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kkurt [141]
3 years ago
10

I need help pls pls pls

Mathematics
1 answer:
Amiraneli [1.4K]3 years ago
3 0

Answer: The answer may be 26%

Step-by-step explanation: Have a brilliant day mate- Lily (^_^)

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Tom was using wire of the following thicknesses .33 mm, .275 mm, .25 mm, and .3 mm for some electrical work. Order the wire from
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A rectangle has a width of 5 inches and a length of 9 inches. What will be the new dimensions of the rectangle after it is
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the third one

Step-by-step explanation:

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2 years ago
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What will most likely happen to a characteristic that has a high survival value in a population?
Levart [38]
The characteristic will spread and be more common throughout populations.
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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
seraphim [82]

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

-14 / ∝ = -2.301

∝ = -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
Expand ( x - 1/x^2)^4
emmainna [20.7K]

Answer:

We want to expand the expression:

(x - \frac{1}{x^2} )^4

We can just do it by brute force, this is:

First, rewrite our expression as the product of two square factors:

(x - \frac{1}{x^2} )^4 = (x - \frac{1}{x^2} )^2*(x - \frac{1}{x^2} )^2

Now we can expand each one these two factors:

(x - \frac{1}{x^2} )^2 = (x - \frac{1}{x^2} )*(x - \frac{1}{x^2} ) = x^2 + \frac{1}{x^4} -2*x*\frac{1}{x^2}

That can be simplified to

x^2 - \frac{2}{x} + \frac{1}{x^4}

Now we can replace that in our original expression to get:

(x^2 - \frac{2}{x} + \frac{1}{x^4})*(x^2 - \frac{2}{x} + \frac{1}{x^4})

Now we can expand that last product, to get:

(x^2)^2 + 2*(x^2)*(-\frac{2}{x} ) + 2*(x^2)*(\frac{1}{x^4}) + 2*(\frac{-2}{x})*(\frac{1}{x^4}) + (\frac{-2}{x}  )^2 + (\frac{1}{x^4})^2

We can simplify that to:

x^4 - 4x + 2x^2 - \frac{4}{x^5}  + \frac{4}{x^2} + \frac{1}{x^8}

That is the expanded expression.

6 0
2 years ago
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