Answer:
Travis needs to slow down
Step-by-step explanation:
Stop eating cookies, Travis.
The new mean and standard deviation is 26 and 15, when each score in data set is multiplied by 5 and then 7 is added.
According to the question,
Original mean is 10 and original standard deviation is 5 . In order to find to new mean and standard deviation when each score in data set is multiplied by 5 and then 7 is added.
First "change of scale" when every score in a data set is multiplied by a constant, its mean and standard deviation is multiplied by a same constant.
Mean: 10*3 = 30
Standard deviation: 5*3 = 15
Secondly "change of origin" when every score in a data set by a constant, its mean get added or subtracted by the same constant and standard deviation remains constant.
Applying change of origin in the above mean and standard deviation
Mean: 30 - 4 = 26
Standard deviation: Remains same = 15
Hence, the new mean and standard deviation is 26 and 15, when each score in data set is multiplied by 5 and then 7 is added.
Learn more about Mean and standard deviation here
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The prime factors of 12 are : 2 x 2 x 3.....and 3 is the largest
4 1/3 is a mixed number because it has a whole number and a fraction.
$0.15x>$5.00
x>33.3333
So if she sends or receives more than 33 text messages, she will be paying more than $5.00.
Answer:
t = 460.52 min
Step-by-step explanation:
Here is the complete question
Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.
Solution
Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.
inflow = 0 (since the incoming water contains no dye)
outflow = concentration × rate of water inflow
Concentration = Quantity/volume = Q/200
outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.
So, Q' = inflow - outflow = 0 - Q/100
Q' = -Q/100 This is our differential equation. We solve it as follows
Q'/Q = -1/100
∫Q'/Q = ∫-1/100
㏑Q = -t/100 + c

when t = 0, Q = 200 L × 1 g/L = 200 g

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

㏑0.01 = -t/100
t = -100㏑0.01
t = 460.52 min