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evablogger [386]
2 years ago
13

if a sum of money increased by 1/5 of itself every year and in 5 years amounts to Rs.8546,find the sum

Mathematics
1 answer:
weqwewe [10]2 years ago
4 0

Answer:

Answer :

Given  : If a sum of money amounts to Rs 2200 in 5 years

Let sum of money =  x

And as given : The interest is 3838 of the principal . So we get

x ( 3838 ) = 2200 - x

3x  =  17600 - 8x

11x  =  17600

x   =  1600

So, Sum of money  =  Rs . 1600

Then interest =  2200 -  1600 =  Rs . 600

Let Rate of interest  = r  %

And we know Simple interest  = P × r × t100P × r × t100 , So we get

⇒1600 × r × 5100 =  600⇒16× r × 5100 =  6⇒80r  =  600⇒r = 60080 = 152 =  712%⇒1600 × r × 5100 =  600⇒16× r × 5100 =  6⇒80r  =  600⇒r = 60080 = 152 =  712%

So

Step-by-step explanation:

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Hershel had 100 baseball cards that he labeled from 1-100. He started with number
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Hey there!

This question is basically asking for the least common multiple, or LCM of the three numbers.

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Thus, your answer is 70.

Hope this helps :)

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Un número racional positivo siempre es menor que un número racional negativo​
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Answer:

This sentence is true.

Step-by-step explanation:

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3 years ago
A jar holds 42 yellow and 14 purple crayons. A crayon is pulled out of the jar at random and then placed back in the jar.
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2 years ago
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The time intervals between successive barges passing a certain point on a busy waterway have an exponential distribution with me
lisov135 [29]

Answer:

a) <u>0.4647</u>

b) <u>24.6 secs</u>

Step-by-step explanation:

Let T be interval between two successive barges

t(t) = λe^λt where t > 0

The mean of the exponential

E(T) = 1/λ

E(T) = 8

1/λ = 8

λ = 1/8

∴ t(t) = 1/8×e^-t/8   [ t > 0]

Now the probability we need

p[T<5] = ₀∫⁵ t(t) dt

=₀∫⁵ 1/8×e^-t/8 dt

= 1/8 ₀∫⁵ e^-t/8 dt

= 1/8 [ (e^-t/8) / -1/8 ]₀⁵

= - [ e^-t/8]₀⁵

= - [ e^-5/8 - 1 ]

= 1 - e^-5/8 = <u>0.4647</u>

Therefore the probability that the time interval between two successive barges is less than 5 minutes is <u>0.4647</u>

<u></u>

b)

Now we find t such that;

p[T>t] = 0.95

so

t_∫¹⁰ t(x) dx = 0.95

t_∫¹⁰ 1/8×e^-x/8 = 0.95

1/8 t_∫¹⁰ e^-x/8 dx = 0.95

1/8 [( e^-x/8 ) / - 1/8 ]¹⁰_t  = 0.95

- [ e^-x/8]¹⁰_t = 0.96

- [ 0 - e^-t/8 ] = 0.95

e^-t/8 = 0.95

take log of both sides

log (e^-t/8) = log (0.95)

-t/8 = In(0.95)

-t/8 = -0.0513

t = 8 × 0.0513

t = 0.4104 (min)

so we convert to seconds

t = 0.4104 × 60

t = <u>24.6 secs</u>

Therefore the time interval t such that we can be 95% sure that the time interval between two successive barges will be greater than t is <u>24.6 secs</u>

6 0
2 years ago
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