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3 years ago

What decimal is equivalent to 10/3?

Mathematics

2 answers:

fiasKO [112]3 years ago

8 0

Answer:

3.33333333 . . .

Step-by-step explanation:

worty [1.4K]3 years ago

5 0

Answer:

3.33333333

Step-by-step explanation:

Its a forever number, i forgot the term i think its called a terminal number, or nonterminal i forget. You divide 10 / 3 and get 3.33333333333333333333 forever

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Vanessa bought a house for $268,500. She has a 30 year mortgage with a fixed rate of 6.25%. Vanessa’s monthly payments are $1,59

Genrish500 [490]

Answer:

Option a - $9,314.45

Step-by-step explanation:

Cost of the house = $268,500

Time of repayment = 30 years

Repayment is done monthly, so number of repayments = 30 X 12 = 360

Monthly Payment = $1595.85

Rate of interest per payment period = $\frac{.0625}{12}$

So, Present value of monthly payments = 1595.85 X $\frac{(1+\frac{.0625}{12})^{360}-1}{(1+\frac{.0625}{12})^{360}*(\frac{.0625}{12})}$

= $259,185.55

So, Vanessa's down payment = $268,500 - $259,185.55 = $9,314.45

Hope it helps.

Thank you !!

8 0

3 years ago

Find the domain and range of the function graphed below.

geniusboy [140]

I don’t see any graph attached to the question

7 0

3 years ago

If 4 bushels of rye weigh 76 lb, how much do 8.5 bushels of rye weigh?

mina [271]

76/4=19 lb is weighht of 1 bushel of rye
19*8.5=161.5 lb

8.5 bushels of rye weigh 161.5 lb

5 0

3 years ago

Read 2 more answers

Working for a car company, you have been assigned to find the average miles per gallon (mpg) for acertain model of car. you take

Orlov [11]

Answer:

a) The 95% confidence interval for the mean mpg, for the certain model of car is (23.3, 30.1). This means that we are 95% sure that the true mean mpg of the model of the car is between 23.3 mpg and 30.1 mpg.

b) Increasing the confidence level, the value of T would increase, thus increasing the margin of error and making the interval wider.

c) 37 cars would have to be sampled.

Step-by-step explanation:

Question a:

We have the sample standard deviation, and thus, the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 15 - 1 = 14

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 14 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$ . So we have T = 2.1448

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.1448\frac{6.2}{\sqrt{15}} = 3.4$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 26.7 - 3.4 = 23.3 mpg.

The upper end of the interval is the sample mean added to M. So it is 26.7 + 3.4 = 30.1 mpg.

The 95% confidence interval for the mean mpg, for the certain model of car is (23.3, 30.1). This means that we are 95% sure that the true mean mpg of the model of the car is between 23.3 mpg and 30.1 mpg.

b. What would happen to the interval if you increased the confidence level from 95% to 99%? Explain

Increasing the confidence level, the value of T would increase, thus increasing the margin of error and making the interval wider.

c. The lead engineer is not happy with the interval you constructed and would like to keep the width of the whole interval to be less than 4 mpg wide. How many cars would you have to sample to create the interval the engineer is requesting?

Width is twice the margin of error, so a margin of error of 2 would be need. To solve this, we have to consider the population standard deviation as $\sigma = 6.2$ , and then use the z-distribution.