Answer:
A) attached below
B) 0.61
C) 0.47
Step-by-step explanation:
Given data:
Total number of lizards infected = 38
Of the 15 species B lizards 40% survived
For specie C one more survived than died
Out of the 24 lizards that died 1/3 were species A
<u>A) contingency table </u>
attached below
<u>B) Determine the proportion of these lizards in this study that were either specie A or Specie B </u>
P ( A or B ) = ( 8 + 15 ) / 38 = 0.605 ≈ 0.61
<u>C) determine the probability that specie C lizard did not study </u>
P ( not surviving | C ) = 7 / 15 = 0.466 ≈ 0.47
Answer: 1.25
Step-by-step explanation:
Given: A college-entrance exam is designed so that scores are normally distributed with a mean
= 500 and a standard deviation
= 100.
A z-score measures how many standard deviations a given measurement deviates from the mean.
Let Y be a random variable that denotes the scores in the exam.
Formula for z-score = 
Z-score = 
⇒ Z-score = 
⇒Z-score =1.25
Therefore , the required z-score = 1.25
Answer:
y = 8/p+q+4.
Step-by-step explanation: