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Mrac [35]
1 year ago
13

A new movie is released each year for 12 years to go along with a popular book series. Each movie is 3 minutes longer than the l

ast to go along with a plot twist. The first movie is 65 minutes long. Use an arithmetic series formula to determine the total length of all 12 movies. 244.5 minutes
Mathematics
1 answer:
frosja888 [35]1 year ago
5 0

Answer:

S_n=978\ minutes

Step-by-step explanation:

Arithmetic sequences have the following form

a_n=a_1+d(n-1)

Where

a_1 is the first term of the series

d is the common difference between consecutive terms.

a_n is the nth term.

The formula to find the sum Sn of the first n terms of an arithmetic series is:

S_n=\frac{a_1+a_n}{2}*n

In this case n=12, d=3, a_1=65

a_{12}=65+3(12-1)

a_{12}=98

Then:

S_n=\frac{65+98}{2}*12

S_n=978\ minutes

You can also find the sum through the following series

S_n=\sum_{n=1}^{n=12}a_1+d(n-1)

S_n=\sum_{n=1}^{n=12}65+3(n-1)

S_n=\sum_{n=1}^{n=12}65+3(n-1)\\\\S_n=978

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Answer:

180

Step-by-step explanation:

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Suppose two different methods are available for eye surgery. The probability that the eye has not recovered in a month is 0.002
umka21 [38]

Answer:

0.4007

Step-by-step explanation:

Let's define the following events:

A: method A is used

B: method B is used

NR: the eye has not recovered in a month

R: the eye is recovered in a month

The probability that the eye has not recovered in a month is 0.002 if method A is used, i.e., P(NR|A) = 0.002, so P(R|A) = 0.998.

When method B is used, the probability that the eye has not recovered in a month is 0.005, i.e., P(NR|B) = 0.005, so P(R|B) = 0.995.

40% of eye surgeries are done with method A, i.e., P(A) = 0.4

60% of eye surgeries are done with method B, i.e., P(B) = 0.6

If an eye is recovered in a month after surgery is done in the hospital, what is the probability that method A was performed? We are looking for P(A|R), then, by Bayes' Formula

P(A|R) = P(R|A)P(A)/(P(R|A)P(A) + P(R|B)P(B)) = 0.998*0.4/(0.998*0.4 + 0.995*0.6) = 0.4007

4 0
2 years ago
A machine is set to fill the small-size packages of m&m candies with 56 candies per bag. a sample revealed: three bags of 56
luda_lava [24]

In statistics, the amount of degrees of freedom is the quantity of values in the final computation of a statistic that are free to differ. In this case, you can get the answer by adding the number of bags and subtracting 1.

So in computation, this would look like: 3 + 2 + 1 + 2 - 1 = 7

Therefore, 7 is the degrees of freedom.

7 0
1 year ago
What is 63.4 divided by 20
Zanzabum
317 is the answer you are looking for.
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A store sells 8 colors of balloons with at least 28 of each color. How many different combinations of 28 balloons can be chosen?
Len [333]

Answer:

(a) Selection = 6724520

(b) At\ most\ 12 = 6553976

(c) At\ most\ 8 = 6066720

(d) At\ most\ 12\ red\ and\ at\ most\ 8\ blue =  5896638

Step-by-step explanation:

Given

Colors = 8

Balloons = 28 --- at least

Solving (a): 28 combinations

From the question, we understand that; a combination of 28 is to be selected. Because the order is not important, we make use of combination.

Also, because repetition is allowed; different balloons of the same kind can be selected over and over again.

So:

n => 28 + 8-1= 35

r = 28

Selection = ^{35}^C_{28

Selection = \frac{35!}{(35 - 28)!28!}

Selection = \frac{35!}{7!28!}

Selection = \frac{35*34*33*32*31*30*29*28!}{7!28!}

Selection = \frac{35*34*33*32*31*30*29}{7!}

Selection = \frac{35*34*33*32*31*30*29}{7*6*5*4*3*2*1}

Selection = \frac{33891580800}{5040}

Selection = 6724520

Solving (b): At most 12 red balloons

First, we calculate the ways of selecting at least 13 balloons

Out of the 28 balloons, there are 15 balloons remaining (i.e. 28 - 13)

So:

n => 15 + 8 -1 = 22

r = 15

Selection of at least 13 =

At\ least\ 13 = ^{22}C_{15}

At\ least\ 13 = \frac{22!}{(22-15)!15!}

At\ least\ 13 = \frac{22!}{7!15!}

At\ least\ 13 = 170544

Ways of selecting at most 12  =

At\ most\ 12 = Total - At\ least\ 13 --- Complement rule

At\ most\ 12 = 6724520- 170544

At\ most\ 12 = 6553976

Solving (c): At most 8 blue balloons

First, we calculate the ways of selecting at least 9 balloons

Out of the 28 balloons, there are 19 balloons remaining (i.e. 28 - 9)

So:

n => 19+ 8 -1 = 26

r = 19

Selection of at least 9 =

At\ least\ 9 = ^{26}C_{19}

At\ least\ 9 = \frac{26!}{(26-19)!19!}

At\ least\ 9 = \frac{26!}{7!19!}

At\ least\ 9 = 657800

Ways of selecting at most 8  =

At\ most\ 8 = Total - At\ least\ 9 --- Complement rule

At\ most\ 8 = 6724520- 657800

At\ most\ 8 = 6066720

Solving (d): 12 red and 8 blue balloons

First, we calculate the ways for selecting 13 red balloons and 9 blue balloons

Out of the 28 balloons, there are 6 balloons remaining (i.e. 28 - 13 - 9)

So:

n =6+6-1 = 11

r = 6

Selection =

^{11}C_6 = \frac{11!}{(11-6)!6!}

^{11}C_6 = \frac{11!}{5!6!}

^{11}C_6 = 462

Using inclusion/exclusion rule of two sets:

Selection = At\ most\ 12 + At\ most\ 8 - (12\ red\ and\ 8\ blue)

Only\ 12\ red\ and\ only\ 8\ blue\ = 170544+ 657800- 462

Only\ 12\ red\ and\ only\ 8\ blue\ = 827882

At\ most\ 12\ red\ and\ at\ most\ 8\ blue = Total - Only\ 12\ red\ and\ only\ 8\ blue

At\ most\ 12\ red\ and\ at\ most\ 8\ blue =  6724520 - 827882

At\ most\ 12\ red\ and\ at\ most\ 8\ blue =  5896638

3 0
1 year ago
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