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Sloan [31]
2 years ago
6

This we havent learned but teacher gave us for end of year est plz help

Mathematics
1 answer:
chubhunter [2.5K]2 years ago
3 0

Answer:

15

Step-by-step explanation:

Finding total number of pupils(y)

3/y x 360°=27°

3x360°/y=27°

1080°/y=27°

1080°=27°y

1080°/27°=27°y/27°

40=y

Number of pupils who said that Maths was their favorite (z)

z/40x360°=135°

360°z/40=135°

360°z=135°x40

360°z=5400°

360°z/360°=5400°/360°

z=15

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Can somebody please help me I dont know how to do this and I really could use some help
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Step-by-step explanation:

wait ta minute

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5 0
2 years ago
In a village, there was an epidemic and the population decreased by 10%. If there were only 1800 people left in the village, wha
tester [92]

\bold{\boxed{\huge{\blue{\underline{Answer }}}}}

\bold{\underline{ Given :- }}

\sf{The\: population\: is\: decreased \:by\: 10 percent  }

\sf{There\: were \: only \: 1800 \: people \: left}

\bold{\underline{ To\: Find :- }}

\sf{we \: have \: to \: find\: original \: population? }

\bold{\underline{ Let's \: Begin :- }}

\sf{Let \: the \: total \:  population\: be \: x }

\sf{\underline{ Therefore }}

\sf{According \: to \: the \: question }

\sf{ x - 10/100 = 1800 }

\sf{ 100x - 10/100 = 1800}

\sf{ 100x - 10 = 1800 × 100}

\sf{100x - 10 = 180000}

\sf{ 100x = 180000 + 10}

\sf{ x = 180010/100}

\sf{ x = 1800.1 }

\bold{\pink{Hence , The\: original \:population \:of \: the\: village \:is \:1800.1 }}

7 0
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 
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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OLEGan [10]

Answer:

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