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kondaur [170]
3 years ago
14

Mikey has a bag that contains 8 green jelly beans, 7 red jelly beans and 2 yellow jelly beans. Which best describes the likeliho

od that Mikey will reach in without looking and select a yellow jelly bean?
A. Likely
B. Impossible
C. Certain
D. Unlikely
Mathematics
1 answer:
igor_vitrenko [27]3 years ago
3 0

Answer:

unlikely

Step-by-step explanation:

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Nani needs to buy 8 cups of fresh pineapple for a fruit cocktail. The table shows the unit prices (per cup) of pineapple at diff
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She could purchase from stores A, D and E.

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She needs 8 cups so I multiplied each price by 8;

Store A 0.49(8)=$3.92

Store B 0.51(8)=$4.08

Store C 0.55(8)=$4.40

Store D 0.48(8)=$3.84

Store E 0.45(8)=$3.60

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It has been reported that men are more likely than women to participate in online auctions. In a recent survey, 65% of repondent
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Answer:

Hence, the Female and not participated is \frac{28}{100}=0.28\%

Step-by-step explanation:

Let Total respondents =100

Participated in online action (P)=65

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Male and participated (M\; \text{and}\;P)=38

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Male and not participated is 45-38=7

Female and not participated is 55-27=28

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Irina’s basketball team has a 1 in 6 chance of winning the first-place trophy. She rolled a six-sided number cube repeatedly, as
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Suppose that 5 out of 13 people are to be chosen to go on a mission trip. In how many ways can these 5 be chosen if the order in
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5 People can be chosen  in 1287 ways if the order in which they are chosen is not important.

Step-by-step explanation:

Given:

Total number of students= 13

Number of Students to be selected= 5

To Find :

The number of ways in which the 5 people can be selected=?

Solution:

Let us use the permutation and combination to solve this problem

nCr=\frac{(n)!}{(n-r)!(r)!}

So here , n =13  and r=5 ,  

So after putting the value  of n and r , the equation will be

13C_5=\frac{(13)!}{(13-5)!(5)!}

13C_5=\frac{(13 \times12 \times11 \times10 \times9 \times8\times7 \times6 \times5 \times4 \times3 \times2 \times1)}{(8 \times7 \times6 \times5 \times4 \times3 \times2 \times1)(5 \times4 \times3 \times2 \times1)}

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13C_5= 1287

8 0
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