Answer:
-0.25, 3/12, 1/4
Step-by-step explanation:
Absolute value is a numbers distance away from zero.
0.4 does not work
-0.25 has and absolute value of 0.25
3/12=1/4=0.25
1/4=0.25
The curve for f(x) and g(x) have the same shape but only differs in their position relative to the y and x-axis. With respect to the y-axis, it shifted 5 units upward. With respect to the x-axis, it shifted 9 units to the right. Since the right hand side is equal to y, we add +5 to the right. Similarly, since the left hand side is equal to x, we add +9 to the left. If we transpose this to the right,it becomes -9. So the equation for g(x) is: <span>g(x)=(1/2)^x−9+5</span>
Answer: 6/12 are white, 3/12 are colored and 3/12 are albino.
Step-by-step explanation: If the horses are white and their parents are ccww (albino) and CCWw (white horse), according to Mendel's premises, they both must be CcWw, since the crossing provides one C from one parent and other c from the other parent, one W and the other w. Using Mendel's chess and the principle of independent segregation, the crossing between CcWw results in the following fenotypical ratio:
1/16 CCWW (lethal)
2/16 CCWw (white)
2/16 CcWW (lethal)
4/16 CcWw (white)
1/16 CCww (normal)
2/16 Ccww (normal)
2/16 ccWw (albino)
1/16 ccWW (lethal)
1/16 ccww (albino)
Excluding the 4 individuals that have the lethal locus, we have 6/12 that are white (2/12 + 4/12) and 3/12 (1/12 + 2/12) that are colored. Also, there are 3/12 of albino individuals as well.
Using the binomial distribution, it is found that there is a:
a) The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.
b) The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.
c) The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.
-----------
For each chipmunk, there are only two possible outcomes. Either they will live to be 4 years old, or they will not. The probability of a chipmunk living is independent of any other chipmunk, which means that the binomial distribution is used to solve this question.
Binomial probability distribution
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem:
- 0.96516 probability of a chipmunk living through the year, thus
Item a:
- Two is P(X = 2) when n = 2, thus:
The probability that two randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.93153 = 93.153%.
Item b:
- Six is P(X = 6) when n = 6, then:
The probability that six randomly selected 3-year-old male chipmunks will live to be 4 years old is 0.80834 = 80.834%.
Item c:
- At least one not living is:
The probability that at least one of six randomly selected 3-year-old male chipmunks will not live to be 4 years old is 0.19166 = 19.166%. This probability is not unusual, as it is greater than 5%.
A similar problem is given at brainly.com/question/24756209
Answer:
x= 361
Step-by-step explanation: