:

∑\left \ {{7} \atop {j=1}} \right. (\frac{1}{2})^{j-1}

5)

−3+(−1)+1+3+5

This is an arithmetic sequence where the first term is -3 and the common difference is 2. So  

a = 1

This can be derived as

-1 - (-3) = -1 + 3 = 2

1 - (-1) = 1 + 1 = 2

3 - 1 = 2

5 - 3 = 2

Hence the common difference d = 2

The nth term is:

a + (n - 1) d

= -3 + (n−1)2

= -3 + 2(n−1)

= -3 + 2n - 2

= 2n - 5

So the answer that represents the series in sigma notation is:

∑\left \ {{5} \atop {j=1}} \right. (2j−5)

6 0
3 years ago
The length of a rectangular field is 7 m less than 4 times the width. The perimeter is 136 m. Find the width and length
cestrela7 [59]

Answer:

Width: 15 Length: 53

Step-by-step explanation:

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3 years ago
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