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Ratling [72]
3 years ago
14

Solve. 7 x 3/12 plz answer. and tell me, is it 21/12 ?

Mathematics
1 answer:
finlep [7]3 years ago
4 0
7 *  \frac{3}{12} = \frac{7*3}{12}= \frac{21}{12}
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The ages of the winners of a cycling tournament are approximately bell-shaped. The mean age is 27.2 years, with a standard devia
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Answer:

a. z-score = -1.2

b. The age 23 years old is -1.2 standard deviations below the mean

c. The age is not unusual

Step-by-step explanation:

a. Transformation to z-score

Mathematically;

z-score = (x - μ)/σ

where σ and μ are the standard deviation and the mean respectively.

From the question, x = 23, μ = 27.2 and σ = 3.5

Let’s put these into the equation;

z-score = (23-27.2)/3.5 = -1.2

b. Interpretation of result

The interpretation is that 23 years old is -1.2 standard deviations below the mean

c. Determination

The age is not unusual. This is because any z-value below -2 or above 2 is considered unusual

-1.2 is above -2 and thus it is not unusual

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8 0
3 years ago
PLEASE HELP ME MATH URGENT
slega [8]
S answer is no for the question
7 0
2 years ago
It is generally claimed that the average body temperature for healthy human adults is 98.6°F. To test this claim a simple random
Rzqust [24]

Answer:

Explained below.

Step-by-step explanation:

The information provided is as follows:

\mu=98.6^{o}F\\\bar x=98.1^{o}F\\s=0.9^{o}F\\n=9

(1)

A single mean test is to be performed in this case.

As the population standard deviation is not provided, a one-sample <em>t</em>-test will be used.

The correct option is b.

(2)

The null hypothesis is:

<em>H</em>₀: The average temperature in the population is 98.6°F, i.e. <em>μ </em>= 98.6°F.

The correct option is b.

(3)

The alternative hypothesis is:

<em>Hₐ</em>: The average temperature in the population is less than 98.6°F, i.e. <em>μ </em>< 98.6°F.

The correct option is c.

(4)

The standard deviation of the sample mean is as follows:

SD_{\bar x}=\frac{s}{\sqrt{n}}=\frac{0.9}{\sqrt{9}}=0.3^{o}F

Thus, the value of SD is 0.3°F.

(5)

Compute the value of test statistic as follows:

t=\frac{\bar x-\mu}{SD_{\bar x}}

  =\frac{98.1-98.6}{0.3}\\\\=-1.666666667\\\\\approx -1.67

Thus, the value of test statistic is -1.67.

(6)

The degrees of freedom of the test are:

df = n - 1

   = 9 - 1

   = 8

Thus, the degrees of freedom of the test is 8.

8 0
3 years ago
5. Joe and Mike went video game shopping and it cost $69.00. Mike had a
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Answer:

it would be 20 I'm not 100 percent tho

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