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ahrayia [7]
3 years ago
8

The number of passengers using a railway fell from 190,205 to 174,989 during a 5-year period. Find the annual percentage decreas

e over this period.
Mathematics
1 answer:
lakkis [162]3 years ago
5 0

Answer:

1.65%

Step-by-step explanation:

We will assume the percentage was the same over the 5 years. The number of passengers using a railway is expressed by the formula

                           P(t)  = P_{o}(1 - r)^{t}

Where

P(t) is the number of passengers after 5 years

                  P(t) = 174,989

P₀ is the number of the passengers at the beginning of the period

                 P₀ = 190,205

r is the rate in percentage    r  = ?

t is time duration    t = 5 years

Substituting these given values into the formula,

                 174,989 = 190,205(1 - r)^{5}

                 (1 - r)^{5} = \frac {174,989}{190,205}

                 1 - r = \sqrt[5]{ \frac {174,989}{190,205}}

                 1 - r = \sqrt[5]{0.920}

                 1 - r = 0.9835

                 r = 1 - 0.9835

                 r = 0.0165

                r = 1.65%

The annual percentage decrease over the period of 5 years is 1.65%

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Answer:

The value of T₂₀ - T₁₅ is <u>-20</u>.

Step-by-step explanation:

<u>Given</u> :

  • >> If for an A.P, d = -4

<u>To</u><u> </u><u>Find</u> :

  • >> T₂₀ - T₁₅

<u>Using Formula</u> :

General term of an A.P.

\star{\small{\underline{\boxed{\sf{\red{ T_n = a  + (n - 1)d}}}}}}

  • >> Tₙ = nᵗʰ term
  • >> a = first term
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<u>Solution</u> :

Firstly finding the A.P of T₂₀ by substituting the values in the formula :

{\dashrightarrow{\pmb{\sf{ T_n = a  + (n - 1)d}}}}

{\dashrightarrow{\sf{ T_{20} = a  + (20 - 1) d}}}

{\dashrightarrow{\sf{ T_{20} = a  + (19)d}}}

{\dashrightarrow{\sf{ T_{20} = a  + 19  \times d}}}

{\dashrightarrow{\sf{ T_{20} = a  + 19d}}}

{\star \: {\underline{\boxed{\sf{\pink{ T_{20} = a  + 19d}}}}}}

Hence, the value of T₂₀ is a + 19d.

\rule{190}1

Secondly, finding the A.P of T₁₅ by substituting the values in the formula :

{\dashrightarrow{\pmb{\sf{ T_n = a  + (n - 1)d}}}}

{\dashrightarrow{\sf{ T_{15}= a  + (15 - 1) d}}}

{\dashrightarrow{\sf{ T_{15}= a  + (14) d}}}

{\dashrightarrow{\sf{ T_{15}= a  + 14 \times d}}}

{\dashrightarrow{\sf{ T_{15}= a  + 14d}}}

{\star{\underline{\boxed{\sf \pink{ T_{15}= a  + 14d}}}}}

Hence, the value of T₁₅ is a + 14d

\rule{190}1

Now, finding the difference between T₂₀ - T₁₅ :

{\dashrightarrow{\pmb{\sf{T_{20} -  T_{15}}}}}

{\dashrightarrow{\sf{(a + 19d) -  (a + 14d)}}}

{\dashrightarrow{\sf{a + 19d -  a  -  14d}}}

{\dashrightarrow{\sf{a - a + 19d -  14d}}}

{\dashrightarrow{\sf{0+ 19d -  14d}}}

{\dashrightarrow{\sf{19d -  14d}}}

{\dashrightarrow{\sf{5 \times  - 4}}}

{\dashrightarrow{\sf{ - 20}}}

{\star \: \underline{\boxed{\sf{\pink{T_{20} -  T_{15} =  - 20}}}}}

Hence, the value of T₂₀ - T₁₅ is -20.

\underline{\rule{220pt}{3.5pt}}

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Total area of the sector = area of the shaded + area of the unshaded

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