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34kurt
3 years ago
8

A flashlight battery manufacturer makes a model of battery whose mean shelf life is three years and four months, with a standard

deviation of three months. The distribution is approximately normal. One production run of batteries in the factory was 25,000 batteries. How many of those batteries can be expected to last between three years and one month and three years and seven months?
Mathematics
1 answer:
iren2701 [21]3 years ago
8 0

Answer:

17000 batteries

Step-by-step explanation:

Three years and one month is equivalent to the mean minus one standard deviation.

Three years and seven months is equivalent to the mean plus one standard deviation.

For a normal distribution, we know that 68% of population is between mean ± 1 sd, then can be expected that 25000*68% = 17000 of batteries last between three years and one month and three years and seven months

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andrew-mc [135]

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Step-by-step explanation:

To find the the unit rate, you will find the amount  one mug will cost.

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GCF and LCM:
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5 0
3 years ago
the company also discovered that it costs 30 to 2 widgets, 118 to produce 4 and 766 to produce 10 widgets. Find the total cost o
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We will first write cost function. As we are given three points here so we will use them to write quadratic function

y = ax^{2} +bx+c----------------------------------------(1)

So we will plug point (2,30) in it first ( for 2 widgets cost is 30)

So simply plug 30 in y place and 2 in x place in equation (1) as shown

30 = a(2)^{2} +b(2)+c

30 = 4a +2b +c ----------------------------------------------(2)

Similarly plug next point (4,118) now in equation (1)

118 = a(4)^{2} +b(4)+c

118 = 16a +4b +c ----------------------------------------------(3)

Similarly plug third point (10,766) now in equation (1)

766 = a(10)^{2} +b(10)+c

766 = 100a +10b +c ----------------------------------------------(4)

Now we have to solve system of linear equations (2),(3) and (4) for a,b,c

We can use method of elimination

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118 = 16a +4b +c

-(30 = 4a +2b +c)

----------------------------------------------------

88 = 12a +2b -----------------------------------(5)

Similarly subtract equation (4)-equation (3)

766 =100a +10b +c

-( 118 = 16a +4b +c)

--------------------------------------

648 = 84a + 6b ------------------------(6)

Now to eliminate b in equations (5) and (6) multiply equation (5) by -3 and then add to equation (6)

-3×(88 = 12a +2b)

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---------------------------------

384 = 48a

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\frac{384}{48} = \frac{48a}{48}

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now plug 8 in a place, -4 in b place in either of equations (2),(3) or (4) and solve for c

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y = 288 -24 +6

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