I think you divide.. find the mean of all of the numbers?
Let 
....(1)
Multiply both sides of equation (1) by 10.
.... (2)
Multiply both sides of equation (1) by 100.
... (3)
Subtract equation (2) from equation (3)
Hence, the required fraction form is 
.... Answer.
<span>At least 75% of the data will fall within 2 standard deviations of the mean.
This is tricky problem. Usually when you're dealing with standard deviation, you have a bell curve, or something close to a bell curve and for such a data distribution, there will be approximately 95% of the data within 2 standard deviations of the mean. But if you don't know that you have a bell curve, you have to fall back to Chebyshev’s Theorem, which states that at least 75% of the data points will fall within 2 standard deviations of the mean for any set of numbers.</span>
She bikes 761.25 miles in all to and from her job in 29 weeks.
Step-by-step explanation:
Distance from her house = 2.625 miles
As this is the distance of one side, she covers same distance for coming back.
Total distance for one day = 2.625 + 2.625 = 5.25 miles
She goes to work for 5 days per week.
29 weeks = 
Total distance covered in 29 weeks;
Total distance = Distance for one day * Days in 29 weeks
She bikes 761.25 miles in all to and from her job in 29 weeks.
Keywords: multiplication, distance
Learn more about multiplication at:
#LearnwithBrainly
Answer:
Step-by-step explanation:
Answer:
a) y-8 = (y₀-8) , b) 2y -5 = (2y₀-5)
Explanation:
To solve these equations the method of direct integration is the easiest.
a) the given equation is
dy / dt = and -8
dy / y-8 = dt
We change variables
y-8 = u
dy = du
We replace and integrate
∫ du / u = ∫ dt
Ln (y-8) = t
We evaluate at the lower limits t = 0 for y = y₀
ln (y-8) - ln (y₀-8) = t-0
Let's simplify the equation
ln (y-8 / y₀-8) = t
y-8 / y₀-8 =
y-8 = (y₀-8)
b) the equation is
dy / dt = 2y -5
u = 2y -5
du = 2 dy
du / 2u = dt
We integrate
½ Ln (2y-5) = t
We evaluate at the limits
½ [ln (2y-5) - ln (2y₀-5)] = t
Ln (2y-5 / 2y₀-5) = 2t
2y -5 = (2y₀-5)
c) the equation is very similar to the previous one
u = 2y -10
du = 2 dy
∫ du / 2u = dt
ln (2y-10) = 2t
We evaluate
ln (2y-10) –ln (2y₀-10) = 2t
2y-10 = (2y₀-10)
