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Marat540 [252]
1 year ago
12

Assume that females have pulse rates that are normally distributed with a mean of p= 75.0 beats per minute and a standard deviat

ion of o= 125 beats per minutea. If 1 adult female is randomly selected, find the probability that her pulse rate is less than 82 beats per minuteThe probability is
Mathematics
1 answer:
Ivan1 year ago
4 0

We have a normally distributed population with a mean of 75 and a standard deviation of 125. The probability of finding a female with a pulse rate smaller than 82 is given by the area under the curve of the corresponding normal distribution from negative infinte to 82. In order to find this area first we need to pass from our normal distribution to one with a mean value of 0 and a standard deviation of 1. In order to do this we do the following calculation:

\frac{x-p}{o}=\frac{x-75}{125}

So we take x=82:

\frac{82-75}{125}=0.056

This value is known as the z value. It indicates the value of 82 in the [0,1] normal distribution. The following step is to look for this value in a z-table. Since 0.05 is the closest smaller value to 0.056 in the table we choose it:

So for a z value of 0.05 the table give us the number 0.5199. This means that the area under the curve between negative infinite and 82 beats per minute is 0.5199*total area under the curve. This means that the probability to find a female with a pulse under 82 beats is 0.5199. As a percentage it would be 51.99%.

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Let $M$, $N$, and $P$ be the midpoints of sides $\overline{TU}$, $\overline{US}$, and $\overline{ST}$ of triangle $STU$, respect
AysviL [449]

Answer:

<NPM + <NZM = 146°

Step-by-step explanation:

Given:

In triangle STU: M, N and P are the midpoints of the line TU, US and ST respectively.

Line UZ  is the altitude of the triangle STU

<TSU =71°, <TSU = 36°, <TUS = 73°

From the diagram:

N is the midpoint of line SU and M is the mid point of line UT.

∴ Line NM is parallel to line ST

P is the mid point of  line ST and M is the mid point of line UT

∴Line PM is parallel to line SU

N is the mid point of  line SU and P is the mid point of line ST

∴Line NP is parallel to line UT

Δ SPN = Δ STU = 36°

<SPN + <NPM + <MPT (Sum of angles in a triangle = 180°)

<UST = <MPT = 71°

36° + <NPM + 71° = 180°

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<NPM= 180°-107°

<NPM= 73°

In ΔSNZ, line SN = line NZ

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< NSZ = <NZS = 71°

<MTZ = <MZT = 36°

<NZS + <NZM <MZT = 180°  (Sum of angles in a triangle = 180°)

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<NZM = 180° - 107°

<NZM = 73°

<NPM + <NZM =73° + 73°

<NPM + <NZM = 146°

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3 years ago
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