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2 years ago

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A rental car company charges $63.41 per day to rent a car and $0.05 for every mile driven. Ella wants to rent a car, knowing tha

t: She plans to drive 275 miles. She has at most $200 to spend. What is the maximum number of days that Ella can rent the car while staying within her budget?

1 answer:

zvonat [6]2 years ago

6 0

Answer:

2 days

Step-by-step explanation:

275 miles times 0.05 = $13.75 for mileage

200-13.75=186.25 left

186.25/63.41(daily fee)= 2.93

She can rent it for 2 days

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Coles has cut the price of Cadbury's chocolate bars by 20%. Their normal price is $3.45. A shareholder who has a Coles-Myer disc

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The first discounted price is given through the equation,
discounted price = ($3.45)(1 - 0.20) = $2.76
Then, due to the privilege of being a Coles-Myer discount card holder, another 7.5% discount is given. Such that the final price would be,
final price = ($2.76)(1 - 0.075) = $2.553
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3 years ago

If you have a cubic polynomial of the form y = ax^3 + bx^2 + cx + d and lets say it passes through the points (2,28), (-1, -5),

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Step-by-step explanation:

<u>Step 1: Solve using the first point</u>

<em>(2, 28)</em>

$28 = a(2)^3 + b(2)^2 + c(2) + d$

$28 = 8a + 4b + 2c + d$

<u>Step 2: Solve using the second point</u>

<em>(-1, -5)</em>

$-5 = a(-1)^3 + b(-1)^2 + c(-1) + d$

$-5 = -a + b - c + d$

<u>Step 3: Solve using the third point</u>

<em>(4, 220)</em>

$220 = a(4)^3 + b(4)^2 + c(4) + d$

$220 = 64a + 16b + 4c + d$

<u>Step 4: Solve using the fourth point</u>

<em>(-2, -20)</em>

$-20 = a(-2)^3 + b(-2)^2 + c(-2) + d$

$-20 = -8a + 4b - 2c + d$

<u>Step 5: Combine the first and fourth equations</u>

<u /> $28 - 20 = 8a - 8a + 4b + 4b + 2c - 2c + d + d$

$8 = 8b + 2d$

$8 - 8b = 8b - 8b + 2d$

$(8 -8b)/2 = 2d/2$

$4 - 4b = d$

<u>Step 6: Solve for c in the second equation</u>

$-5 + 5 = -a + b - c + d + 5$

$0 + c = -a + b - c + c + d + 5$

$c = -a + b + d + 5$

<u>Step 7: Substitute d with the stuff we got in step 5</u>

$c = -a + b + (4 - 4b) + 5$

$c = -a + b + 4 - 4b + 5$

$c = -a - 3b + 9$

<u>Step 8: Substitute d and c into the first equation</u>

<u /> $28 = 8a + 4b + 2(-a - 3b + 9) + (4 - 4b)$

$28 = 8a + 4b - 2a - 6b + 18 + 4 - 4b$

$28 - 22 = 6a - 6b + 22 - 22$

$6 / 6 = (6a - 6b) / 6$

$1 + b = a - b + b$

$1 + b = a$

<u>Step 9: Substitute a, b, and c into the third equation</u>

$220 = 64(1 + b) + 16b + 4(-(1 + b) - 3b + 9) + (4 - 4b)$

$220 = 64 + 64b + 16b + 4(-1 - b - 3b + 9) + 4 - 4b$

$220 - 100 = 60b + 100 - 100$

$120 / 60 = 60b / 60$

$2 = b$

<u>Step 10: Find a using b = 2</u>

$a = b + 1$

$a = (2) + 1$

$a = 3$

<u>Step 11: Find c using a = 3 and b = 2</u>

$c = -a - 3b + 9$

$c = -(3) - 3(2) + 9$

$c = -3 - 6 + 9$

$c = 0$

<u>Step 12: Find d using b = 2</u>

$d = 4 - 4b$

$d = 4 - 4(2)$

$d = 4 - 8$

$d = -4$

Answer: $a = 3, b = 2, c = 0,d = -4$

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Answer:

46.67% probability that they flew with airline B.

Step-by-step explanation:

We have these following probabilities:

A 40% probability that a traveler chooses airline A.

A 35% probability that a traveled chooses airline B.

A 25% probability that a traveler chooses airline C.

If a passenger chooses airline A, a 10% probability that he arrives late.

If a passenger chooses airline B, a 15% probability that he arrives late.

If a passenger chooses airline C, a 8% probability that he arrives late.

If a randomly selected traveler is on a flight from Seattle which arrives late to Sydney, what is the probability that they flew with airline B?

This can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

So

What is the probability that the traveler flew with airline B, given that he was late?

P(B) is the probability that he flew with airline B.

So $P(B) = 0.35$

P(A/B) is the probability of being late when traveling with airline B. So P(A/B) = 0.15.

P(A) is the probability of being late. This is the sum of 10% of 40%(airline A), 15% of 35%(airline B) and 8% of 25%(airline C).

So

$P(A) = 0.1*0.4 + 0.15*0.35 + 0.08*0.25 = 0.1125$

Probability

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.35*0.15}{0.1125} = 0.4667$

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