<h3>
Answer: 20 </h3>
=====================================================
Reasoning:
The empirical rule says that about 68% of the population is within 1 standard deviation of the mean. This applies to the normal distribution.
The mean in this case is 72 and the standard deviation is 7.
One standard deviation below the mean is 72-7 = 65, while one standard deviation above the center point is 72+7 = 79. This is visually shown as the closest tickmarks to the center 72. Each tickmark represents a unit of standard deviation (eg: 85 is two standard deviations above the mean, since it's 2 tickmarks above 72).
Saying "within 7 points of the mean" therefore is the same as saying "the score is between 65 and 79, inclusive of both endpoints".
So you're teacher is asking how many students got between 65 and 79. As mentioned at the very top, roughly 68% of the population fits this description.
Meaning,
68% of 30 = 0.68*30 = 20.4 = 20 students got a quiz score within seven points of the mean.
The length of the image of the segment CD is 10 units if the segment CD has endpoints at C(0, 3) and D(0, 8).
<h3>What is a distance formula?</h3>
It is defined as the formula for finding the distance between two points. It has given the shortest path distance between two points.
The distance formula can be given as:

We have two points C(0, 3) and D(0, 8)
The distance between these two points:
d = √(8-3)²
d = 5 units
The length of the image of segments CD = dilation factor×length of CD
= 2×5
= 10 units
Thus, the length of the image of the segment CD is 10 units if the segment CD has endpoints at C(0, 3) and D(0, 8).
Learn more about the distance formula here:
brainly.com/question/18296211
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Answer:
1.061
Step-by-step explanation:
I'm pretty sure anyway.
The centre of this circle is at 12
1
2
and has radius 12
1
2
, so the equation is (−12)2+2=12
(
x
−
1
2
)
2
+
y
2
=
1
2
.
Answer:
The expected number of experiment is 198
Step-by-step explanation:
Solution
Given that:
We need to carry out a test
where,
H0 : p =0.5
and
H1 : p ≠ 0.5
n = the number of flip coin which is = 200
x = this is the number of heads declared = 106
So,
xₙ = x/n = 106/200 = 0.53 = p
Thus,
D₂ =√n (xₙ - 0.5)/√0.5 * (1-0.5)
=√200 * (0.53 - 0.5)/ √0.5 * (1-0.5)
= 0.848528137
D₂ = 0.8485
Now,
p ( z> D₂ ) = p ( z > 0.8485)
=0.198072
Thus,
By applying R,
1 - pnorm (0.8485, 0,1)
That is (1- pnorm (D₂, 0, 1)
Hence,
p ( z> D₂ )≈ 0.198072
So,
We find The expected number of experiment such that the estimator √n (xₙ - 0.5)/√0.5 * (1-0.5)i s larger than the value D₂ when the total is 1000 times or attained in the first experiment Thus
1000 * p ( z> D₂ )
= 1000 * 0.198072
=198.072
=198
Note: Kindly find an attached copy of the complete question below