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Daniel [21]
3 years ago
15

The distribution of the amount of money spent by students for textbooks in a semester is approximately normal in shape with a me

an of $235 and a standard deviation of $20. According to the standard deviation rule, almost 2.5% of the students spent more than what amount of money on textbooks in a semester?
Mathematics
1 answer:
faust18 [17]3 years ago
4 0

Answer:

Almost 2.5% of the students spent more than $275 in a semester.

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 = 235

Standard deviation = 20

According to the standard deviation rule, almost 2.5% of the students spent more than what amount of money on textbooks in a semester?

95% of the measures are within 2 standard deviation of the mean. The other 5% are more than 2 standard deviations of the mean. Since the normal distribution is symmetric, 2.5% of those are below two standard deviations of the mean and 2.5% are more than two standard deviations above the mean.

235 + 2*20 = $275

Almost 2.5% of the students spent more than $275 in a semester.

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3 years ago
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Answer:

x = 7 m and x = −7 m

Step-by-step explanation:

Its a modulus problem

concept

|x|  = x when x>=0

|x| = -x when x < 0

____________________________________

Now given

|x| − 2 = 5

adding 2 both sides

|x| − 2  + 2 = 5 + 2

|x|  = 7

now

x = 7  when x >= 0

x = -7 when x<0

Thus, correct answer is x = 7 m and x = −7 m

7 0
3 years ago
A fitness club has two options, one for members and one for nonmembers. Members pay a one-time registration fee of $12 plus $8 p
Liula [17]

Answer:

See below.

Step-by-step explanation:

Let's look at the cost for members (C1) first.  Let x be the number of visits.

C1(x) = 12 + 8x

For non-members (C2), we can do the same.

C2(x) = 10x

You can graph these two equations.

x        C1         C2

0       12          0

1        20        10

2       28        20

3       36        30

4       44        40

5       52        50

6       60        60

7       68        70

Let's make the two equations equal, to find out where the benefit is the same.

12 + 8x = 10x

2x = 12

x = 6

Up to 5 visits, the non-member cost is better.  At 6 visits, there's the same price.  For more than 6 visits, the member cost is better.

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3 years ago
Solve The Equation <br> 4x×9y=7<br> 4x-9y=9
hoa [83]

Answer:

\large\boxed{x=\dfrac{9}{8}-\dfrac{\sqrt{109}}{8},\ y=-\dfrac{1}{2}-\dfrac{\sqrt{109}}{18}}\\or\\\boxed{x=\dfrac{9}{8}+\dfrac{\sqrt{109}}{2},\ y=-\dfrac{1}{2}+\dfrac{\sqrt{109}}{18}}

Step-by-step explanation:

\left\{\begin{array}{ccc}4x\times9y=7&(1)\\4x-9y=9&(2)\end{array}\right\\\\(2)\\4x-9y=9\qquad\text{subtract}\ 4x\ \text{from both sides}\\-9y=-4x+9\qquad\text{change the signs}\\9y=4x-9\qquad\text{substitute it to (1)}\\\\4x(4x-9)=7\qquad\text{use the distributive property}\ a(b+c)=ab+ac\\(4x)(4x)+(4x)(-9)=7\\(4x)^2-36x=7\\(4x)^2-2(4x)(4.5)=7\qquad\text{add}\ 4.5^2\ \text{to both sides}\\(4x)^2-2(4x)(4.5)+4.5^2=7+4.5^2\qquad\text{use}\ (a-b)^2=a^2-2ab+b^2

(4x-4.5)^2=7+20.25\\(4x-4.5)=27.25\to 4x-4.5=\pm\sqrt{27.25}\\\\4x-\dfrac{45}{10}=\pm\sqrt{\dfrac{2725}{100}}\\\\4x-\dfrac{45}{10}=\pm\dfrac{\sqrt{2725}}{\sqrt{100}}\\\\4x-\dfrac{45}{10}=\pm\dfrac{\sqrt{25\cdot109}}{10}\\\\4x-\dfrac{45}{10}=\pm\dfrac{\sqrt{25}\cdot\sqrt{109}}{10}\\\\4x-\dfrac{45}{10}=\pm\dfrac{5\sqrt{109}}{10}\qquad\text{add}\ \dfrac{45}{10}\ \text{to both sides}\\\\4x=\dfrac{45}{10}\pm\dfrac{5\sqrt{109}}{10}

4x=\dfrac{9}{2}\pm\dfrac{\sqrt{109}}{2}\qquad\text{divide both sides by 4}\\\\x=\dfrac{9}{8}\pm\dfrac{\sqrt{109}}{8}\\\\\text{Put the values of}\ x\ \text{to (2):}\\\\9y=4\left(\dfrac{9}{8}\pm\dfrac{\sqrt{109}}{8}\right)-9\\\\9y=\dfrac{9}{2}\pm\dfrac{\sqrt{109}}{2}-\dfrac{18}{2}\\\\9y=-\dfrac{9}{2}\pm\dfrac{\sqrt{109}}{2}\qquad\text{divide both sides by 9}\\\\y=-\dfrac{1}{2}\pm\dfrac{\sqrt{109}}{18}

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