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ch4aika [34]
3 years ago
14

Order the following numbers written in scientific notation from greatest to least: 3.9 × 105, 6.1 × 103, 9.3 × 105, 1.6 × 103 A.

9.3 × 105, 3.9 × 105, 6.1 × 103, 1.6 × 103 B. 1.6 × 103, 6.1 × 103, 9.3 × 105, 3.9 × 105, C. 3.9 × 105, 9.3 × 105, 1.6 × 103, 6.1 × 103,
Mathematics
2 answers:
Licemer1 [7]3 years ago
5 0

The numbers are:

3.9 × 105 = 409.5

6.1 ×103 = 628.3

9.3 × 105 = 976.5

1.6 × 103 = 164.8

Arranging from greatest to least and in scientific notation form gives:

9.3 × 105 , 6.1 × 103 , 3.9 × 105 , 1.6 × 103

*(Seems like there is no correct answer from the choices given there).

Irina-Kira [14]3 years ago
5 0

Answer:

3.9 × 105 = 409.5

6.1 ×103 = 628.3

9.3 × 105 = 976.5

1.6 × 103 = 164.8

Step-by-step explanation:

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99.5% of all samples of 21 of​ today's women in their 20's have mean heights of at least 65.86 ​inches.

Step-by-step explanation:

We are given that a half-century ago, the mean height of women in a particular country in their 20's was 64.7 inches. Assume that the heights of​ today's women in their 20's are approximately normally distributed with a standard deviation of 2.07 inches.

Also, a samples of 21 of​ today's women in their 20's have been taken.

<u><em /></u>

<u><em>Let </em></u>\bar X<u><em> = sample mean heights</em></u>

The z-score probability distribution for sample mean is given by;

                          Z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean height of women = 64.7 inches

            \sigma = standard deviation = 2.07 inches

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the sample of 21 of​ today's women in their 20's have mean heights of at least 65.86 ​inches is given by = P(\bar X \geq 65.86 inches)

  P(\bar X \geq 65.86 inches) = P( \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } } \geq \frac{65.86-64.7}{\frac{2.07}{\sqrt{21} } } ) = P(Z \geq -2.57) = P(Z \leq 2.57)

                                                                        = <u>0.99492  or  99.5%</u>

<em>The above probability is calculated by looking at the value of x = 2.57 in the z table which has an area of 0.99492.</em>

<em />

Therefore, 99.5% of all samples of 21 of​ today's women in their 20's have mean heights of at least 65.86 ​inches.

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