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koban [17]
3 years ago
13

Task: E = mc2

Mathematics
1 answer:
Afina-wow [57]3 years ago
3 0

Answer:

$c=\left(\frac{E}{m} \right)^{\frac{1}{2} }$

Step-by-step explanation:

E = mc^2

$c^2=\frac{E}{m} $

$c=\sqrt{\frac{E}{m} } $

or

$c=\left(\frac{E}{m} \right)^{\frac{1}{2} }$

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rebecca wants to crochet a scarf that is at least 4.4 feet long. So far, she's completed 2.5 feet. How much more does Rebecca in
vlabodo [156]

Answer:

4.4 < 2.5 +x

She needs to crochet at least 1.9 more ft

Step-by-step explanation:

She wants the scarf to be at least 4.4 ft long

4.4< scarf

The scarf is 2.5 ft long now

She will crochet x more ft

4.4 < 2.5 +x

Subtract 2.5 from each side

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3 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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