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ivolga24 [154]
3 years ago
5

PLEASE HELPPP!!!!!! Write an equation of the line passing through point P that is perpendicular to the given line.

Mathematics
1 answer:
ArbitrLikvidat [17]3 years ago
4 0

Answer:(d): y=ax+b (a≠0); (d1)y=-4x+13

because of (d)⊥(d1) --> a.(-4)=-1 --> a=\frac{1}{4}

(d): y=\frac{1}{4}x+b. (d) pass through  P(-1;-2) so we have

-2=(1/4).(-1)+b ---> b=\frac{-7}{4}

(d) y= \frac{1}{4}x+ \frac{-7}{4}

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b) Range = (76.08, 83.92)

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d) Smaller

e) Greater

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\sigma_m=\dfrac{\sigma}{\sqrt{n}}=\dfrac{10}{\sqrt{25}}=\dfrac{10}{5}=2

where n is te sample size (n=25) and σ is the population standard deviation (σ=10).

Then, the standard error of the classroom average score is 2.

b) The calculations for this range are the same that for the confidence interval, with the difference that we know the population mean.

The population standard deviation is know and is σ=10.

The population mean is M=80.

The sample size is N=25.

The standard error of the mean is σM=2.

The z-value for a 95% confidence interval is z=1.96.

The margin of error (MOE) can be calculated as:

MOE=z\cdot \sigma_M=1.96 \cdot 2=3.92

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 80-3.92=76.08\\\\UL=M+t \cdot s_M = 80+3.92=83.92

The range that we expect the average classroom test score to fall 95% of the time is (76.08, 83.92).

c) We can calculate this by calculating the z-score of X=79.

z=\dfrac{X-\mu}{\sigma}=\dfrac{79-80}{2}=\dfrac{-1}{2}=-0.5

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P(X>79)=P(z>-0.5)=0.69146

The approximate probability that a classroom will have an average test score of 79 or higher is 0.69.

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e) If the population standard deviation is smaller, the standard error for the sample (the classroom) become smaller too. This means that the values are more concentrated around the mean (less spread). This results in a higher probability for every range that include the mean.

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