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3 years ago

Pls help if you can i need the answer right away

Mathematics

1 answer:

Anarel [89]3 years ago

3 0

Answer:

C ($9.65)

Step-by-step explanation:

the cost of the ride starts at $2.75. for each additional mile, you want to add $2.30; the ride is 3 miles long so you must add the starting cost plus $2.30 times 3.

2.75+(2.30x3) = $9.65 aka answer C

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3 years ago

In a cookie jar, there are 2 sugar cookies and 10 chocolate chip cookies. If a cookie is selected randomly, how likely is it tha

Fynjy0 [20]

B, it is unlikely, because there are only 2 sugar cookies, while there are 10 chocolate chip cookies.

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3 years ago

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HELPPPP ME PLEASE!!!!!

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Answer:

10/3 ..don't trust I might be wrong

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3 years ago

Determine the solution to the equation below.

mezya [45]

Answer:

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2 years ago

There are three local factories that produce radios. Each radio produced at factory A is defective withprobability .02, each one

diamong [38]

Answer:

The probability is 0.02667

Step-by-step explanation:

Let's call D1 the event that the first radio is defective and D2 the event that the second radio is defective.

So, if we select both radios any factory, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

P(D2/D1) = P(D2∩D1)/P(D1)

Taking into account that 0.02 is the probability that a radio produced at factory A is defective, P(D2/D1) for factory A is:

$P(D2/D1)_A=\frac{0.02*0.02}{0.02} =0.02$

At the same way, if both radios are from factory B, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

$P(D2/D1)_B=\frac{0.01*0.01}{0.01} =0.01$

Finally, if both radios are from factory C, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

$P(D2/D1)_C=\frac{0.05*0.05}{0.05} =0.05$

So, if the radios are equally likely to have been any factory, the probability to select both radios from any of the factories A, B or C are respectively:

P(A)=1/3

P(B)=1/3

P(C)=1/3

Then, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

$P(D2/D1)=P(A)P(D2/D1)_A+P(B)P(D2/D1)_B+P(C)P(D2/D1)_C$

P(D2/D1) = (1/3)*(0.02) + (1/3)*(0.01) + (1/3)*(0.05)

P(D2/D1) = 0.02667

6 0

3 years ago