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amid [387]
2 years ago
13

Bob Rohrman is offering a 2016 Toyota Camry for $29,000. If you have a $5,000 down payment and are able to borrow the rest at a

2.99% APR on a 48 month loan, how much will your payments be?
Mathematics
1 answer:
vova2212 [387]2 years ago
4 0

Answer:

The monthly payment will be $531.12

Step-by-step explanation:

Consider the provided information.

After paying $5,000 down payment you need to pay:

$29,000-$5,000=$24,000

APR is 2.99% or APR = 0.299%

Therefore, r=\frac{0.0299}{12}

n = 48

We can calculate the monthly payment by using the formula:

P=\frac{r(PV)}{1-(1+r)^{-n}}

Where P is the monthly payment, PV is the present value, r is the rate per period and n is the number of period.

Substitute the respective values in the above formula we get,

P=\frac{\frac{0.0299}{12}(24000)}{1-(1+\frac{0.0299}{12})^{-48}}

P=\frac{59.8}{0.112593}

P\approx531.12

Hence, the monthly payment will be $531.12

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Suppose that a ball is dropped from the upper observation deck of the CN Tower in Toronto, 450 m above the ground. Find the velo
mr Goodwill [35]

Answer:

1350

Step-by-step explanation:

460 X 3

7 0
2 years ago
Help please <br> If steak costs $8.45 per pound, what will be the total cost of 0.8 pounds of steak?
krok68 [10]

Answer:

$6.76

Step-by-step explanation:

Multiply 8.45 by 0.8 and you will get 6.76

5 0
2 years ago
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On average, the number of customers who had items to return for refunds or exchanges at a certain retail store's service desk is
spayn [35]

Answer:

The probability that the service desk will have at least 100 customers with returns or exchanges on a randomly selected day is P=0.78.

Step-by-step explanation:

With the weekly average we can estimate the daily average for customers, assuming 7 days a week:

M=756/7=108

We can model this situation with a Poisson distribution, with parameter λ=108. But because the number of events is large, we use the normal aproximation:

P(\lambda)\approx N(\lambda,\lambda)

Then we can calculate the z value for x=100:

z=\frac{x-\mu}{\sigma}=\frac{100-108}{\sqrt{108}}=\frac{-8}{10.4} =-0.77

Now we calculate the probability of x>100 as:

P(x>100)=P(z>-0.77)=0.78

The probability that the service desk will have at least 100 customers with returns or exchanges on a randomly selected day is P=0.78.

5 0
2 years ago
Suppose that the weight of seedless watermelons is normally distributed with mean 6.1 kg. and standard deviation 1.9 kg. Let X b
horrorfan [7]

Answer:

a. X\sim N(\mu = 6.1, \sigma = 1.9) b. 6.1 c. 0.6842 d. 0.4166 e. 0.1194 f. 8.5349

Step-by-step explanation:

a. The distribution of X is normal with mean 6.1 kg. and standard deviation 1.9 kg. this because X is the weight of a randomly selected seedless watermelon and we know that the set of weights of seedless watermelons is normally distributed.

b. Because for the normal distribution the mean and the median are the same, we have that the median seedless watermelong weight is 6.1 kg.

c. The z-score for a seedless watermelon weighing 7.4 kg is (7.4-6.1)/1.9 = 0.6842

d. The z-score for 6.5 kg is (6.5-6.1)/1.9 = 0.2105, and the probability we are seeking is P(Z > 0.2105) = 0.4166

e. The z-score related to 6.4 kg is z_{1} = (6.4-6.1)/1.9 = 0.1579 and the z-score related to 7 kg is z_{2} = (7-6.1)/1.9 = 0.4737, we are seeking P(0.1579 < Z < 0.4737) = P(Z < 0.4737) - P(Z < 0.1579) = 0.6821 - 0.5627 = 0.1194

f. The 90th percentile for the standard normal distribution is 1.2815, therefore, the 90th percentile for the given distribution is 6.1 + (1.2815)(1.9) = 8.5349

7 0
2 years ago
Ann, Emily, and Bob collect stickers. They have a total of 70 stickers in the ratio 2:3:5, respectively. How many stickers does
Papessa [141]

Answer:

The stickers are divided by groups of

2/10, 3/10 and 5/10

2/10 * 70 = 14

3/10 * 70 = 21

5/10 * 70 = 35

14 + 21 + 35 = 70

So, the stickers are allotted to the 3 children in a group of

14 to one child, 21 to another child, and 35 to the third child.

Step-by-step explanation:

6 0
2 years ago
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