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Elenna [48]
3 years ago
15

Olivia walks up to a tank of water that can hold up to 50 gallons. When it is active, a drain empties water form the tank at a c

onstant rate. When Olivia first sees the tank, it contains 38 gallons of water. Four minutes later, the tank contains 18 gallons of water.
a.) At what rate is the amount of water in the tank changing? Use a signed number and include the unit of measurement in your answer.

b.) How many more minutes will it take for the tank to drain completely?

c.) How many minutes before Olivia arrived was the water tank completely full?
Mathematics
2 answers:
agasfer [191]3 years ago
8 0

Answer:

C

Step-by-step explanation:

I got the answer C because if the estimated value of the tank is 50 and it gets drained it will only have 18 gallon in it

faltersainse [42]3 years ago
5 0

5%

3.36m inutos

2.24mi nutos

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There are 10 employees in a particular division of a company. Their salaries have a mean of 570,000, a median of $55,000,and a s
harkovskaia [24]

Answer:

a) $160,000

b) $55,000

c) $332264.804

Step-by-step explanation:

We are given that there are 10 employees in a particular division of a company and their salaries have a mean of $70,000, a median of $55,000, and a standard deviation of $20,000.

And also the largest number on the list is $100,000 but By accident, this number is changed to $1,000,000.

a) Value of mean after the change in value is given by;

     Original Mean = $70,000

       \frac{\sum X}{n} = $70,000  ⇒ \sum X = 70,000 * 10 = $700,000

   New \sum X after change = $700,000 - $100,000 + $1,000,000 = $1600000

  Therefore, New mean = \frac{1600000}{10} = $160,000 .

b) Median will not get affected as median is the middle most value in the data set and since $1,000,000 is considered to be an outlier so median remain unchanged at $55,000 .

c) Original Variance = 20000^{2} i.e.  20000^{2} = \frac{\sum x^{2} - n*xbar }{n -1}

    Original \sum x^{2} = (20000^{2} * (10-1)) + (10 * 70,000) = $3,600,700,000

    New \sum x^{2} = $3,600,700,000 - 100,000^{2} + 1,000,000^{2} = 9.936007 * 10^{11}  

    New Variance = \frac{new\sum x^{2} - n*new xbar }{n -1} = \frac{9.936007 *10^{11}  - 10*160000 }{10 -1} = 1.103999 * 10^{11}    Therefore, standard deviation after change = \sqrt{1.103999 * 10^{11} } = $332264.804 .

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4 years ago
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