Will give Brainliest if you answer all

Over the first five years of owning her car, Gina drove about 12,200 miles the first year, 16,211 miles the second year, 12,050

attashe74 [19]

The answer is mean = 13,027; median = 12,200; no mode

Let's rearrange values from the lowest to the highest:
11350, 12050, 12200, 13325, 16211

The mean is the sum of all values divided by the number of values:
(11350 + 12050 + 12200 + 13325 + 16211)/5 ≈ 13027

The median is the middle value. If there is an odd number of data, then the median is the value in the middle. In the data set 11350, 12050, 12200, 13325, 16211, the median (the middle value) is 12200

The mode is the value that occurs most frequently. Since none of the number does not occur most frequently, there is no mode.

7 0

3 years ago

Read 2 more answers

Whats 2 + 3 + 5 + 6 please help

Ratling [72]

Answer:

16

Step-by-step explanation:

2+3=5+5=10+6=16 i cant tell if this is a joke but oh well ‍♀️

4 0

3 years ago

Read 2 more answers

Let X denote the distance (m) that an animal moves from its birth site to the first territorial vacancy it encounters. Suppose t

worty [1.4K]

Answer:

a) $P(X \leq 100) = 1- e^{-0.01342*100} =0.7387$

$P(X \leq 200) = 1- e^{-0.01342*200} =0.9317$

$P(100\leq X \leq 200) = [1- e^{-0.01342*200}]-[1- e^{-0.01342*100}] =0.1930$

b) $P(X>223.547) = 1-P(X\leq 223.547) = 1-[1- e^{-0.01342*223.547}]=0.0498$

c) $m = \frac{ln(0.5)}{-0.01342}=51.65$

d) $a = \frac{ln(0.05)}{-0.01342}=223.23$

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

$P(X=x)=\lambda e^{-\lambda x}$

Solution to the problem

For this case we have that X is represented by the following distribution:

$X\sim Exp (\lambda=0.01342)$

Is important to remember that th cumulative distribution for X is given by:

$F(X) =P(X \leq x) = 1-e^{-\lambda x}$

Part a

For this case we want this probability:

$P(X \leq 100)$

And using the cumulative distribution function we have this:

$P(X \leq 100) = 1- e^{-0.01342*100} =0.7387$

$P(X \leq 200) = 1- e^{-0.01342*200} =0.9317$

$P(100\leq X \leq 200) = [1- e^{-0.01342*200}]-[1- e^{-0.01342*100}] =0.1930$

Part b

Since we want the probability that the man exceeds the mean by more than 2 deviations

For this case the mean is given by:

$\mu = \frac{1}{\lambda}=\frac{1}{0.01342}= 74.516$

And by properties the deviation is the same value $\sigma = 74.516$

So then 2 deviations correspond to 2*74.516=149.03

And the want this probability:

$P(X > 74.516+149.03) = P(X>223.547)$

And we can find this probability using the complement rule:

$P(X>223.547) = 1-P(X\leq 223.547) = 1-[1- e^{-0.01342*223.547}]=0.0498$

Part c

For the median we need to find a value of m such that:

$P(X \leq m) = 0.5$

If we use the cumulative distribution function we got:

$1-e^{-0.01342 m} =0.5$

And if we solve for m we got this:

$0.5 = e^{-0.01342 m}$

If we apply natural log on both sides we got:

$ln(0.5) = -0.01342 m$

$m = \frac{ln(0.5)}{-0.01342}=51.65$

Part d

For this case we have this equation:

$P(X\leq a) = 0.95$

If we apply the cumulative distribution function we got:

$1-e^{-0.01342*a} =0.95$

If w solve for a we can do this:

$0.05= e^{-0.01342 a}$

Using natural log on btoh sides we got:

$ln(0.05) = -0.01342 a$

$a = \frac{ln(0.05)}{-0.01342}=223.23$

5 0

3 years ago

Look at the pic and answer

lapo4ka [179]

Answer:D

Step-by-step explanation:

5 0

3 years ago

What two consecutive odd integers have a sum of 88?

Sindrei [870]

Answer:

What two consecutive odd integers have a sum of 88? x + x + 2 = 88 • 2x + 2 = 88 • 2x = 86 • x = 43, x + 2 = 45 • 43 + 45 = 88 • Ans: 43, 45 • They equal 88 and are both odd.

3 0

3 years ago