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Andrew [12]
4 years ago
8

What is the quote that of 2^4/2^-4

Mathematics
1 answer:
Margaret [11]4 years ago
3 0

Answer:

2^8

Step-by-step explanation:

To find the quotient of:

\frac{2^4}{2^{-4}}

Solution:

We will apply quotient rule of exponents to find the quotient of the given expression.

By Quotient rule: \frac{a^x}{a^y}=a^{(x-y)}

Thus the quotient for the expression can be given as:

\frac{2^4}{2^{-4}}

⇒ 2^{(4-(-4))}

⇒ 2^{(4+4)}

⇒ 2^8 (Answer)

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5 0
3 years ago
The population of a small town is decreasing by 2% per year. The population in the year 2017 was 3,850. A. By how many people wi
lesya692 [45]

Answer:

A. 77

B. 3,773

Step-by-step explanation:

According to the problem, computation of given data are as follows,

Population in 2017 = 3,850

Decrease rate = 2% per year

(A) We can calculate number of population decrease from 2017 to 2018 by using following formula,

Number of population decrease = Population in 2017 × Decrease rate

= 3,850 × 2%

= 77

(B). Total population in 2018 = 3,850 - 77

= 3,773

7 0
3 years ago
2/3 x-4 = -2<br><br>A -3 <br>B -4<br>C -9<br>D 3​
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Answer:

i think its d

Step-by-step explanation:

cause i solved it and got the same answer

5 0
3 years ago
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

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3 years ago
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