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Art [367]
3 years ago
7

In March 2007, Business Week reported that at the top 50 business schools, students studied an average of 14.6 hours. You wonder

whether the amount UMSL students study is different from this 14.6 hour benchmark. Set up the hypotheses used in this situation.
Mathematics
1 answer:
Rasek [7]3 years ago
5 0

Answer:

The hypotheses used in this situation

H_0:\mu = 14.6

H_a:\mu \neq 14.6

Step-by-step explanation:

We are given that  Business Week reported that at the top 50 business schools, students studied an average of 14.6 hours.

Mean = \mu = 14.6

Claim : The amount UMSL students study is different from this 14.6 hour benchmark.

The hypotheses used in this situation

H_0:\mu = 14.6

H_a:\mu \neq 14.6

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Which factorization is equivalent to this expression?
grigory [225]

Answer:

10(3x+7)

Step-by-step explanation:

we have the expression:

30x + 70

To factor this expression we need to indentify the components that are common in both terms.

At first glance there is nothing in common, but we can notice that 30 and 70 are multiples of 10, that is:

10*3=30\\10*7=70

so we can substitute this into the expression:

10*3x+10*7

and now that we have the common term (the number 10) we can factorize it, that is, take out the common term and include a parentheses:

10(3x+7)

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What is 10 times as much as 1.0
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The accompanying data contains the depth​ (in kilometers) and​ magnitude, measured using the Richter​ Scale, of all earthquakes
Sunny_sXe [5.5K]

Answer:

Depth:

μ =20.2025 km

M = 15.625 km

Range = 47.15 km

σ ≈ 15.92 km

Q₁ = 5.7375 km

Q₃ =  34.6675 km

Magnitude:

μ = 2.08375

M = 1.465

Range, R = 5.17

σ = 1.801485 ≈ 1.8

Q₁ = 0.5625

Q₃ = 3.925

Step-by-step explanation:

The given data are;

Depth {}                                 Magnitude

0.76 {}                                    0.84

4.93 {}                                    0.47

8.16 {}                                     0.35

33.58 {}                                  1.32

21.2 {}                                     1.61

35.03 {}                                  4.57

10.05 {}                                   5.52

47.91 {}                                    1.99

For the Depth, we have;

The mean, μ = (0.76+4.93+8.16+33.58+21.2+35.03+10.05+47.91)/8 =20.2025 km

The median, M = The (n + 1)/2th term after arranging the term in increasing order as follows;

0.76, 4.93, 8.16, 10.05, 21.2, 33.58, 35.03, 47.91 , the median is therefore;

(8 + 1)/2th term or the 4.5th term which is 10.05 + (21.2 - 10.05)/2 = 15.625 km

The Range = The highest - The lowest value = 47.91 - 0.76 = 47.15 km

The Standard deviation of, σ, is given as follows;

\sigma =\sqrt{\dfrac{\sum \left (x_i-\mu  \right )^{2} }{N}}

Where;

x_i = The individual data point = (0.76, 4.93, 8.16, 10.05, 21.2, 33.58, 35.03, 47.91 )

N = The total number of data point = 8

Substituting, (using Microsoft Excel) we get;

\sigma =\sqrt{\dfrac{\sum \left (x_i-20.2025  \right )^{2} }{8}} \approx 15.92 \ km

Q₁ = The first quartile = The (n + 1)/4th =  term arranged in increasing order

Q₁ = The (8 + 1)/4th term = The 2.25th term = 4.93 + (8.16 - 4.93)×0.25) = 5.7375 km

Q₃ = The first quartile = The 3×(n + 1)/4th =  term arranged in increasing order

Q₃ = The 3×(8 + 1)/4th term = The 6.75th term = 33.58 + 3×(35.03 - 33.58)×0.25) = 34.6675 km

For the magnitude, we have, using the same formulas and procedures as above;

μ = 2.08375

M = 1.465

Range, R = 5.17

σ = 1.801485 ≈ 1.8

Q₁ = 0.5625

Q₃ = 3.925

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100|2\\.\ 50|2\\.\ 25|5\\.\ \ 5|5\\.\ \ 1|\\\\100=2\cdot2\cdot5\cdot5\\\\126|2\\.\ 63|3\\.\ 21|3\\.\ \ 7|7\\.\ \ 1|\\\\126=2\cdot3\cdot3\cdot7
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3 years ago
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