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NeTakaya
2 years ago
7

What is the probability that a random sample of second grade students from the city results in a mean reading rate of more than

words per minute?
Mathematics
1 answer:
-BARSIC- [3]2 years ago
6 0

The probability that a random sample of 28 second grade students from the city results in a mean reading rate of more than 96 wpm is 0.01715.

<h3>What is normal distribution?</h3>

A normal distribution is a probability distribution that has been symmetrical around this mean, with most observations clustering around the central peak and probabilities tapering off evenly in both directions.

Now, according to the question,

The formula for normal distribution is;

z = (x - μ)/(σ/√s)

x is the sample size (= 96)

μ = standard mean (= 92)

σ = standard deviation (= 10)

s = random sample (= 28)

Substituting all the values in the equation;

z = (96 - 92)/(10/√28)

z = 2.117

Now, calculate the probability;

P(X > 96) = 1 - P(X ≤ 96)

               = 1 - P(X ≤ 2.117)

               = 1 - 0.98285

P(X > 96) =  0.01715

Therefore, the probability that a random sample of 28 second grade students from the city results in a mean reading rate of more than 96 wpm is 0.01715.

To know more about the normal distribution, here

brainly.com/question/4079902

#SPJ4

The complete question is-

The reading speed of a second grade students in a large city is approximately normal with a mean of 92 words per minute and a standard deviation of 10 wpm. What is the probability that a random sample of 28 second grade students from the city results in a mean reading rate of more than 96 wpm?

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What are the coordinates of the orthocenter of △JKL with vertices at J(−4, −1) , K(−4, 8) , and L(2, 8) ?
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The orthocenter is (-4,8)


<u>EXPLANATION</u>


The orthocenter is  the point of intersection of all the altitudes of a triangle.

If it is a right angle triangle then the orthocenter is the point at which the <em>right angle is made.</em>

If it is not a right angle triangle, then you have to find the equations of any two of the altitudes and solve simultaneously.

So you must first check to see if it is a right angle triangle that way you do not have to do bunch work.


The given triangle has vertices J(-4,-1), (-4,8), and L(2,8).


Slope_{JK}=\frac{8--1}{-4--4}


Slope_{JK}=\frac{8+1}{-4+4}


Slope_{JK}=\frac{9}{0}


Since the denominator is zero, the slope is undefined.

If the slope of a straight line is undefined, then that line is <em>parallel to the y-axis</em>. In other words, it is a <em>vertical line</em>.


Slope_{JL}=\frac{8--1}{2--4}


Slope_{JL}=\frac{8+1}{2+4}


Slope_{JL}=\frac{9}{6}


Slope_{JL}=\frac{3}{2}



Slope_{KL}=\frac{8-8}{2--4}


Slope_{KL}=\frac{0}{2+4}


Slope_{KL}=\frac{0}{6}


Slope_{KL}=0


if a straight line has a slope of zero, that line is parallel to the <em>x-axis</em> or it is <em>horizontal.</em>


From the slopes we can see that side KL of the triangle is horizontal and side JK is vertical. This two sides will meet at right angles at K.

See graph.

Therefore the coordinates of K, (-4,8) is the orthocenter of the given triangle.







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