______________________________
<h3>A = m ÷ t × 100</h3><h3>= $100 ÷ $350 × 100</h3><h3>= <u>28.6%</u></h3><h3>THE GOAL WAS SURPASSED BY 28.6%</h3>
______________________________
Answer:
in what time will 24000 amount to rs.30000 at 10% p.a?
Answer:
The sum of the probabilities is greater than 100%; and the distribution is too uniform to be a normal distribution.
Step-by-step explanation:
The sum of the probabilities of a distribution should be 100%. When you add the probabilities of this distribution together, you have
22+24+21+26+28 = 46+21+26+28 = 67+26+28 = 93+28 = 121
This is more than 100%, which is a flaw with the results.
A normal distribution is a bell-shaped distribution. Graphing the probabilities for this distribution, we would have a bar up to 22; a bar to 24; a bar to 21; a bar to 26; and bar to 28.
The bars would not create a bell-shaped curve; thus this is not a normal distribution.
Since ![[0,4]=[0,1]\cup(1,4]](https://tex.z-dn.net/?f=%5B0%2C4%5D%3D%5B0%2C1%5D%5Ccup%281%2C4%5D)
, we can rewrite the integral as
Now there is no ambiguity about the definition of f(t), because in each integral we are integrating a single part of its piecewise definition:
Both integrals are quite immediate: you only need to use the power rule
to get
![\displaystyle \int_0^11-3t^2\;dt = \left[t-t^3\right]_0^1,\quad \int_1^4 2t\; dt = \left[t^2\right]_1^4](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E11-3t%5E2%5C%3Bdt%20%3D%20%5Cleft%5Bt-t%5E3%5Cright%5D_0%5E1%2C%5Cquad%20%5Cint_1%5E4%202t%5C%3B%20dt%20%3D%20%5Cleft%5Bt%5E2%5Cright%5D_1%5E4)
Now we only need to evaluate the antiderivatives:
![\left[t-t^3\right]_0^1 = 1-1^3=0,\quad \left[t^2\right]_1^4 = 4^2-1^2=15](https://tex.z-dn.net/?f=%5Cleft%5Bt-t%5E3%5Cright%5D_0%5E1%20%3D%201-1%5E3%3D0%2C%5Cquad%20%5Cleft%5Bt%5E2%5Cright%5D_1%5E4%20%3D%204%5E2-1%5E2%3D15)
So, the final answer is 15.
This is a division problem:
Each box holds 20 bags.
800 bags / 20 bags per box = # of boxes
40 boxes are needed.
