Answer:
Yes, we can put 1 kg of the soul in the container.
Step-by-step explanation:
Consider the provided information.
2/5 kg of soul fills 1/3 of a container.
It means if we put 2/5 kg of the soul 3 times in the container then the container will be full.
![\frac{2}{5}+\frac{2}{5}+\frac{2}{5}=\frac{6}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B5%7D%2B%5Cfrac%7B2%7D%7B5%7D%2B%5Cfrac%7B2%7D%7B5%7D%3D%5Cfrac%7B6%7D%7B5%7D)
It means
kgs of soul fill the container.
is greater than 1 so it means we can put 1 kg of the soul in the container.
Answer:
![11\text{ cm}^2](https://tex.z-dn.net/?f=11%5Ctext%7B%20cm%7D%5E2)
Step-by-step explanation:
Given: In a parallelogram ABCD, diagonals intersect at O and ar(ABCD) is
.
We need to find the area of triangle AOB.
We know that each diagonal divide the parallelogram in two equal parts and diagonals bisect each other.
It means both diagonals divide the parallelogram in 4 equal parts.
![ar(AOB)=\dfrac{1}{4}\times ar(ABCD)](https://tex.z-dn.net/?f=ar%28AOB%29%3D%5Cdfrac%7B1%7D%7B4%7D%5Ctimes%20ar%28ABCD%29)
![ar(AOB)=\dfrac{1}{4}\times 44](https://tex.z-dn.net/?f=ar%28AOB%29%3D%5Cdfrac%7B1%7D%7B4%7D%5Ctimes%2044)
![ar(AOB)=11](https://tex.z-dn.net/?f=ar%28AOB%29%3D11)
Hence, the values of ar(AOB) is
.
Answer:
b
Step-by-step explanation:
5 because increased by means to add 2+3
and quotient means the answer to a division problem 10/2 so ur answer is 5
I got D.
There's a few ways to solve it; I prefer using tables, but there are functions on a TI-84 that'll do it for you too. The logic here is, you have a standard normal distribution which means right away, the mean is 0 and the standard deviation is 1. This means you can use a Z table that helps you calculate the area beneath a normal curve for a range of values. Here, your two Z scores are -1.21 and .84. You might notice that this table doesn't account for negative values, but the cool thing about a normal distribution is that we can assume symmetry, so you can just look for 1.21 and call it good. The actual calculation here is:
1 - Z-score of 1.21 - Z-score .84 ... use the table or calculator
1 - .1131 - .2005 = .6864
Because this table calculates areas to the RIGHT of the mean, you have to play around with it a little to get the bit in the middle that your graph asks for. You subtract from 1 to make sure you're getting the area in the middle and not the area of the tails in this problem.