Answer:
yes
Step-by-step explanation:
because 7 is equal to 3 so the answer is correct
Answer:
0.86.
Step-by-step explanation:
Probability (None of the viruses will do any damage) = (1 - 0.3)(1-0.5)(1-0.6)
= 0.14.
Therefore the probability that at least one will do damage = 1 - 0.14 = 0.86.
Answer:
17 Units
Step-by-step explanation:
Since the y coordinate is the same for both points we only need to know the change in the x coordinates to find the distance between these two points.
5+12=17
So these two points are 17 units apart.
Answer: C, 2 3/5
Step-by-step explanation:
- Let's start by making these fractions easily subtractable
7 1/5 = 6 6/5 (1 is equal to 5/5 so we take that from the 7 and add it to the fraction)
4 3/5 remains the same because this is what we are subtracting
7 1/5 - 4 3/5 will be subsituted for 6 6/5 - 4 3/5=? because it's equivilant
I like to break this into two parts to make it less complicated, fraction then number
- Fraction: 6/5-3/5=3/5
- Whole Number 6-4=2
- Combine back into a mixed number
This will give you the proper fraction for the result of your equation
2+3/5=2 3/5
2 3/5
Answer:
a) 0.076
b) 0.5
c) 0.084
d) 0.25
e) 0.00024
f) 0.156
Step-by-step explanation:
a) In an ordinary deck there are 4 aces. The probability of getting an ace is given by:
![P=\frac{n}{N}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7Bn%7D%7BN%7D)
n: options for an ace = 4
N: total number of cards = 52
![P=\frac{4}{52}=0.076](https://tex.z-dn.net/?f=P%3D%5Cfrac%7B4%7D%7B52%7D%3D0.076)
The probability is 7.6%
b) On the first 100 positive integers, there are 50 odd integer numbers. Then you have:
![P=\frac{50}{100}=0.5](https://tex.z-dn.net/?f=P%3D%5Cfrac%7B50%7D%7B100%7D%3D0.5)
The probability of getting an odd number is 50%
c) From a total of 365 day in the year, May has 31 day. Then you have:
![P=\frac{31}{365}=0.084](https://tex.z-dn.net/?f=P%3D%5Cfrac%7B31%7D%7B365%7D%3D0.084)
The probability is 8.4%
d) In this case you have two events. To get an odd number with the sum of the number of two dices you need and odd number in one dice and a pair number in the other dice.
The probability is the product of the occurrence of both events. The probability for an odd number is 1/2 and for a pair number is 1/2.
![P=P_{odd}P_{pair}=(\frac{1}{2})(\frac{1}{2})=\frac{1}{4}=0.25](https://tex.z-dn.net/?f=P%3DP_%7Bodd%7DP_%7Bpair%7D%3D%28%5Cfrac%7B1%7D%7B2%7D%29%28%5Cfrac%7B1%7D%7B2%7D%29%3D%5Cfrac%7B1%7D%7B4%7D%3D0.25)
The probability is 25%
e) To obtain Head 6 times in a row, the probability is:
![P=\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}=0.00024](https://tex.z-dn.net/?f=P%3D%5Cfrac%7B1%7D%7B2%7D%2A%5Cfrac%7B1%7D%7B2%7D%2A%5Cfrac%7B1%7D%7B2%7D%2A%5Cfrac%7B1%7D%7B2%7D%2A%5Cfrac%7B1%7D%7B2%7D%2A%5Cfrac%7B1%7D%7B2%7D%3D0.00024)
Probability of 0.024%
f) To get Head 4 times from 5 flights, you first calculate the probability for 1 head in five flips:
![P=(\frac{1}{2})^5=\frac{1}{32}](https://tex.z-dn.net/?f=P%3D%28%5Cfrac%7B1%7D%7B2%7D%29%5E5%3D%5Cfrac%7B1%7D%7B32%7D)
There are 5 forms of getting 4 Heads for 5 flips, then you have:
![P=5(\frac{1}{32})=\frac{5}{32}=0.156](https://tex.z-dn.net/?f=P%3D5%28%5Cfrac%7B1%7D%7B32%7D%29%3D%5Cfrac%7B5%7D%7B32%7D%3D0.156)
Te probability is 15.6%