Answer:
0.47178 = 47.178%
Step-by-step explanation:
Put simply, this question is asking the probability of having rain atleast twice in the next 5 days. The easiest way that comes to mind is reverse count - find the probability of it not happening. There are two cases for this:
Raining only once:
This has a .7^4 * 0.3 * 5 chance
Raining no times:
This has a .7^5 chance
Adding these together you get 0.52822
This is the inverse probability, so to find the actual probability, subtract it from 1.
P = 0.47178
Answer:
B) Infinitely many solutions.
Step-by-step explanation:
-x + 4y = 24
4y = x + 24
y = 1/4x + 6
It is the same equation twice.
-4 and 4 on the number line have an absolute value of 4
Hope it helps.
Answer:
Tristan has a 17/24 lb fraction of a pound of grapes now.
Step-by-step explanation:
Given
- The number of grapes Tristan had at home = 1/3 lb
- The number of grapes Tristan had at the store = 3/8 lb
To determine
What fraction of a pound of grapes does Tristan have now?
In order to get the total fraction of a pound of grapes, all we need is to add the number of grapes Tristan had at home and the number of grapes Tristan had at the store.
i.e.
Current number grapes = 1/3 + 3/8
Least Common Multiple of 3, 8: 24
Adjust fractions based on L.C.M
= 8/24 + 9/24
Apply the fraction rule: ![\frac{a}{c}+\frac{b}{c}=\frac{a+b}{c}](https://tex.z-dn.net/?f=%5Cfrac%7Ba%7D%7Bc%7D%2B%5Cfrac%7Bb%7D%7Bc%7D%3D%5Cfrac%7Ba%2Bb%7D%7Bc%7D)
= (8 + 9) / 24
= 17/24 lb
Therefore, Tristan has a 17/24 lb fraction of a pound of grapes now.
Answer:
we conclude that:
![x^2+6x+9>2x^2+14\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:1](https://tex.z-dn.net/?f=x%5E2%2B6x%2B9%3E2x%5E2%2B14%5Cquad%20%3A%5Cquad%20%5Cbegin%7Bbmatrix%7D%5Cmathrm%7BSolution%3A%7D%5C%3A%26%5C%3A1%3Cx%3C5%5C%3A%5C%5C%20%5C%3A%5Cmathrm%7BInterval%5C%3ANotation%3A%7D%26%5C%3A%5Cleft%281%2C%5C%3A5%5Cright%29%5Cend%7Bbmatrix%7D)
Hence, (1, 5) is the solution in interval notation.
Please also check the attached graph.
Step-by-step explanation:
Given the inequality expression
![\left(x+3\right)^2>\:2\left(x^2+7\right)](https://tex.z-dn.net/?f=%5Cleft%28x%2B3%5Cright%29%5E2%3E%5C%3A2%5Cleft%28x%5E2%2B7%5Cright%29)
as
(x + 3)² = x² + 6x + 9
2(x² + 7) = 2x² + 14
so
![\:x^2+6x+9\:>\:2x^2+14](https://tex.z-dn.net/?f=%5C%3Ax%5E2%2B6x%2B9%5C%3A%3E%5C%3A2x%5E2%2B14)
rewriting in the standard form
![-x^2+6x-5>0](https://tex.z-dn.net/?f=-x%5E2%2B6x-5%3E0)
Factor -x² + 6x - 5: - (x - 1) (x - 5)
![-\left(x-1\right)\left(x-5\right)>0](https://tex.z-dn.net/?f=-%5Cleft%28x-1%5Cright%29%5Cleft%28x-5%5Cright%29%3E0)
Multiply both sides by -1 (reverse the inequality)
![\left(-\left(x-1\right)\left(x-5\right)\right)\left(-1\right)](https://tex.z-dn.net/?f=%5Cleft%28-%5Cleft%28x-1%5Cright%29%5Cleft%28x-5%5Cright%29%5Cright%29%5Cleft%28-1%5Cright%29%3C0%5Ccdot%20%5Cleft%28-1%5Cright%29)
Simplify
![\left(x-1\right)\left(x-5\right)](https://tex.z-dn.net/?f=%5Cleft%28x-1%5Cright%29%5Cleft%28x-5%5Cright%29%3C0)
so
![1](https://tex.z-dn.net/?f=1%3Cx%3C5)
Therefore, we conclude that:
![x^2+6x+9>2x^2+14\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:1](https://tex.z-dn.net/?f=x%5E2%2B6x%2B9%3E2x%5E2%2B14%5Cquad%20%3A%5Cquad%20%5Cbegin%7Bbmatrix%7D%5Cmathrm%7BSolution%3A%7D%5C%3A%26%5C%3A1%3Cx%3C5%5C%3A%5C%5C%20%5C%3A%5Cmathrm%7BInterval%5C%3ANotation%3A%7D%26%5C%3A%5Cleft%281%2C%5C%3A5%5Cright%29%5Cend%7Bbmatrix%7D)
Hence, (1, 5) is the solution in interval notation.
Please also check the attached graph.