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tatyana61 [14]
3 years ago
14

The scale on a map is 1 inch = 2.5 miles. Two towns on the map are 6 inches apart. What is the actual distance between the towns

?
Mathematics
2 answers:
Vedmedyk [2.9K]3 years ago
8 0
1in/2.5mi
6 • 2.5 = 15
15 mi apart
makkiz [27]3 years ago
6 0
The total distance between the towns is 15 miles.


6 x 2.5 = 15
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Bag contains 2 red marbles 2Green marbles and 4 blue marbles if we choose a marble than other marble without putting the first o
Murrr4er [49]

Answer:

The probability that the first marvel will be red and the second will be green is 7.14%.

Step-by-step explanation:

Since a bag contains 2 red marbles, 2 green marbles and 4 blue marbles, if we choose a marble and then other marble without putting the first one back in the bag, to determine what is the probability that the first marvel will be red and the second will be green, the following calculation must be performed:

2 + 2 + 4 = 8

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0.0714 = X

Therefore, the probability that the first marvel will be red and the second will be green is 7.14%.

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2 years ago
You deposit 2000 in account A, which pays 2.25% annual interest compounded monthly. You deposit another 2000 in account b, which
stellarik [79]
To model this situation, we are going to use the compound interest formula: A=P(1+ \frac{r}{n} )^{nt}
where
A is the final amount after t years 
P is the initial deposit 
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n is the number of times the interest is compounded per year
t is the time in years 

For account A: 
We know for our problem that P=2000 and r= \frac{2.25}{100} =0.0225. Since the interest is compounded monthly, it is compounded 12 times per year; therefore, n=12. Lets replace those values in our formula:
A=2000(1+ \frac{0.0225}{12} )^{12t}

For account B:
P=2000, r= \frac{3}{100} =0.03, n=12. Lest replace those values in our formula:
A=2000(1+ \frac{0.03}{12} )^{12t}

Since we want to find the time, t, <span>when  the sum of the balance in both accounts is at least 5000, we need to add both accounts and set that sum equal to 5000:
</span>2000(1+ \frac{0.0225}{12} )^{12t}+2000(1+ \frac{0.03}{12} )^{12t}=5000

Now that we have our equation, we just need to solve for t:
2000[(1+ \frac{0.0225}{12} )^{12t}+(1+ \frac{0.03}{12} )^{12t}]=5000
(1+ \frac{0.0225}{12} )^{12t}+(1+ \frac{0.03}{12} )^{12t}= \frac{5000} {2000}
(1.001875)^{12t}+(1.0025 )^{12t}= \frac{5}&#10;{2}
ln(1.001875)^{12t}+ln(1.0025 )^{12t}=ln( \frac{5} {2})
12tln(1.001875)+12tln(1.0025 )=ln( \frac{5} {2})
t[12ln(1.001875)+12ln(1.0025 )]=ln( \frac{5} {2})
t= \frac{ln( \frac{5}{2} )}{12ln(1.001875)+12ln(1.0025 )}
17.47

We can conclude that after 17.47 years <span>the sum of the balance in both accounts will be at least 5000.</span>
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Answer:

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No, since it a repeating(terminating) decimal, so it is not a rational number.

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