<h2>
Hello!</h2>
The answer is:
The third option:
2.7 times as much.
<h2>
Why?</h2>
To calculate how many more juice will the new can hold, we need to calculate the old can volume to the new can volume.
So, calculating we have:
Old can:
Since the cans have a right cylinder shape, we can calculate their volume using the following formula:

Where,

We are given the old can dimensions:

So, calculating the volume, we have:

We have that the volume of the old can is:

New can:
We are given the new can dimensions, the diameter is increased but the height is the same, so:

Calculating we have:

Now, dividing the volume of the new can by the old can volume to know how many times more juice will the new can hold, we have:

Hence, we have that the new can hold 2.7 more juice than the old can, so, the answer is the third option:
2.7 times as much.
Have a nice day!
95x = total distance traveled
x represents time
Answer:
0.0157
Step-by-step explanation:
From the information given:
The sample size = 70
The expected no. of days of year that are birthday of exactly 4 people is:![P = \bigg [ \dfrac{1}{365} \bigg]^4](https://tex.z-dn.net/?f=P%20%3D%20%5Cbigg%20%5B%20%5Cdfrac%7B1%7D%7B365%7D%20%5Cbigg%5D%5E4)
The expected number of days with 4 birthdays = 
![\sum \limits ^{365}_{i=1} E(x_i) = 365 \times \bigg[ \ ^{70}C_{4} \times ( \dfrac{1}{365})^4 ( 1 - \dfrac{1}{365})^{70-4} \bigg]](https://tex.z-dn.net/?f=%5Csum%20%5Climits%20%5E%7B365%7D_%7Bi%3D1%7D%20%20E%28x_i%29%20%3D%20365%20%5Ctimes%20%5Cbigg%5B%20%20%5C%20%5E%7B70%7DC_%7B4%7D%20%5Ctimes%20%28%20%5Cdfrac%7B1%7D%7B365%7D%29%5E4%20%28%201%20-%20%5Cdfrac%7B1%7D%7B365%7D%29%5E%7B70-4%7D%20%5Cbigg%5D)
![\sum \limits ^{365}_{i=1} E(x_i) = 365 \times \bigg[ \ \dfrac{70!}{4!(70-4)!} \times ( \dfrac{1}{365})^4 ( 1 - \dfrac{1}{365})^{66} \bigg]](https://tex.z-dn.net/?f=%5Csum%20%5Climits%20%5E%7B365%7D_%7Bi%3D1%7D%20%20E%28x_i%29%20%3D%20365%20%5Ctimes%20%5Cbigg%5B%20%20%5C%20%5Cdfrac%7B70%21%7D%7B4%21%2870-4%29%21%7D%20%5Ctimes%20%28%20%5Cdfrac%7B1%7D%7B365%7D%29%5E4%20%28%201%20-%20%5Cdfrac%7B1%7D%7B365%7D%29%5E%7B66%7D%20%5Cbigg%5D)
![\sum \limits ^{365}_{i=1} E(x_i) = 365 \times \bigg[ \ 916895 \times 5.6342 \times 10^{-11} \times 0.8343768898 \bigg]](https://tex.z-dn.net/?f=%5Csum%20%5Climits%20%5E%7B365%7D_%7Bi%3D1%7D%20%20E%28x_i%29%20%3D%20365%20%5Ctimes%20%5Cbigg%5B%20%20%5C%20916895%20%5Ctimes%205.6342%20%5Ctimes%2010%5E%7B-11%7D%20%5Ctimes%200.8343768898%20%5Cbigg%5D)
= 0.0157
Therefore, the required probability = 0.0157