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prohojiy [21]
3 years ago
12

If ⅓y = 9; then y =?​

Mathematics
2 answers:
Semmy [17]3 years ago
7 0
The answer is 27. We plug in 9 for y and get the equation 1/3(9). 1/3 times 9 equals 27.
Svetach [21]3 years ago
4 0

<u>Answer:</u>

<h2>y = 27</h2>

<u>Eexplanation:</u>

1/3y = 9

y/3 = 9

y = 9 × 3

y = 27

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A total of 907 went to a fishing tournament. Of these people, 626 arrived before noon. Alina estimates that fewer that 300 peopl
larisa [96]
She estimated by rounding 907 -> 900 and 626 -> 630 and when you subtract them, you get 270, which is how she estimated >300 people would arrive in the afternoon
5 0
4 years ago
Read 2 more answers
Dose anyone know this answer?
LiRa [457]
So I can show you on a piece of paper

It’s reflections
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3 0
3 years ago
| 1. Add the polynomials.<br> (-14x2 + 6x – 5) + (x3 + 2x2 + 5)
docker41 [41]

Answer:

So I'm not completely sure what the x is standing for in this. Is it standing at a multiplication sign like -14 TIMES 2? Or is it -14x TIMES 2?

Step-by-step explanation:

7 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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AlexFokin [52]
Since there is no semicircle, the angle is 0.
8 0
3 years ago
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