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Gnoma [55]
3 years ago
11

The grades received by 200 students follow a normal distribution. The mean of the grades is 70%, and the standard deviation is 7

%. The number of students who received a grade greater than 70% is about ___, and the number of students who got a grade higher than 84% is about ___.
Mathematics
2 answers:
alexandr402 [8]3 years ago
7 0
It may help you to consider the standard normal curve.  If the mean here is 70%, then 0.5 of the area under this curve.  In other words, half the students received a grade > 70%.  That would be 100 students.

Note that 84 is 2 std dev above the mean.

Recall the Empirical Rule:  95% of scores lie within 2 std. dev of the mean.
Measuring from the mean, that would be 47.5% above the mean and 47.5% below the mean.  The area to the right of the mean is 0.5.  Subtract 0.475 from that to obtain the fraction of students who rec'd a grade greater than 2 std dev from the mean:  It's 0.025.

Then the # of students who rec'd a grade greater than 84% would be
0.025(200 students) = 5 students.
ICE Princess25 [194]3 years ago
3 0
The correct is 100 and 0
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Please find the exact length of the midsegment of trapezoid JKLM with vertices J(6, 10), K(10, 6), L(8, 2), and M(2, 2). Thank y
I am Lyosha [343]

Answer:

the exact length of the midsegment of trapezoid JKLM  = \mathbf{ = 3 \sqrt{5} } i.e 6.708 units on the graph

Step-by-step explanation:

From the diagram attached below; we can see a graphical representation showing the mid-segment of the trapezoid JKLM. The mid-segment is located at the line parallel to the sides of the trapezoid. However; these mid-segments are X and Y found on the line JK and LM respectively from the graph.

Using the expression for midpoints between two points to determine the exact length of the mid-segment ; we have:

\mathbf{ YX = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} }

\mathbf{ YX = \sqrt{(8-5)^2+(8-2)^2} }

\mathbf{ YX = \sqrt{(3)^2+(6)^2} }

\mathbf{ YX = \sqrt{9+36} }

\mathbf{ YX = \sqrt{45} }

\mathbf{ YX = \sqrt{9*5} }

\mathbf{ YX = 3 \sqrt{5} }

Thus; the exact length of the midsegment of trapezoid JKLM  = \mathbf{ = 3 \sqrt{5} } i.e 6.708 units on the graph

8 0
3 years ago
Find the area of quadrilateral ABCD
andreev551 [17]

Answer:

<em>A ≈ 28.5</em>

Step-by-step explanation:

a, b, c

P = a + b + c

Semiperimeter s = \frac{a+b+c}{2}

A = \sqrt{s(s-a)(s-b)(s-c)}

~~~~~~~~~~~~~~~

P_{ABC} = 4.3 + 2.89 + 6.81 = 14

s = 14 ÷ 2 = 7

A_{ABC} = \sqrt{7(7-4.3)(7-2.89)(7-6.81)} = √14.75901 ≈ 3.84

P_{BCD} = 8.59 + 7.58 + 6.81 = 22.98

s = 22.98 ÷ 2 = 11.49

A_{BCD} = \sqrt{11.49(11.49-8.59)(11.49-7.58)(11.49-6.81)} = √609.7343148 ≈ 24.6928

A_{ABCD} = 3.84 + 24.6928 ≈ <em>28.5</em>

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Answer:

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Step-by-step explanation:

1. 0.00000402

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