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3 years ago

Select the correct graph.

Mathematics

1 answer:

ASHA 777 [7]3 years ago

8 0

Answer:

The correct graph is the second one from the first row.

Step-by-step explanation:

Since his initial income was $20,000 per year, then the graph line must cross the "y" axis in that value. Since his average increase in pay on those eight years was $4,000, then for each one year step on the "x" axis the function must increase approximately "4,000" on the "y" axis. The only graph that follow this rule is the second graph on the first row. Since it starts at 20 on the "y" axis and it doubles to "40" only after 5 years, therefore increasing 4 per year.

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5k^3-8-4k^2+5k^2-2+3k^3

lara31 [8.8K]

<h3>I'll teach you how to solve 5k^3-8-4k^2+5k^2-2+3k^3</h3>

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5k^3-8-4k^2+5k^2-2+3k^3

Group like terms:

5k^3+3k^3-4k^2+5k^2-8-2

Add similar elements:

5k^3+3k^3+k^2-8-2

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8k^3+k^2-8-2

Subtract the numbers:

8k^3+k^2-10

Your Answer Is 8k^3+k^2-10

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3 years ago

The blood platelet counts of a group of women have a bell-shaped distribution with a mean of 257.62 and a standard deviation of

Effectus [21]

Answer:

(a) Approximately 68 % of women in this group have platelet counts within 1 standard deviation of the mean, or between 195.5 and 319.7.

(b) Approximately 99.7% of women in this group have platelet counts between 71.3 and 443.9.

Step-by-step explanation:

We are given that the blood platelet counts of a group of women have a bell-shaped distribution with a mean of 257.62 and a standard deviation of 62.1

Let X = <u><em>the blood platelet counts of a group of women</em></u>

So, X ~ Normal( $\mu=257.62, \sigma^{2} =62.1^{2}$ )

Now, the empirical rule states that;

68% of the data values lie within the 1 standard deviation of the mean.

95% of the data values lie within the 2 standard deviations of the mean.

99.7% of the data values lie within the 3 standard deviations of the mean.

(a) The approximate percentage of women with platelet counts within 1 standard deviation of the mean, or between 195.5 and 319.7 is 68% according to the empirical rule.

(b) The approximate percentage of women with platelet counts between 71.3 and 443.9 is given by;

z-score of 443.9 = $\frac{X-\mu}{\sigma}$

= $\frac{443.9-257.62}{62.1}$ = 3

z-score of 71.3 = $\frac{X-\mu}{\sigma}$

= $\frac{71.3-257.62}{62.1}$ = -3

So, approximately 99.7% of women in this group have platelet counts between 71.3 and 443.9.

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