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White raven [17]
3 years ago
14

The equation M = 0.6666 log x – 3.2 represents the magnitude of an earthquake depending on the energy it produces. What does 0.6

666 represent?
xxxA.the growth factor
B. the value of M when x equals 1
C.the magnitude of the earthquake
D.the energy produced by the earthquake

It's A
Mathematics
1 answer:
lesya692 [45]3 years ago
5 0

Answer:

A

Step-by-step explanation:

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James is 10 years old right now.
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Simplify.<br><br> 2(x+5)+6x<br><br> Please answere
marissa [1.9K]

Answer:

8x+10

Step-by-step explanation:

2(x+5)+6x

Distribute

2x+10 +6x

Combine like terms

8x+10

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3 years ago
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The surface area of each face of a cube is (x
Assoli18 [71]

Answer:

length of each edge = (x + 3)

Step-by-step explanation:

The surface area of each face of the cube = (x^{2} + 6x + 9) m^{2}. But the surface of a cube has the shape of a square. So that,

Area of a square = length x length

                            = l^{2}

By factorizing the area,

x^{2} + 6x + 9 = x^{2} + 3x + 3x + 9

                  = (x^{2} + 3x) (3x + 9)

                  = x(x +3) + 3(x + 3)

                  = (x + 3)(x + 3)

                  = (x + 3)^{2}

x^{2} + 6x + 9 = (x + 3)^{2}

Thus,

(x + 3)^{2} = l^{2}

find the square root of both sides

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3 0
3 years ago
A social psychologist wants to know whether students' performance on a problem solving task is lowered by having other people lo
Orlov [11]

Answer:

Yes, there is sufficient evidence to conclude that performance on the problem solving task is lowered by having onlookers.

Step-by-step explanation:

Null hypothesis: The performance on the problem solving task is lowered by having onlookers.

Alternate hypothesis: The performance on the problem solving task is not lowered by having onlookers.

Test statistic (t) = (sample mean - population mean) ÷ (sample sd/√n)

sample mean = 89.8

population mean = 89.1

sample sd = √sample variance = √84.6 = 9.198

n = 21

Degree of freedom = n-1 = 21-1 = 20

Assuming a 5% significance level

The test is a two-tailed test. Critical values corresponding to 20 degrees of freedom and 5% significance level are -2.086 and 2.086

Test statistic (t) = (89.8 - 89.1) ÷ (9.198/√21) = 0.7 ÷ 2.007 = 0.349

Conclusion:

Fail to reject the null hypothesis because the test statistic 0.349 falls within the region bounded by the critical values.

There is sufficient evidence to conclude that performance on the problem solving task is lowered by having onlookers.

4 0
3 years ago
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