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xz_007 [3.2K]
2 years ago
12

What is the equation of the line that passes through the point (4,-6) and has a slope of -5/2?

Mathematics
2 answers:
guapka [62]2 years ago
7 0

Answer:

y = (-5/2)X + 4

Step-by-step explanation:

we know that the equation of a line is y = mx + b

m = -5/2

so

y = -5/2x + b

to find b:

-6 = -10 + b

b = 4

so

y = (-5/2)X + 4

nadezda [96]2 years ago
7 0

Answer:

Use the point-slope form of a linear equation:

y-y_1=m(x-x_1)

where:

  • m = slope
  • (x_1,y_1) = point on line

Given:

  • m=-\dfrac52
  • (x_1,y_1)=(4,-6)

Substitute given information into the equation:

\implies y-y_1=m(x-x_1)

\implies y-(-6)=-\dfrac52(x-4)

\implies y+6=-\dfrac52x+10

\implies y=-\dfrac52x+4

Therefore, the equation of the line in different formats is:

\textsf{slope-intercept form: }y=-\dfrac52x+4

\textsf{standard form: }5x+2y=8

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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
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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;

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