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3 years ago

Evaluate: 54-75+81-(-27)+53

Mathematics

2 answers:

romanna [79]3 years ago

4 0

Hi the answer is 344

solmaris [256]3 years ago

4 0

Answer:

140

Step-by-step explanation:

54-75=-21

-21+81=60

60-(-27)=87 or 60+27=87

87+53=140

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Christine deposits $700 into an account that pays simple interest at a rate of 2 per year. How much interest will she be paid in

xxMikexx [17]

Answer:

$56

Step-by-step explanation:

A=P(1+i.n)

A is final amount

P is amount deposited

i is interest rate

n is number os years

A = $700(1+2%×4)

=$756

$756-$700=$56

7 0

3 years ago

3t - 7 = 5t, then 6t= ?

netineya [11]

Answer:

t = -5/2
-21

Step-by-step explanation:

3t - 7 = 5t

~Subtract 3t to both sides

-7 = 2t

~Divide 2 to both sides

-7/2 = t

~Find 6t with what we found for t

6(-7/2)

-21

Best of Luck!

3 0

3 years ago

Read 2 more answers

Sarah earned a 4% commission on all of her sales in March. Her total sales were 80,000 in March, How many money did she earn. Fr

o-na [289]

Answer:

Amount of commission = 3,200

Step-by-step explanation:

Given:

Total sales value = 80,000

Rate of commission = 4% on sales

Find:

Amount of commission

Computation:

Amount of commission = Total sales value x Rate of commission

Amount of commission = 80,000 x 4%

Amount of commission = 3,200

8 0

3 years ago

A Survey of 85 company employees shows that the mean length of the Christmas vacation was 4.5 days, with a standard deviation of

GenaCL600 [577]

Answer:

The 95% confidence interval for the population's mean length of vacation, in days, is (4.24, 4.76).

The 92% confidence interval for the population's mean length of vacation, in days, is (4.27, 4.73).

Step-by-step explanation:

We have the standard deviations for the sample, which means that 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 = 85 - 1 = 84

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 84 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$ . So we have T = 1.989.

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.989\frac{1.2}{\sqrt{85}} = 0.26$

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 4.5 - 0.26 = 4.24 days

The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.26 = 4.76 days

The 95% confidence interval for the population's mean length of vacation, in days, is (4.24, 4.76).

92% confidence interval:

Following the sample logic, the critical value is 1.772. So

$M = T\frac{s}{\sqrt{n}} = 1.772\frac{1.2}{\sqrt{85}} = 0.23$

The lower end of the interval is the sample mean subtracted by M. So it is 4.5 - 0.23 = 4.27 days

The upper end of the interval is the sample mean added to M. So it is 4.5 + 0.23 = 4.73 days

The 92% confidence interval for the population's mean length of vacation, in days, is (4.27, 4.73).

8 0

3 years ago

Ryder Industries is considering a project that will produce cash inflows of $92,000 a year for five years. What is the internal

RSB [31]

Answer:

20.02%

Step-by-step explanation:

Formula : $NVP = 0 =-P_0 + \frac{P_1}{(1+IRR)} + \frac{P_2}{(1+IRR)^2} + . . . +\frac{P_n}{(1+IRR)^n}$

$P_0 = 275000$

n = 1,2,3,4,5

Substitute the values in the formula :

$0 =-275000 + \frac{92000}{(1+IRR)} + \frac{92000}{(1+IRR)^2} + \frac{92000}{(1+IRR)^3}+\frac{92000}{(1+IRR)^4}+\frac{92000}{(1+IRR)^5}$

$275000 = \frac{92000}{(1+IRR)} + \frac{92000}{(1+IRR)^2} + \frac{92000}{(1+IRR)^3}+\frac{92000}{(1+IRR)^4}+\frac{92000}{(1+IRR)^5}$

Solving for IRR using calculator

IRR = 20.02

Hence the internal rate of return if the initial cost of the project is $275,000 is 20.02%

4 0

3 years ago