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Ganezh [65]
3 years ago
10

Find the median first quarrile, third quartile, interquartile range, and an outliers for each set of data

Mathematics
1 answer:
Igoryamba3 years ago
3 0
The outlier is 68

The median is 102
First quartile: 87
Third Quartile: 115
The interquartile range is 87-115
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Pls help me also i like mr.bubz
Serggg [28]

you Answer is:

x  =  − 27

6 0
3 years ago
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Solve. <br> (2/3) (1/4) =
stellarik [79]

You multiply the numerators (top numbers) and separately multiply the (bottom numbers) denominators.

numerator (top number): 2 * 1 = 2

denominator (bottom number): 3 * 4 = 12

Now, we have 2/12 which can be simplified to 1/6.

<h2>1/6</h2>
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4 years ago
What is nintetty-nine hundredths as a decimal
guapka [62]
It is still 99 Hundredths as a decimal.
8 0
3 years ago
You buy a car for $25,000. It depreciates at the rate of 23% per year.
Juliette [100K]

The worth of the car in 10 years is $1,831.67 using an exponential equation approach.

What is an exponential equation?

An exponential equation is the one with exponents such X^3(the 3 is the exponent)

The exponential equation required here is the one where the future value would be lower than current value because the car reduces in value year-in-year-out.

FV=PV*(1-r)^N

FV=future worth of the car

PV=today's value=$25,000

r=depreciation rate=-23%

N=number of years=10

The fact that r is negative means the car is depreciating not appreciating.

FV=$25,000*(1-23%)^10

FV=$1,831.67

Find further explanation on exponential equation below in the link:

brainly.com/question/11832081

#SPJ1

7 0
2 years ago
In an experiment, college students were given either four quarters or a $1 bill and they could either keep the money or spend it
gavmur [86]

Answer:

a) P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341

b) P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659

c)  A. A student given a $1 bill is more likely to have kept the money.

Because the probability 0.659 is atmoslt two times greater than 0.341

Step-by-step explanation:

Assuming the following table:

                                                     Purchased Gum      Kept the Money   Total

Students Given 4 Quarters              25                              14                      39

Students Given $1 Bill                       15                               29                    44

Total                                                   40                              43                     83

a. find the probability of randomly selecting a student who spent the money, given that the student was given a $1 bill.

For this case let's define the following events

B= "student was given $1 Bill"

A="The student spent the money"

For this case we want this conditional probability:

P(A|B) =\frac{P(A and B)}{P(B)}

We have that P(A)= \frac{40}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{15}{83}

And if we replace we got:

P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341

b. find the probability of randomly selecting a student who kept the money, given that the student was given a $1 bill.

For this case let's define the following events

B= "student was given $1 Bill"

A="The student kept the money"

For this case we want this conditional probability:

P(A|B) =\frac{P(A and B)}{P(B)}

We have that P(A)= \frac{43}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{29}{83}

And if we replace we got:

P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659

c. what do the preceding results suggest?

For this case the best solution is:

A. A student given a $1 bill is more likely to have kept the money.

Because the probability 0.659 is atmoslt two times greater than 0.341

3 0
4 years ago
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