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tresset_1 [31]
2 years ago
11

In their stamp collections, Marlee has 62 stamps, Xavier has 56 stamps, Nicki has 48 stamps, and Cameron has 89 stamps. Estimate

how many stamps they would have if they combined their collections by rounding to the nearest ten first and then adding the rounded numbers.
Mathematics
1 answer:
nalin [4]2 years ago
5 0

Answer:260 is the answer

Step-by-step explanation:

marlee=60 xavier=60 nicki=50 cameron=90 60+60=120 120+50=170 170+90 =260

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If the temperature decreased 7 and a half degreen in 3 hours and now is -14 dedgree celsius, what was the original temperature?
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1.Mr. Redding sells vehicles to 40% of the people that come to the sales lot. If 165 people came to the lot last month, how many
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He would have sold 66 vehicles
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Compare 2x+8>18 and 2x+8 greaterthan or equal to 18
Sergio039 [100]
2x + 8 > 18                      2x+ 8 ≥ 18
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6 0
3 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 
r is the interest rate in decimal form 
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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3 years ago
Answer fast please .
MArishka [77]
I think it's D. The area of A is k2 times the area of rectangle B
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