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1 year ago

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Dave, Frank, and Kathy can process 1,300 catalog orders per day when all three of them are working. Dave and Frank can do 780 or

ders per day and Frank and Kathy can do 850 orders per day. How many can each do in a day?

1 answer:

Natalija [7]1 year ago

3 0

(1) x+y+z=1300
(2) x+y=780 => x = 780 - y
(3) y+z=850 => z = 850 - y

Replace x and z with y in function (1)

780 - y + y + 850 - y = 1300
1630 - y = 1300
y = 330

Therefore, x = 780 - y = 780 - 330 = 450
z = 850 - y = 850 - 330 = 520

Dave can do 450 catalog orders per day
Frank can do 330 catalog orders per day
Kathy can do 520 catalog orders per day

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Answer:

$A_1 = 24m^2$ -- Swing set

$A_2 = 16m^2$ -- Slide

$A_3 = 5m^2$ --- Pathway

Step-by-step explanation:

Missing part of the question:

Fill in the table above to show the area of each section (see attachment for table)

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$Area = Length * Width$

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$A_2 = 8m * 2m$

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$A_3 = 5m * 1m$

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$Area = 24m^2 + 16m^2 + 5m^2$

$Area = 45m^2$

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1 year ago

The number 2^1993 + 3^1993 is a multiple of 5. What is the units digit of the quotient (2^1993 + 3^1993)/5?

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Answer: The unit digit of the quotient is 1.

Step-by-step explanation:

Since the number 2^1993 + 3^1993 is a multiple of 5, this means that no matter the value of the answer to the equation, the last digit will be 5 (we call the last digit of any number its "unit digit").

Since the unit digit of 2^1993 + 3^1993 is 5, if the unit digit is divided by 5 i.e 5/5, it will give us 1.

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The measures in the table describe the weights of animals that visited a vet on one day, in pounds. Mean Median Mode Mean Absolu

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Answer: On average, the weight of a pet visiting this vet on this day is about 2.4 pounds away from 12.9 pounds.

If the MAD of weights for another day was 1.5, then that day's weights would be less variable than the weights of pets seen on this day.

Step-by-step explanation:

Given: The measures in the table describe the weights of animals that visited a vet on one day, in pounds.

Mean = 12.9

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Mode = 12.0

Mean Absolute Deviation = 2.4

We know that the mean absolute deviation (MAD) of a data set is the mean distance between each and every data value and the mean.It tells about the variation in a data set.

Therefore, On average, the weight of a pet visiting this vet on this day is about 2.4 pounds away from 12.9 pounds.

Also, If the MAD of weights for another day was 1.5, and since 1.5< 2.4.

Then that day's weights would be less variable than the weights of pets seen on this day.

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Step-by-step explanation:

Given,

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Fudgin [204]

Answer:

A. $301

B. $721

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Let $x be the amount of money they raised.

Rowena tried to put the $1 bills into two equal piles and found one left over at the end, then

$x=2q_1+1$

Polly tried to put the $1 bills into three equal piles and found one left over at the end, then

$x=3q_2+1$

Frustrated, they tried 4, 5, and 6 equal piles and each time had $1 left over, then

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Finally Rowena put all the bills evenly into 7 equal piles, and none were left over, then

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$x=60n+1, \ n\in N$

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Now, the number $x=60n+1$ should be divisible by 7 and must be greater than 500.

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When n = 9,

$x=60\cdot 9+1=541$ is not divisible by 7.

When n = 10,

$x=60\cdot 10+1=601$ is not divisible by 7.

When n = 11,

$x=60\cdot 11+1=661$ is not divisible by 7.

When n = 12,

$x=60\cdot 12+1=721$ is divisible by 7.

B. The least amount of money they could have raised is $721

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1 year ago

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