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2 years ago

Solve for e 9e-7=7e-11

Mathematics

2 answers:

Tresset [83]2 years ago

5 0

Answer:

e = -2

Step-by-step explanation:

from the question we are required to solve for the value of e. to solve for e is very simple but always take note of the positive and negative sign for accuracy in your result.

9e-7=7e-11

collect the like terms

9e- 7 - 7e = -11

9e - 7e = -11 + 7

2e = -4

divide both sides by the coefficient of e which is 2

2e/2 = -4/2

e = -2

to check if your answer is correct put the value of e into the question and see if your right hand side equals your left hand side

9e-7=7e-11

9(-2) - 7 = 7(-2) - 11

-18 - 7 = -14 - 11

-25 = -25

Scilla [17]2 years ago

3 0

Answer:

e = -2

Step-by-step explanation:

9e-7=7e-11

Subtract 7e from both sides

9e-7e -7=7e -7e-11

2e -7 = -11

Add 7 to each side

2e -7+7 = -11+7

2e = -4

Divide each side by 2

2e/2 = -4/2

e =-2

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What is the greatest common factor of 16x4y3 and 12x2y7?

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2 years ago

Statistics In the manual “How to Have a Number One the Easy Way,” it is stated that a song “must be no longer than three minutes

dimaraw [331]

Answer:

We conclude that the sample is from a population of songs with a mean greater than 210 seconds.

Step-by-step explanation:

We are given that a simple random sample of 40 current hit songs results in a mean length of 252.5 seconds.

Assume that the standard deviation of song lengths is 54.5 sec.

Let $\mu$ = population mean length of the songs

So, Null Hypothesis, $H_0$ : $\mu$ $\leq$ 210 seconds {means that the sample is from a population of songs with a mean smaller than or equal to 210 seconds}

Alternate Hypothesis, $H_A$ : $\mu$ > 210 seconds {means that the sample is from a population of songs with a mean greater than 210 seconds}

The test statistics that will be used here is One-sample z-test statistics because we know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean length of songs = 252.5 seconds

$\sigma$ = population standard deviation = 54.5 seconds

n = sample of current hit songs = 40

So, the test statistics = $\frac{252.5-210}{\frac{54.5}{\sqrt{40} } }$

= 4.932

The value of z-test statistics is 4.932.

Now, at 0.05 level of significance, the z table gives a critical value of 1.645 for the right-tailed test.

Since the value of our test statistics is more than the critical value of z as 4.932 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the sample is from a population of songs with a mean greater than 210 seconds.

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