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sleet_krkn [62]
2 years ago
15

Hhhhhhhhhhheeeellllpppppp me

Mathematics
2 answers:
Norma-Jean [14]2 years ago
7 0
Answer is A i’m pretty sure
kolezko [41]2 years ago
3 0

Answer:

ooo acellus its A

Step-by-step explanation:

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HELP ME I NEED TO FIND THE SLOPEEE!!
N76 [4]

Answer:

y=3/4 x+1

Step-by-step explanation:

first you start with y=mx+b find the y intercept first which is 1 next you just count down the block which is 3 then count to the right till you find a spot where it lands which would get you y=3/4+1

5 0
2 years ago
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A test driver has to drive a car on a 1-mile track for two rounds in a way, that his average speed is 60 mph. On his first round
Volgvan

to average 60 mph for 2 laps his total speed needs to equal 60 x 2 = 120

his first lap was 40 so his 2nd lap needs to be 120-40 = 80 mph

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3 years ago
I am not sure how to slove
notsponge [240]

First .... the limit of the given function is NOT EQUAL to 5 (it diverges so it has no limit).  There is a typo.  The 14 should be positive.

\lim_{x \to\ 7} \bigg(\dfrac{x^2-9x+14}{x-7}\bigg)=5

The precise definition of a limit is:

\text{If for every } \epsilon>0\text{ there exists a }\delta >0\text{ such that}\\|f(x)-L|

Given:

f(x) = \dfrac{x^2-9x+14}{x-7}\\\\L=5\\\\a=7\\\\\\|f(x)-L|

⇒ \epsilon = \delta

When ε = 0.1,     δ = 0.1

When ε = 0.01,   δ = 0.01

5 0
2 years ago
A coin is tossed two times. what is the probability of getting tails two times in a row?
goldfiish [28.3K]
It's a conditional probability: (We know that the probability of getting 1 tail or 1 head is equal to 1/2)

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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

8 0
2 years ago
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