What is 4/5 out of 60

The Weeks' family goes out to eat and the total bill was $72.48. If they want to leave an 18% tip, how much extra money should t

Viktor [21]

Answer:

Step-by-step explanation:

13.05

6 0

3 years ago

The number of defects in the first five cars to come through a new production line are 9, 7, 10, 4, and 6, respectively. If the

garri49 [273]

Answer:

D. I and III only

Step-by-step explanation:

Number of first fie defects are given as 9,7,10,4 and 6.

First we have to arrange the above data in ascending order which is 4,6,7,9,10

Now if we consider the defect in sixth car to be 3 then our data will look like: 3,4,6,7,9,10

So the mean of above data would be $\frac{3+4+6+7+9+10}{6}$ = 6.5

and Median of the above data would be = $\frac{3^{rd} obs + 4^{th} obs}{2}$ = $\frac{6+7}{2}$ = 6.5

Hence mean and median number of defects are same if the sixth car has 3 defects.

Now if we consider the defect in sixth car to be 12 then our data will look like: 4,6,7,9,10,12

So the mean of above data would be $\frac{4+6+7+9+10+12}{6}$ = 8

and Median of the above data would be = $\frac{3^{rd} obs + 4^{th} obs}{2}$ = $\frac{7+9}{2}$ = 8

Hence mean and median number of defects are same if the sixth car has 12 defects.

But Now if we consider the defect in sixth car to be 7 then our data will look like: 4,6,7,7,9,10

So the mean of above data would be $\frac{4+6+7+7+9+10}{6}$ = 7.167

and Median of the above data would be = $\frac{3^{rd} obs + 4^{th} obs}{2}$ = 7

Hence mean and median number of defects are not same if the sixth car has 7 defects.

Therefore option D is correct.

8 0

3 years ago

See attachment to help, please. Only correct answers.

LekaFEV [45]

I believe the answer is B.) 120 because 170 is much wider than the CFE so it needs to be lower than that but not to much lower which is where 120 would come in.

4 0

4 years ago

An urn initially contains 5 white and 7 black balls. Each time a ball is selected, its color is noted and it is replaced in the

iragen [17]

Answer:

A. P("Select 2 black balls and then select 2 white balls")=35/768

Step-by-step explanation:

A. First, we have to get an idea of what our experiment talks about:

If I take a black ball, then we note its color and put it back to the urn with 2 more balls of the same color.

So, every time that we take a ball the quantity of balls in the urn changes with the probability that we had before replacing. The first case (that the first ball you select is black) would go like this:

P("Select a black ball first try") = 7 (number of black balls)/12 (total of balls in the urn)

P(B₁)=7/12

Then, the amount of balls in the urn rise to 14 (12 + 2 balls added) and now the quantity of black balls it´s 9 (7 + 2 added). The urn currently has 9 black balls and 5 white balls, of course:

P("Select black ball second try after getting a black ball before") = 9 (number of black balls)/ 14 (total of balls in the urn)

P(B₁B₂)=9/14

<u>Notice that while we keep taking black balls, the quantity of white ball keeps constant</u> and, for the probability that we take a white ball, we just consider the change in the number of balls in the urn

P(B₁B₂W₃)=5/16

And now, we consider the new white balls that are put in the urn (2 white balls, for a total of 18 and 7 white balls):

P(B₁B₂W₃W₄)=7/18

Because all these probabilities are chained and everything comes from the first ball we choose <u>(if you use a tree diagram, they will be in the same branch</u>), we multiply the probabilities to find the probability that all happens in the same experiment:

P(B₁)*P(B₁B₂)*P(B₁B₂W₃)*P(B₁B₂W₃W₄) = 7/12 * 9/14 * 5/16 * 7/18 = 2205/48384

P(B₁)*P(B₁B₂)*P(B₁B₂W₃)*P(B₁B₂W₃W₄)=35/768

And this is the final answer

7 0

3 years ago

1

ladessa [460]

The midpoint is (1,2)

7 0

4 years ago