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3 years ago

what is the tax on a bicycle that costs 235.00 if the tax rate is 5%? what is the total cost of the bicycle

Mathematics

1 answer:

anyanavicka [17]3 years ago

6 0

235 + 0.05(235) = 235 + 11.75 = 246.75 <== total cost (with the 5% tax being 11.75)

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Mrs. Wong has placed together 60% of the pieces of a jigsaw puzzle She has placed 570 pieces of the puzzle together How many pie

Ivenika [448]

570/x = 60/100. 60% of the total amount of jigsaw puzzles is 570, we could use that information to form the equation above, where x is the total amount of pieces. The total amount of jigsaw puzzles is 950.

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3 years ago

While on vacation, a family of five rented a car to get them to all the places they wanted to see for five days. The car costs a

Sloan [31]

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3 years ago

6 2/3 miles ➕ 4 11/16 =

mihalych1998 [28]

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Step-by-step explanation:

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4 years ago

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Answer: $y=-4x-6$

Step-by-step explanation:

Given

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3 years ago

A company is interviewing potential employees. Suppose that each candidate is either qualified, or unqualified with given probab

stira [4]

Answer:

$P(C=1|T=1)=q(\sum_{i=15}^{20}\binom{20}{i} p^i(1-p)^{20-i})( \sum_{i=15}^{20}\binom{20}{i}[qp^i(1-p)^{20-i} + (1-q)p^{20-i}(1-p)^i])^{-1}$

Step-by-step explanation:

Hi!

Lets define:

C = 1 if candidate is qualified

C = 0 if candidate is not qualified

A = 1 correct answer

A = 0 wrong answer

T = 1 test passed

T = 0 test failed

We know that:

$P(C=1)=q\\P(A=1 | C=1) = p\\P(A=0 | C=0) = p$

The test consist of 20 questions. The answers are indpendent, then the number of correct answers X has a binomial distribution (conditional on the candidate qualification):

$P(X=x | C=1)=f_1(x)=\binom{20}{x}p^x(1-p)^{20-x}\\P(X=x | C=0)=f_0(x)=\binom{20}{x}(1-p)^xp^{20-x}$

The probability of at least 15 (P(T=1))correct answers is:

$P(X\geq 15|C=1)=\sum_{i=15}^{20}f_1(i)\\P(X\geq 15|C=0)=\sum_{i=15}^{20}f_0(i)\\$

We need to calculate the conditional probabiliy P(C=1 |T=1). We use Bayes theorem:

$P(C=1|T=1)=\frac{P(T=1|C=1)P(C=1)}{P(T=1)}\\P(T=1) = qP(T=1|C=1) + (1-q)P(T=1|C=0)$

$P(T=1)=q\sum_{i=15}^{20}f_1(i) + (1-q)\sum_{i=15}^{20}f_0(i)\\P(T=1)=\sum_{i=15}^{20}\binom{20}{i}[qp^i(1-p)^{20-i} + (1-q)p^{20-i}(1-p)^i)]$

$P(C=1|T=1)=q(\sum_{i=15}^{20}\binom{20}{i} p^i(1-p)^{20-i})( \sum_{i=15}^{20}\binom{20}{i}[qp^i(1-p)^{20-i} + (1-q)p^{20-i}(1-p)^i])^{-1}$

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3 years ago