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zepelin [54]
3 years ago
6

In an accounting class of 200 students, the mean and standard deviation of scores was 70 and 5, respectively. Use the empirical

rule to determine the number of students who scored less than 65 or more than 75?
a. About 64
b. About 136
c. About 32
d. About 68
Mathematics
1 answer:
lina2011 [118]3 years ago
5 0

Answer:

a. About 64

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 70

Standard deviation = 5

Use the empirical rule to determine the number of students who scored less than 65 or more than 75?

65 = 70 - 5

So 65 is one standard deviation below the mean

75 = 70 + 5

So 75 is one standard deviation above the mean.

By the Empirical Rule, 68% of the students scored between 65 and 75. The other 100-68 = 32% scored less than 65 or more than 75.

Out of 200, that is

0.32*200 = 64

So the correct answer is:

a. About 64

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S
zzz [600]

Answer:

x = 4

Step-by-step explanation:

A trapezium is a quadrilateral (has four sides and four angles) with one pair of parallel sides. A trapezium is said to be isosceles if the bases are parallel, the two other sides known as the legs are equal and diagonals of the trapezium are equal to each other.

In trapezoid ABCD, BC and AD are the diagonals.

Hence BC = AD (property of isosceles trapezoid)

9x + 1 = 12x - 11

Simplifying:

12x - 9x = 1 + 11

3x = 12

Dividing both sides by 2:

x = 4

8 0
3 years ago
Help me please, I don't really get this at all.
Varvara68 [4.7K]

Answer:

b.

Step-by-step explanation:

Seats to minutes = 2 to 11:

2/11 = s/m.

When  s = 16:

2/11 = 16/m

m = 11*16 / 2

= 176/2

= 88  minutes.

When s =  19:

2/11 = 19 / m

m = 11*19 / 2

m  = 209/2

= 104.5 minutes.

6 0
4 years ago
Read 2 more answers
Suppose X, Y, and Z are random variables with the joint density function f(x, y, z) = Ce−(0.5x + 0.2y + 0.1z) if x ≥ 0, y ≥ 0, z
kompoz [17]

a.

f_{X,Y,Z}(x,y,z)=\begin{cases}Ce^{-(0.5x+0.2y+0.1z)}&\text{for }x\ge0,y\ge0,z\ge0\\0&\text{otherwise}\end{cases}

is a proper joint density function if, over its support, f is non-negative and the integral of f is 1. The first condition is easily met as long as C\ge0. To meet the second condition, we require

\displaystyle\int_0^\infty\int_0^\infty\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz=100C=1\implies \boxed{C=0.01}

b. Find the marginal joint density of X and Y by integrating the joint density with respect to z:

f_{X,Y}(x,y)=\displaystyle\int_0^\infty f_{X,Y,Z}(x,y,z)\,\mathrm dz=0.01e^{-(0.5x+0.2y)}\int_0^\infty e^{-0.1z}\,\mathrm dz

\implies f_{X,Y}(x,y)=\begin{cases}0.1e^{-(0.5x+0.2y)}&\text{for }x\ge0,y\ge0\\0&\text{otherwise}\end{cases}

Then

\displaystyle P(X\le1.375,Y\le1.5)=\int_0^{1.5}\int_0^{1.375}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy

\approx\boxed{0.12886}

c. This probability can be found by simply integrating the joint density:

\displaystyle P(X\le1.375,Y\le1.5,Z\le1)=\int_0^1\int_0^{1.5}\int_0^{1.375}f_{X,Y,Z}(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz

\approx\boxed{0.012262}

7 0
3 years ago
Can you help plz!!!!
Zolol [24]
Okay so the table represents B, in one hour of work, the person will make 18.25 (you get this by dividing 36.50 by two because that is how much they made in two hours, we need to know how much they make in one)
Therefore, the equation for A, which has to be less than B, would be either be Y=18x or Y=18.1x
8 0
3 years ago
HELP ASAP!!<br> What is the unit rate for the situation in terms of centimeter/minute?
vesna_86 [32]
The unit rate is 5cm per 1min

5:1
4 0
3 years ago
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