Answer:
L = the length of the field
W = the width of the field
The First Question:
The system is: L - 12 = w
2L + 2W = 76
Plug in L - 12 for W and you will get
2L + 2L - 24 = 76
4L -24 = 76
4L = 100
L = 25
To find W do L - 12 = w
25 - 12 = w
w = 13
Answer:
To see if multiple ratios are proportional, you could write them as fractions, reduce them, and compare them. If the reduced fractions are all the same, then you have proportional ratios.
Step-by-step explanation:
The excluded values are the values that the variable cannot take on because it makes the overall expression undefined.
For example, we can't have
in the expression
because
is undefined.
For both of these problems you do the same sort of thing, with some factorization along the way:
![\dfrac{9p^2}{4p^2-12p}=\dfrac{9p^2}{4p(p-3)}](https://tex.z-dn.net/?f=%5Cdfrac%7B9p%5E2%7D%7B4p%5E2-12p%7D%3D%5Cdfrac%7B9p%5E2%7D%7B4p%28p-3%29%7D)
If
or
, the denominator is 0 and we have an undefined expression, so these are the excluded values.
![\dfrac6{m^2+m-42}=\dfrac6{(m+7)(m-6)}](https://tex.z-dn.net/?f=%5Cdfrac6%7Bm%5E2%2Bm-42%7D%3D%5Cdfrac6%7B%28m%2B7%29%28m-6%29%7D)
If
or
, we get an undefined expression.
Answer:
Once.
Step-by-step explanation:
When the equation is graphed the x-axis is intersected/touched only once.
It crosses the x-axis at coordinates (-1 , 0)
When graphed it looks like this:
Using conditional probability, it is found that there is a 0.7873 = 78.73% probability that Mona was justifiably dropped.
Conditional Probability
In which
- P(B|A) is the probability of event B happening, given that A happened.
is the probability of both A and B happening.
- P(A) is the probability of A happening.
In this problem:
- Event A: Fail the test.
- Event B: Unfit.
The probability of <u>failing the test</u> is composed by:
- 46% of 37%(are fit).
- 100% of 63%(not fit).
Hence:
![P(A) = 0.46(0.37) + 0.63 = 0.8002](https://tex.z-dn.net/?f=P%28A%29%20%3D%200.46%280.37%29%20%2B%200.63%20%3D%200.8002)
The probability of both failing the test and being unfit is:
![P(A \cap B) = 0.63](https://tex.z-dn.net/?f=P%28A%20%5Ccap%20B%29%20%3D%200.63)
Hence, the conditional probability is:
![P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.63}{0.8002} = 0.7873](https://tex.z-dn.net/?f=P%28B%7CA%29%20%3D%20%5Cfrac%7BP%28A%20%5Ccap%20B%29%7D%7BP%28A%29%7D%20%3D%20%5Cfrac%7B0.63%7D%7B0.8002%7D%20%3D%200.7873)
0.7873 = 78.73% probability that Mona was justifiably dropped.
A similar problem is given at brainly.com/question/14398287