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Tpy6a [65]
3 years ago
8

Each time Kristine gets paid, she spends $20 and saves the rest. If the amount Kristine earns is represented by x and the amount

she saves is represented by y, which graph models her savings?
A graph titled Kristine's Savings has money earned on the x-axis and money saved on the y-axis. A line goes through points (5, 100) and (10, 200).
A graph titled Kristine's Savings has money earned on the x-axis and money saved on the y-axis. A line goes through points (20, 1) and (40, 2).
A graph titled Kristine's Savings has money earned on the x-axis and money saved on the y-axis. A line goes through points (0, 20) and (10, 30).
A graph titled Kristine's Savings has money earned on the x-axis and money saved on the y-axis. A line goes through points (20, 0) and (25, 5).
Mathematics
1 answer:
kenny6666 [7]3 years ago
5 0

Answer:

A graph titled Kristine's Savings has money earned on the x-axis and money saved on the y-axis. A line goes through points (20, 0) and (25, 5).

Step-by-step explanation:

If Kristine spends $20 and saves the rest when she gets paid, the amount she saves should always be 20 less than the amount she earned.

The points (20, 0) and (25, 5) accurately represent this, where x is how much she earns and y is how much she saves.

(20, 0) represents her earning $20 and saving $0, because she spends $20 every time she gets paid.

(25, 5) represents her earning $25 and saving $5, because she spent $20.

So, the correct graph is A graph titled Kristine's Savings has money earned on the x-axis and money saved on the y-axis. A line goes through points (20, 0) and (25, 5).

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A company purchases a small metal bracket in containers of 5,000 each. Ten containers have arrived at the unloading facility, an
MariettaO [177]

Answer:

Do the data from this shipment indicate statistical control: No

Step-by-step explanation:

Calculating the mean of the sample, we have;

Mean (x-bar) = sum of individual sample/number of sample

                     = (0+0+0+0.004+0.008+0.020+0.004+0+0+0.008)/10

                     = 0.044/10

                    = 0.0044

Calculating the lower control limit (LCL) using the formula;

LCL= (x-bar) - 3*√(x-bar(1-x-bar))/n

      = 0.0044 - 3*√(0.0044(1-0.0044))

       = 0.0044- (3*0.0042)

        = 0.0044 - 0.01256

        = -0.00816 ∠ 0

Calculating the upper control limit (UCL) using the formula;

UCL = (x-bar) + 3*√(x-bar(1-x-bar))/n

      = 0.0044 + 3*√(0.0044(1-0.0044))

       = 0.0044+ (3*0.0042)

        = 0.0044 + 0.01256

       =0.01696∠ 0

Do the data from this shipment indicate statistical control: No

Since the value 0.02 from the 6th shipment is greater than the upper control limit (0.01696), we can conclude that  the data from this shipment do not indicate statistical control.

8 0
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Answer:

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Step-by-step explanation:

p=6

(6+2)= 8

8x8= 64

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