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nirvana33 [79]
4 years ago
12

The scatterplot shows the ages and values of cars that residents in a neighborhood own. A graph titled Values of Cars has age of

car (years) on the x-axis and Value of car (thousands of dollars) on the y-axis. Points are grouped together and decrease. Point (10, 15) is above and to the right of the cluster. Which statement about the scatterplot is true? The point (10, 15) shows there is no relationship between the age of a car and the value of a car. By excluding (10, 15), a better description can be given for the data set. By including (10, 15), the description of the data set can be understated. Although (10, 15) is an extreme value, it should be part of the description of the relationship between the age of a car and the value of the car.
Mathematics
2 answers:
kiruha [24]4 years ago
8 0

Answer:

C.By including (10, 15), the description of the data set can be understated.

inysia [295]4 years ago
5 0

Answer:C

Step-by-step explanation:

Just c

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In 2014, the enrollment at Luling High School was 625 students. In 2015, the enrollment grew to 650. By what percentage did the
Inessa05 [86]

Answer:

\%Increase = 4\%

Step-by-step explanation:

Given

Initial\ Population = 625

Final\ Population = 650

Required

Determine the percentage population increase

This is calculated using the following formula:

\%Increase = \frac{Final - Initial}{Initial} * 100\%

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4 0
3 years ago
) Determine the probability that a bit string of length 10 contains exactly 4 or 5 ones.
yanalaym [24]

Answer: 0.4512

Step-by-step explanation:

A bit string is sequence of bits (it only contains 0 and 1).

We assume that the  0 and 1 area equally likely to any place.

i.e. P(0)= P(1)= \dfrac{1}{2}

The length of bits : n = 10

Let X = Number of getting ones.

Then , X \sim Bin(n=10,\ p=\dfrac{1}{2})

Binomial distribution formula : P(X=x)=^nC_x p^x q^{n-x} , where p= probability of getting success in each event and q= probability of getting failure in each event.

Here , p=q=\dfrac{1}{2}

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=\dfrac{10!}{4!6!}(\dfrac{1}{2})^{10}+\dfrac{10!}{5!5!}(\dfrac{1}{2})^{10}

=(\dfrac{1}{2})^{10}(\dfrac{10!}{4!6!}+\dfrac{10!}{5!5!})

=(\dfrac{1}{2})^{10}(210+252)

=(0.0009765625)(462)

=0.451171875\approx0.4512

Hence, the  probability that a bit string of length 10 contains exactly 4 or 5 ones is 0.4512.

3 0
4 years ago
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