Answer:
a) 0.06 = 6% probability that a person has both type O blood and the Rh- factor.
b) 0.94 = 94% probability that a person does NOT have both type O blood and the Rh- factor.
Step-by-step explanation:
I am going to solve this question treating these events as Venn probabilities.
I am going to say that:
Event A: Person has type A blood.
Event B: Person has Rh- factor.
43% of people have type O blood
This means that 
15% of people have Rh- factor
This means that 
52% of people have type O or Rh- factor.
This means that 
a. Find the probability that a person has both type O blood and the Rh- factor.
This is
With what we have
0.06 = 6% probability that a person has both type O blood and the Rh- factor.
b. Find the probability that a person does NOT have both type O blood and the Rh- factor.
1 - 0.06 = 0.94
0.94 = 94% probability that a person does NOT have both type O blood and the Rh- factor.
Answer:
<u>28.5 and 32.5</u>
Step-by-step explanation:
<em>create an equation to solve:</em>
x + (x+4) =61
2x + 4 = 61 combine like terms
2x = 57 subtract 4 from both sides
x = 28.5 divide 2 from both sides
28.5 is the smaller number
the larger number is 4 more so:
28.5 + 4 = 32.5
32.5 is the larger number
check your work:
28.5 + 32.5 = 61
Answer:
The soda costs 2.20 and the sandwich costs 7.70
Step-by-step explanation:
To find this, set the soda cost as x. We now know that the sandwich cost is 3.5x. Add these together and set equal to 9.90
x + 3.5x = 9.90
4.5x = 9.90
x = 2.20
This is the cost of the soda. Now we can multiply that by 3.5 to get the sandwich cost.
3.5 * 2.20 = 7.70
Answer:19/26
Step-by-step explanation:
Total number of students=26
Number of students that play basket ball=8
Number of students that play
baseball=11
Probability of playing both=6
Probability of playing either of the 2 games=8/26+11/26=19/26
X + x + 24 = 58
x = (58-24)/2=17
the first number is 17 and the second number is the sum of 17 and 28. Finding it has been left as an exercise for the reader.
