Answer:
45
Step-by-step explanation:
10+10+10+10+5=45
Answer: 64
Step-by-step explanation:
If EFGH is a parallelogram, its diagonals bisect each other meaning EJ and JG are congruent.
EJ = JG
4w + 4 = 2w + 18
Subtract 2w from both sides
2w + 4 = 18
Subtract 4 from both sides
2w = 14
Divide both sides by 2
w = 7
EG = EJ + JG
EG = 4w + 4 + 2w + 18
EG = 6w + 22
Now that we know what 'w' is, we can substitute it for 7
EG = 6 (7) + 22
EG = 42 + 22 = 64
The measure of EG is 64.
Hope that helped!
Let's first denote:
F - a favorable response
F' - an unfavorable response
S - successful
We know that:
![P(F) = 0.6\\\\P(F')=0.4\\\\P(F\cap S)=0.5\\\\P(F'\cap S)=0.3](https://tex.z-dn.net/?f=P%28F%29%20%3D%200.6%5C%5C%5C%5CP%28F%27%29%3D0.4%5C%5C%5C%5CP%28F%5Ccap%20S%29%3D0.5%5C%5C%5C%5CP%28F%27%5Ccap%20S%29%3D0.3)
So, from the conditional probability, we can calculate:
![P(S|F) = \dfrac{P(S\cap F)}{P(F)} = \dfrac{0.5}{0.6}=\dfrac{5}{6}\approx\boxed{0.83}](https://tex.z-dn.net/?f=P%28S%7CF%29%20%3D%20%5Cdfrac%7BP%28S%5Ccap%20F%29%7D%7BP%28F%29%7D%20%3D%20%5Cdfrac%7B0.5%7D%7B0.6%7D%3D%5Cdfrac%7B5%7D%7B6%7D%5Capprox%5Cboxed%7B0.83%7D)
Answer E.
Answer:
0.13591.
Step-by-step explanation:
We re asked to find the probability of randomly selecting a score between 1 and 2 standard deviations below the mean.
We know that z-score tells us that a data point is how many standard deviation above or below mean.
To solve our given problem, we need to find area between z-score of -2 and -1 that is
.
We will use formula
to solve our given problem.
![P(-2](https://tex.z-dn.net/?f=P%28-2%3Cz%3C-1%29%3DP%28z%3C-2%29-P%28z%3C-1%29)
Using normal distribution table, we will get:
![P(-2](https://tex.z-dn.net/?f=P%28-2%3Cz%3C-1%29%3D0.15866-0.02275)
![P(-2](https://tex.z-dn.net/?f=P%28-2%3Cz%3C-1%29%3D0.13591)
Therefore, the probability of randomly selecting a score between 1 and 2 standard deviations below the mean would be 0.13591.
Answer:
Probability would be 8/18 or 4.44%
Step-by-step explanation:
If its spun twice it acts as if it has 18 sections meaning you multiply the number of reds and number of squares total by two in this situation and put them over each other!