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const2013 [10]
3 years ago
9

An industrial producer is searching for a supplier for ball bearings. Its three most important supplier criteria are price, qual

ity, and delivery reliability. The firm has decided that quality and delivery reliability should carry the same weight, and that each of them are twice as important as price. If the weights sum to 100%, what would a supplier with ratings of 40, 90, and 75 in the three respective categories score as a weighted total?
Mathematics
1 answer:
Bess [88]3 years ago
8 0

Answer:

74%

Step-by-step explanation:

Let the weight of price = y%

The weight of quality = 2y%

The weight of delivery reliability = 2y%

Since the sum total of the weights is  100, it implies that:

y% + 2y% + 2y% = 100%

5y%  = 100%

y = 100/5

y = 20

So the respective weights are as below:

Price = 20%

Quality = 40%

Delivery reliability  = 40%

Weight score = score x category weight

For the supplier:

Price score = 40

Quality score = 90

Delivery reliability  score= 75

Overall weighted score = (Price score  x Price weight) + (Quality score x Quality weight) + (Delivery reliability score x delivery reliability weight)----- (1)

Substituting into (1), we have :

Overall weighted score = (40 x  20%) + (90 x 40%) + (75 x 40%)

                                       =     8 +36 +30

                                       = 74%

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Step-by-step explanation:

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2 years ago
Which equation represents the line that passes through (-6, 7) and (-3, 6)
Tju [1.3M]
Plug the given x and y values (x, y) into the equations and solve
It look like the second equation is the answer.
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3 years ago
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A school group wants to rent part of a bowling alley to have a party. The bowling alley costs $500 to rent, plus an additional c
Firlakuza [10]

Answer:

If there are N students, the total cost will be:

C(N) = $500 + $5*N.

And we want the amount that each student pays to be equal or less than $15.

then:

C(N)/N ≤ $15.

Then the inequality that represents this situation is:

($500 + $5*N)/N ≤ $15.

Now, let's solve this:

($500 + $5*N) ≤ $15*N

$500 ≤ $15*N - $5*N

$500 ≤ $10*N

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So the minimum number of students needed is 50.

7 0
3 years ago
URGENT NEED HELP ON THIS
zvonat [6]

Answer:

I think option no B

hope it will help you

5 0
2 years ago
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If the inspection division of a county weights and measures department wants to estimate the mean amount of
Amanda [17]

If inspection department wants to estimate the mean amount with 95% confidence level with standard deviation 0.05 then it needed a sample size of 97.

Given 95% confidence level, standard deviation=0.05.

We know that margin of error is the range of values below and above the sample statistic in a confidence interval.

We assume that the values follow normal distribution. Normal distribution is a probability that is symmetric about the mean showing the data  near the mean are more frequent in occurence than data far from mean.

We know that margin of error for a confidence interval is given by:

Me=z/2* ST/\sqrt{N}

α=1-0.95=0.05

α/2=0.025

z with α/2=1.96 (using normal distribution table)

Solving for n using formula of margin of error.

n=(z/2ST/Me)^{2}

n=(1.96*0.05)^{2} /(0.01)^{2}

=96.4

By rounding off we will get 97.

Hence the sample size required will be 97.

Learn more about standard deviation at brainly.com/question/475676

#SPJ4

The given question is incomplete and the full question is as under:

If the inspection division of a county weights and measures department wants to estimate the mean amount of soft drink fill in 2 liters bottles to within (0.01 liter with 95% confidence and also assumes that standard deviation is 0.05 liter. What is the sample size needed?

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