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Blababa [14]
3 years ago
13

A.) How many ways can you select 1 flavor of ice cream, 1 topping and 1 drink if there are 5 flavors of ice cream, 3 toppings an

d 4 drinks to choose from?
B.) How many ways can you select a 4-digit passcode without repeating any digit?
C.) How many ways can you select 3 toppings for a pizza with 8 toppings to choose from?
D.) What is the probability of rolling a single die and it landing on a prime number? (Remember, the number 1 is not considered prime.)
E.) How many items are in the sample space for rolling a die and spinning a spinner with 4 different colored sections (red, yellow, green and blue)?
F.) What is the probability of rolling the die and spinning the 4-sectioned spinner described in part E, and getting the number 5 and the color red?
G.) If you wanted to purchase a size T-shirt that "most" people could wear, would you purchase the mean size, the median size or the mode size?
Mathematics
1 answer:
jeka943 years ago
7 0
A) you can combine any of the 5 ice-creams with any of the 3 toppings and any of the 4 drinks.

This means a total of 5*3*4=60 ways.

B) assuming that the first digit can be 0, we have 10 choices for the 1. digit, 9 for the 2., 8 for the third and 7 choices for the fourth digit.

In total, there are 10*9*8*7=5040 passwords we can create.

C) Selection of 3 objects out of 8, can be done in

C(8, 3)=8!/(3!5!)=(8*7*6)/3!=8*7=56 many ways.

D.

P(prime)=n(prime)/n{1, 2, 3, 4, 5, 6}=n{2, 3, 5}/6=3/6=1/2=0.5

E.

any of the 6 outcomes of rolling a die, can be combined with any of the 4 colors.

for example (1, red), (1, yellow), (1, green), (1, blue).

so in total, there are 6*4=24 items in the sample space.

F.
(5, red) is just one of the 24 items in the sample space, so its probability is 1/24= 0.0417

G.

the mode, as it is the size which appears the most.


Answers:

A.60

B.5040

C.56

D. 0.5

E.24

F.1/24= 0.0417

G.mode




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Answer:

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Step-by-step explanation:

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(n + 12) × 5 + 7n

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  1. Distribute 5:                     5(n) + 5(12) + 7n
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use the general slicing method to find the volume of The solid whose base is the triangle with vertices (0 comma 0 )​, (15 comma
lyudmila [28]

Answer:

volume V of the solid

\boxed{V=\displaystyle\frac{125\pi}{12}}

Step-by-step explanation:

The situation is depicted in the picture attached

(see picture)

First, we divide the segment [0, 5] on the X-axis into n equal parts of length 5/n each

[0, 5/n], [5/n, 2(5/n)], [2(5/n), 3(5/n)],..., [(n-1)(5/n), 5]

Now, we slice our solid into n slices.  

Each slice is a quarter of cylinder 5/n thick and has a radius of  

-k(5/n) + 5  for each k = 1,2,..., n (see picture)

So the volume of each slice is  

\displaystyle\frac{\pi(-k(5/n) + 5 )^2*(5/n)}{4}

for k=1,2,..., n

We then add up the volumes of all these slices

\displaystyle\frac{\pi(-(5/n) + 5 )^2*(5/n)}{4}+\displaystyle\frac{\pi(-2(5/n) + 5 )^2*(5/n)}{4}+...+\displaystyle\frac{\pi(-n(5/n) + 5 )^2*(5/n)}{4}

Notice that the last term of the sum vanishes. After making up the expression a little, we get

\displaystyle\frac{5\pi}{4n}\left[(-(5/n)+5)^2+(-2(5/n)+5)^2+...+(-(n-1)(5/n)+5)^2\right]=\\\\\displaystyle\frac{5\pi}{4n}\displaystyle\sum_{k=1}^{n-1}(-k(5/n)+5)^2

But

\displaystyle\frac{5\pi}{4n}\displaystyle\sum_{k=1}^{n-1}(-k(5/n)+5)^2=\displaystyle\frac{5\pi}{4n}\displaystyle\sum_{k=1}^{n-1}((5/n)^2k^2-(50/n)k+25)=\\\\\displaystyle\frac{5\pi}{4n}\left((5/n)^2\displaystyle\sum_{k=1}^{n-1}k^2-(50/n)\displaystyle\sum_{k=1}^{n-1}k+25(n-1)\right)

we also know that

\displaystyle\sum_{k=1}^{n-1}k^2=\displaystyle\frac{n(n-1)(2n-1)}{6}

and

\displaystyle\sum_{k=1}^{n-1}k=\displaystyle\frac{n(n-1)}{2}

so we have, after replacing and simplifying, the sum of the slices equals

\displaystyle\frac{5\pi}{4n}\left((5/n)^2\displaystyle\sum_{k=1}^{n-1}k^2-(50/n)\displaystyle\sum_{k=1}^{n-1}k+25(n-1)\right)=\\\\=\displaystyle\frac{5\pi}{4n}\left(\displaystyle\frac{25}{n^2}.\displaystyle\frac{n(n-1)(2n-1)}{6}-\displaystyle\frac{50}{n}.\displaystyle\frac{n(n-1)}{2}+25(n-1)\right)=\\\\=\displaystyle\frac{125\pi}{24}.\displaystyle\frac{n(n-1)(2n-1)}{n^3}

Now we take the limit when n tends to infinite (the slices get thinner and thinner)

\displaystyle\frac{125\pi}{24}\displaystyle\lim_{n \rightarrow \infty}\displaystyle\frac{n(n-1)(2n-1)}{n^3}=\displaystyle\frac{125\pi}{24}\displaystyle\lim_{n \rightarrow \infty}(2-3/n+1/n^2)=\\\\=\displaystyle\frac{125\pi}{24}.2=\displaystyle\frac{125\pi}{12}

and the volume V of our solid is

\boxed{V=\displaystyle\frac{125\pi}{12}}

3 0
3 years ago
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