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joja [24]
3 years ago
6

Sanjeep wishes to take out a $250,000 mortgage. The yearly interest rate on the loan is 5% and the loan is for 25 years. Calcula

te the monthly repayments. Give your answer in dollars and cents. Do not include commas or the dollar sign in your answer.
Mathematics
1 answer:
aksik [14]3 years ago
8 0

Answer:

1041.66 a month

Step-by-step explanation:

250,000  x 0.05 = 12,500

12,500 a year

12,500 / 12 = 1041.66

1041.66 a month

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Sandra rode her bike 5 times as many miles as Barbara. If b, the distance Barbara rode, equals 3.4 miles, what is the correct ex
Ket [755]
3.4 x 5  = 17 thats how you find it

8 0
3 years ago
Read 2 more answers
Which equation is equivalent to y = x² + 24x - 18?
stira [4]
(X+12)^2 -18-(24/2)^2

(X+12)^2 -18-144

(X+12)^2-162

Your answer is 1.
7 0
3 years ago
Simplify the expression -2(6-5g)
boyakko [2]

Answer:

10g-12

Step-by-step explanation:

4 0
3 years ago
Each day a commuter takes a bus to work, the transportation system has a phone app that tells her what time the bus will arrive.
Paraphin [41]

Answer:

Step-by-step explanation:

Hello!

The commuter is interested in testing if the arrival time showed in the phone app is the same, or similar to the arrival time in real life.

For this, she piked 24 random times for 6 weeks and measured the difference between the actual arrival time and the app estimated time.

The established variable has a normal distribution with a standard deviation of σ= 2 min.

From the taken sample an average time difference of X[bar]= 0.77 was obtained.

If the app is correct, the true mean should be around cero, symbolically: μ=0

a. The hypotheses are:

H₀:μ=0

H₁:μ≠0

b. This test is a one-sample test for the population mean. To be able to do it you need the study variable to be at least normal. It is informed in the test that the population is normal, so the variable "difference between actual arrival time and estimated arrival time" has a normal distribution and the population variance is known, so you can conduct the test using the standard normal distribution.

c.

Z_{H_0}= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } }

Z_{H_0}= \frac{0.77-0}{\frac{2}{\sqrt{24} } }= 1.89

d. This hypothesis test is two-tailed and so is the p-value.

p-value: P(Z≤-1.89)+P(Z≥1.89)= P(Z≤-1.89)+(1 - P(Z≤1.89))= 0.029 + (1 - 0.971)= 0.058

e. 90% CI

Z_{1-\alpha /2}= Z_{0.95}= 1.645

X[bar] ± Z_{1-\alpha /2}* (\frac{Sigma}{\sqrt{n} } )

0.77 ± 1.645 * (\frac{2}{\sqrt{24} } )

[0.098;1.442]

I hope this helps!

4 0
3 years ago
On Mercury, you weigh 2 pounds for each 5 pounds you weigh on Earth. You weigh 140 pounds on Earth. How much do you weigh on Mer
MissTica

Answer:

56

Step-by-step explanation:

Givens

Given Ratio = 2/5

Given mass on earth = 140

Ratio

2/5 = x/140 or

0.4 = x / 140

Solution

2/5 = x/140                    Cross multiply

5x = 2*140                     Combine

5x = 280                       Divide by 5

x = 280/5

x = 56

You could also do this by combining the left side of the original proportion into 0.4

0.4 = x / 140               Multiply both sides by 140

0.4*140 = x                 Combine the left

56 = x

5 0
3 years ago
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