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pav-90 [236]
3 years ago
5

Tawnee is designing a playground in the shape of a rectangle. If Tawnee increases the length and width of the playground by a sc

ale factor of 2
what will be the resulting effect on the perimeter of the new playground?
The perimeter of the new playground will be half the perimeter of the original playground
The perimeter of the new playground will be 1/4 the perimeter of the original playground
The perimeter of the new playground will be twice the perimeter of the original playground
O The perimeter of the new playground will be four times the perimeter of the original playground
Mathematics
2 answers:
Sphinxa [80]3 years ago
5 0

Answer:

I think it is the 3rd one The perimeter of the new playground will be twice the perimeter of the original playground

Step-by-step explanation:

give brainliest plz

svetlana [45]3 years ago
4 0

Answer:

We conclude that If Tawnee increases the length and width of the playground by a scale factor of 2, the perimeter of the new playground will be twice the perimeter of the original playground.

Step-by-step explanation:

We know that the perimeter of a rectangle = 2(l+w)

i.e.

P = 2(l+w)

Here

  • 'l' is the length
  • 'w' is the width

Given that the length and width of the playground by a scale factor of 2

A scale factor of 2 means we need to multiply both length and width by 2.

i.e

P = 2× 2(l+w)

P' = 2 (2(l+w))

    = 2P      ∵ P = 2(l+w)

Therefore, we conclude that If Tawnee increases the length and width of the playground by a scale factor of 2, the perimeter of the new playground will be twice the perimeter of the original playground.

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LenaWriter [7]

Answer:

P(A_{1}|B ) =\frac{mp}{1+p(m-1)}

Step-by-step explanation:

For mutually exclusive events as A1, A2, A3, etc, Bayes' theorem states:

P(A|B)= \frac{P(B|A)P(A)}{P(B)}

P(A|B) is a conditional probability: the likelihood of event A occurring given that B is true.

P(B|A) is a conditional probability: the likelihood of event B occurring given that A is true.

P(A) is the probability that A occurs

P(B) is the probability that B occurs

For this problem:

A1 is the probability that the student knows the answer

A2 is the probability that the student guesses the answer

B is the probability that the student answer correctly

P(A_{1})=p \\P(A_{2})=1-p \\P(B|A_{1})=1 \\P(B|A_{2})=\frac{1}{m} \\P(B)= P(A_{1})P(B|A_{1}) + P(A_{2})P(B|A_{2})= p+\frac{1-p}{m} \\

P(B|A₁) means the probability that the answer is correct when he knew the answer

P(B|A₂) means the probability that the answer is correct when he guessed the answer

P(A₁|B) means the probability that he knew the answer when the answer was correct

Replacing everything in the Bayes' theorem you get:

P(A_{1}|B)= \frac{P(B|A_{1})P(A_{1})}{P(B)}=\frac{(1)(p)}{p+\frac{1-p}{m}} =\frac{mp}{mp+1-p} =\frac{mp}{1+p(m-1)}

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Answer:

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Step-by-step explanation:

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6 0
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