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morpeh [17]
3 years ago
10

Lex read 7 fewer books than Leona how many books did Leona read

Mathematics
1 answer:
postnew [5]3 years ago
8 0

Answer:

14

Lex read 7 but Lena had to have 7 books which yea

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Solve the inequality<br>w/2&lt;-42​
exis [7]

Answer:

w<-84

Step-by-step explanation:

solved the equation

4 0
3 years ago
There are 20 nickels in one dollar. Write an equation to find the number of nickels n in any number of dollars d
n200080 [17]

5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5

20 nickles

20•5=1.00

4 0
3 years ago
Evaluate each expression if a = 5, b = 6, C = 4, and d = 3.<br><br> a. 2ab + cd<br> b. abc
PolarNik [594]

Answer:

a(2ab+cd

(2×5×6)+(4×3)

60+12

72

b)abc

5×6×4

120

8 0
3 years ago
How do I solve 34-5x+2(x-2)=15
LenKa [72]
34-5x+2(x-2)=15      1st get rid of () by distributing the 2.
34-5x+2x-4=15         Now collect like terms.
30-3x=15                  Get -3x by itself by subtracting 30 in both sides.
-3x=-15                     You want X by itself so divide by -3 on both sides.
x=5
8 0
3 years ago
If every student in a large Statistics class selects peanut M&amp;M’s at random until they get a red candy, on average how many
kotykmax [81]

Question:

According to Masterfoods, the company that manufactures M&M's, 12% of peanut M&M's are brown, 15% are yellow, 12% are red, 23% are blue, 23% are orange and 15% are green. You randomly select peanut M&M's from an extra-large bag looking for a red candy.

If every student in a large Statistics class selects peanut M&M’s at random until they get a red candy, on average how many M&M’s will the students need to select? (Round your answer to two decimal places.).

Answer:

If every student in the large Statistics class selects peanut M&M’s at random until they get a red candy the expected value of the number of  M&M’s the students need to select is 8.33 M&M's.

Step-by-step explanation:

To solve the question, we note that the statistical data presents a geometric mean. That is the probability of success is of the form.

The amount of repeated Bernoulli trials required before n eventual success outcome or

The probability of having a given number of failures before the first success is recorded.

In geometric distribution, the probability of having an eventual successful outcome depends on the the completion of a certain number of attempts with each having the same probability of success.

If the probability of each of the preceding trials is p and the kth trial is the  first successful trial, then the probability of having k is given by

Pr(X=k) = (1-p)^{k-1}p  

The number of expected independent trials to arrive at the first success for a variable Xis 1/p where p is the expected success of each trial hence p is the probability for the red and the expected value of the number of trials is 1/p or where p = 12 % which is 0.12

1/p = 1/0.12 or 25/3 or 8.33.

4 0
3 years ago
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