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givi [52]
2 years ago
5

Suppose that when Emma was born, her grandparents deposited $20,000 into an account paying 5% interest compounded quarterly. Wha

t will the balance be when Emma turns 18? Round to the nearest cent
Mathematics
1 answer:
neonofarm [45]2 years ago
4 0

Answer: The balance when Emma turns 18 = $ 48,132.38

Step-by-step explanation:

When interest is compounded quarterly, the accumulated amount after t years will be :

A= P(1+r)^t

, where P =  principal, r= rate of interest( in decimal)

Given: P= $ 20,000 , r= 5% =0.05

t = 18 years

A = 20000(1+0.05)^{18}\\\\= 20000(1.05)^{18}\\\\= 20000(2.406619)\\\\=\$\ 48,132.38

The balance when Emma turns 18 = $ 48,132.38

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A display case of DVDs are marked 19 for $10. If Larry has $50, how many DVDs can Larry get? (Assume no tax or other fees.)
Delicious77 [7]

Answer:

95

Step-by-step explanation:

19 DVDs for 10 dollars and he has 50 dollars well 50 dollars is just 5 groups of 10 so just 19 x 5 = 95

5 0
2 years ago
How should I solve this to get the elapsed time
8090 [49]

Answer:

Subtract the hours to get 12-3 = 9

Or you could start at 3 PM, and count your way up one hour at a time, until you get to 12 AM. You should count out 9 hours in total.

An alternative is you could note from 3 PM to 3 AM is a span of 12 hours. Take 3 hours away from 3 AM, and you'll rewind the clock back to 12 AM. So 12-3 = 9. :-)

plz brainlist plz ;-)

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Pattern x and y are generated using the rules. Pattern x: start with 0 and add 6. Pattern y: start with 1 and multiply 2. Patter
timama [110]

Answer:

Pattern x: 0,6,12,18

Pattern y: 1,2,4,8

Step-by-step explanation:

Given:

Pattern x: start with 0 and add 6.

Pattern y: start with 1 and multiply 2.

To find: patterns x and y

Solution:

Pattern x: 0,0+6,0+6+6,0+6+6+6=0,6,12,18

Pattern y: =1,1(2),1(2)(2),1(2)(2)(2)=1,2,4,8

7 0
3 years ago
A veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses
Licemer1 [7]

Answer:

Probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

Step-by-step explanation:

We are given that a veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses with colic is 12 years. The average age of all horses seen at the veterinary clinic was determined to be 10 years. The researcher also determined that the standard deviation of all horses coming to the veterinary clinic is 8 years.

So, firstly according to Central limit theorem the z score probability distribution for sample means is given by;

                    Z = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \bar X = average age of the random sample of horses with colic = 12 yrs

            \mu = average age of all horses seen at the veterinary clinic = 10 yrs

   \sigma = standard deviation of all horses coming to the veterinary clinic = 8 yrs

         n = sample of horses = 60

So, probability that a sample mean is 12 or larger for a sample from the horse population is given by = P(\bar X \geq 12)

   P(\bar X \geq 12) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } \geq \frac{12-10}{\frac{8}{\sqrt{60} } } ) = P(Z \geq 1.94) = 1 - P(Z < 1.94)

                                                 = 1 - 0.97381 = 0.0262

Therefore, probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

4 0
3 years ago
Solve the following equation for b. 1/2HB=d2
geniusboy [140]

Steps to solve:

1/2hb = d²

~Divide 1/2h to both sides

b = 2d²/h

Best of Luck!

6 0
3 years ago
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