Rewrite the equation to isolate the variable.. df= g- 10 for d
1 answer:
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The two-way table is attached.
We know that 63 people took the survey, and 22 of them were left-handed. This means that 63-22=41 of them are right handed.
Out of the 63 total, 37 are left brain dominant; this means that 63-37=26 are right brain dominant.
Of the 26 that are right brain dominant, 21 are right handed; this means 26-21=5 are left handed.
Of the 22 left-handed people, 5 are right brain dominant; this means 22-5 = 17 are left brain dominant.
Of the 37 left brain dominant people, 17 are left handed; this means 37-17=20 are right handed.
We know that 63 people took the survey, and 22 of them were left-handed. This means that 63-22=41 of them are right handed.
Out of the 63 total, 37 are left brain dominant; this means that 63-37=26 are right brain dominant.
Of the 26 that are right brain dominant, 21 are right handed; this means 26-21=5 are left handed.
Of the 22 left-handed people, 5 are right brain dominant; this means 22-5 = 17 are left brain dominant.
Of the 37 left brain dominant people, 17 are left handed; this means 37-17=20 are right handed.
Answer:
The probability that these four will contain at least two more high-risk drivers than low-risk drivers is 0.0488.
Step-by-step explanation:
Denote the different kinds of drivers as follows:
L = low-risk drivers
M = moderate-risk drivers
H = high-risk drivers
The information provided is:
P (L) = 0.50
P (M) = 0.30
P (H) = 0.20
Now, it given that the insurance company writes four new policies for adults earning their first driver’s license.
The combination of 4 new drivers that satisfy the condition that there are at least two more high-risk drivers than low-risk drivers is:
S = {HHHH, HHHL, HHHM, HHMM}
Compute the probability of the combination {HHHH} as follows:
P (HHHH) = [P (H)]⁴
= [0.20]⁴
= 0.0016
Compute the probability of the combination {HHHL} as follows:
P (HHHL) = × [P (H)]³ × P (L)
= 4 × (0.20)³ × 0.50
= 0.016
Compute the probability of the combination {HHHM} as follows:
P (HHHL) = × [P (H)]³ × P (M)
= 4 × (0.20)³ × 0.30
= 0.0096
Compute the probability of the combination {HHMM} as follows:
P (HHMM) = × [P (H)]² × [P (M)]²
= 6 × (0.20)² × (0.30)²
= 0.0216
Then the probability that these four will contain at least two more high-risk drivers than low-risk drivers is:
P (at least two more H than L) = P (HHHH) + P (HHHL) + P (HHHM)
+ P (HHMM)
= 0.0016 + 0.016 + 0.0096 + 0.0216
= 0.0488
Thus, the probability that these four will contain at least two more high-risk drivers than low-risk drivers is 0.0488.