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1 year ago

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The resting heart rates for 80 women aged 46–55 in a simple random sample are normally distributed, with a mean of 71 beats per

minute and a standard deviation of 6 beats per minute. Assuming a 90% confidence level (90% confidence level = z-score of 1.645), what is the margin of error for the population mean?

2 answers:

Jlenok [28]1 year ago

8 0

Given:
Confidence level = 90%
mean = 71 beats per minute
standard deviation = 6 beats per minute

margin of error = z * δ / √n

where : δ - population of the standard deviation, n is the sample size ; z is the appropriate z value.

90% confidence level = 1.645 in z-value

margin of error = 1.645 * (6/√80) = 1.645 * (6/8.94) = 1.645 * 0.671 = 1.104

Valentin [98]1 year ago

7 0

Answer:

1.10/ answer B

Step-by-step explanation:

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First, note that $m\angle A+m\angle C=90^{\circ}.$ Then

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Option B is false.

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By the definition,

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$\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.$

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E.

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$\sin C=\dfrac{AB}{AC},\\ \\\cos (90^{\circ}-C)=\cos A=\dfrac{AB}{AC},\\ \\\tan A=\dfrac{BC}{AB}.$

Then

$\dfrac{\cos(90^{\circ}-C)}{\tan A}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AB}}=\dfrac{AB^2}{AC\cdot BC}\neq \sin C.$

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Answer:

The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

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To calculate the smallest possible value of y, we have to apply calculus.

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Divide through by 2

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Take square roots of both sides

$\sqrt{256=y^2$

$16=y$

$y = 16$

Recall that:

$x = \frac{128}{y}$

$x = \frac{128}{16}$

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$A" = -2 * -512y^{-3}$

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$A" = \frac{1024}{y^3}$

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This means that the calculated values are at minimum.

<em>Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches</em>

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