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2 years ago

Find the arithmetic of the numbers 109, 113, 56, 87, 39, 68, 116, 79, 111, 91, 78, 121, 56

Mathematics

1 answer:

kvasek [131]2 years ago

8 0

56 would be your answer because it’s the lowest number

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Alex buys a car with a down payment along with a monthly payment. if he pays $5,250 for the down payment, pays the monthly payme

svp [43]

Alex will pay $600 per month
13,650 -5,250= 8400
8400÷ 14= 600

4 0

3 years ago

A school purchases jars of marbles.

a_sh-v [17]

Answer:

a. $0.07

Step-by-step explanation:

Profit = Charge -Cost

We want to make a profit of 25 and the cost is 10

25 = Charge -10

Add 10 to each side

25+10 = Charge -10+10

35 = Charge

We need to charge 35 per jar.

Each jar contains 500 marbles

The cost per marble is Cost/ number of marbles

cost per marbles = $35 / 500

= $.07

6 0

3 years ago

Read 2 more answers

Park rangers released 4 fish into a pond,

amid [387]

That’s cool
Not sure how to answer this but I would like to help you just need to be more specific with your question

5 0

2 years ago

For the following amount at the given interest rate compounded continuously, find (a) the future value after 9 years, (b) the

Stella [2.4K]

Answer:

(a) The future value after 9 years is $7142.49.

(b) The effective rate is $r_E=4.759 \:{\%}$ .

(c) The time to reach $13,000 is 21.88 years.

Step-by-step explanation:

The definition of Continuous Compounding is

If a deposit of $P$ dollars is invested at a rate of interest $r$ compounded continuously for $t$ years, the compound amount is

$A=Pe^{rt}$

(a) From the information given

$P=4700$

$r=4.65\%=\frac{4.65}{100} =0.0465$

$t=9 \:years$

Applying the above formula we get that

$A=4700e^{0.0465\cdot 9}\\A=7142.49$

The future value after 9 years is $7142.49.

(b) The effective rate is given by

$r_E=e^r-1$

Therefore,

$r_E=e^{0.0465}-1=0.04759\\r_E=4.759 \:{\%}$

(c) To find the time to reach $13,000, we must solve the equation

$13000=4700e^{0.0465\cdot t}$

$4700e^{0.0465t}=13000\\\\\frac{4700e^{0.0465t}}{4700}=\frac{13000}{4700}\\\\e^{0.0465t}=\frac{130}{47}\\\\\ln \left(e^{0.0465t}\right)=\ln \left(\frac{130}{47}\right)\\\\0.0465t\ln \left(e\right)=\ln \left(\frac{130}{47}\right)\\\\0.0465t=\ln \left(\frac{130}{47}\right)\\\\t=\frac{\ln \left(\frac{130}{47}\right)}{0.0465}\approx21.88$

3 0

2 years ago

Solve this pls<br> -5x - 10 = 10

Vanyuwa [196]

Step-by-step explanation:

-5x = 10 + 10

-5x = 20

X = 20/-5

X = -4

check :

-5 x -4 =20

20 =20

5 0

1 year ago