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sp2606 [1]
3 years ago
7

PLEASE HELP ME

Mathematics
1 answer:
jarptica [38.1K]3 years ago
5 0

Answer:

end behaver

Step-by-step explanation:

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A rectangle has a length of 12 and a width that is half the length. What is the area?
Basile [38]

Answer:

Area = length x width

width = 1/2 length = 12/2 = 6 = width

area = 6x 12 = 72

Hope that answers your question

don't hesitate to comment if you are confused about something

Step-by-step explanation:

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Find the number of possible outcomes in the sample space.
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3 years ago
The radius of the Moon is 1700 km, and the radius of the Earth is 6350 km. A scale model of the Earth and the Moon is made. In t
Reil [10]

Answer:

1) 12.7 cm

2) 8580.24 cm³

Step-by-step explanation:

given,

Radius of moon(Rm) = 1700 Km              Radius of Earth(Re) = 6350 km

Model radius of Moon(Mm) = 3.4 cm        

2) Volume of earth model =\dfrac{(R_e)^3}{(R_m)^3}\times \dfrac{4}{3}\pi (M_m)^3

                                       = \dfrac{(6350)^3}{(1700)^3}\times \dfrac{4}{3}\pi (3.4)^3

                                       = 8580.24 cm³

1) radius of earth = \dfrac{R_e}{R_m}\times M_m

                        =  \dfrac{6350}{1700}\times 3.4

                        = 12.7 cm

3 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

8 0
3 years ago
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