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gizmo_the_mogwai [7]
3 years ago
11

The geometric sequence a i a i ​ a, start subscript, i, end subscript is defined by the formula: a 1 = 8 a 1 ​ =8a, start subscr

ipt, 1, end subscript, equals, 8 a i = a i − 1 ⋅ ( − 1.5 ) a i ​ =a i−1 ​ ⋅(−1.5)a, start subscript, i, end subscript, equals, a, start subscript, i, minus, 1, end subscript, dot, left parenthesis, minus, 1, point, 5, right parenthesis Find the sum of the first 20 2020 terms in the sequence. Choose 1 answer:
Mathematics
1 answer:
frozen [14]3 years ago
5 0

Question:

The geometric sequence ai is defined by the formula: a₁ = 8, aᵢ = aᵢ₋₁(-1.5 ).

Find the sum of the first 20 terms in the sequence. Choose 1 answer:

Answer:

The sun of 20 terms of the progress is -10,637.621536256

Step-by-step explanation:

Given

Geometric Sequence

a₁ = 8

aᵢ = aᵢ₋₁(-1.5 )

First, the common ratio needs to be calculated.

The common ratio is the ratio of a term to its previous term.

In other words,

Ratio = 2nd term ÷ 1st term or 3rd term ÷ 2nd term, ...... Etc.

We can calculate the common ratio from aᵢ = aᵢ₋₁(-1.5 ) by dividing both sides by aᵢ₋₁. This gives

aᵢ / aᵢ₋₁ = aᵢ₋₁(-1.5 ) / aᵢ₋₁

aᵢ / aᵢ₋₁ = -1.5

So, the common ratio, r = -1.5

Now that we've had the common ratio and first term to be -1.5 and 8 respectively, the sum of 20 terms can then be calculated using the sum of n terms formula.

Sₙ = a(1 - rⁿ)/(1 - r)

We're making use of this formula because r is less than 1

Where n = 20

a = first term = 8

r = -1.5

By substituting these values; we get

S₂₀ = 8(1 - (-1.5)²⁰)/(1 - (-1.5))

S₂₀ = 8(1 - (-1.5)²⁰)/(1 + 1.5))

S₂₀ = 8(1 - (-1.5)²⁰)/(1 + 1.5))

S₂₀ = 8(1 - (3325.25673008

))/(2.5)

S₂₀ = 8(1 - 3325.25673008

)/(2.5)

S₂₀ = 8(-3324.25673008

)/(2.5)

S₂₀ = -10,637.621536256

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The normal distribution An automobile battery manufacturer offers a 31/54 warranty on its batteries. The first number in the war
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Answer:

1) if the manufacturer's assumptions are correct, it would reed to replace 0.62% of its batteries free.

2) a standard deviation of 6.0843 results in a 1.07% replacement rate

3) using the revised standard deviation for battery life, 91.9% of the manufacturer's batteries don't get free replacement but qualifies for the prorated credit

Step-by-step explanation:

based on the given data;

x will represent the random variable such that the lifetime of its auto batteries, is normally distributed with a mean of 45 months and a standard deviation of 5.6 months

so

x → N( U = 45, ∝ = 5.6)

Under the warranty, if a battery fails within 31 months of purchase, the manufacturer replaces the battery at no charges to the consumer.

if the battery fails after 31 months but within 54 months, the manufacturer provides a prostrated credit towards the purchase of anew battery

1) If the manufacturer's assumptions are correct,

p(x < 3) = p( [x-u / ∝ ] < [ 31-45 / 5.6] )

= p( z < -2.5 )

using the standard normal table,

value of z = 0.0062 ≈ 0.62%

so if the manufacturer's assumptions are correct, it would reed to replace 0.62% of its batteries free.

2)

The company finds that it is replacing 1.07% of its batteries free of charge. It suspects that its assumption standard deviation of the life of its batteries is incorrect, so a standard deviation of ? results in a 1.07%

so lets say;

p ( x < 31 ) = ( 1.07%) = 0.0107

p ( [x-u / ∝ ] < [ 31-45 / ∝] ) = 0.0107

now from the standard table

-2.301 is 1.07%

so

( 31 - 45 / ∝ ) = -2.301

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∝ = -14 / - 2.301

∝ = 6.0843

therefore a standard deviation of 6.0843 results in a 1.07% replacement rate

3)

Using the revised standard deviation for battery life, what percentage of the manufacturer's batteries don't free replacement but do qualify for the prorated credit?

p( 31 < x < 54 ) = p ( [31 - u / ∝ ] < [ x-u / ∝]  < [ 54 - 45 / ∝] )

= p ( [31 - 45 / 6.0843 ] < [ x-u / ∝]  < [ 54 - 45 / 6.0843] )

= p ( -2.301 < z < 1.4792 )

= p(Z < 1.5) - p(Z < -2.3)

= 0.9393 - 0.0108

= 0.919 ≈ 91.9%

therefore using the revised standard deviation for battery life, 91.9% of the manufacturer's batteries don't get free replacement but qualifies for the prorated credit

8 0
3 years ago
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