Y + 18 = a(x + 3)^2
<span>0 + 18 = a(0 + 3)^2 </span>
<span>18 = 9a </span>
<span>2 = a
</span>
<span>y = 2(x + 3)^2 - 18 </span>
<span>0 = 2(x + 3)^2 - 18 </span>
<span>18 = 2(x + 3)^2 </span>
<span>9 = (x + 3)^2 </span>
<span>± 3 = x + 3 </span>
<span>-3 ± 3 = x </span>
<span>x = 0 or -6
</span>
<span>(0, 0) and (-6, 0)</span>
13.6 is greater bcuz it is 13.60 not 13.06
Answer:
Step-by-step explanation:
Since the incubation times are approximately normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = incubation times of fertilized eggs in days
µ = mean incubation time
σ = standard deviation
From the information given,
µ = 19 days
σ = 1 day
a) For the 20th percentile for incubation times, it means that 20% of the incubation times are below or even equal to 19 days(on the left side). We would determine the z score corresponding to 20%(20/100 = 0.2)
Looking at the normal distribution table, the z score corresponding to the probability value is - 0.84
Therefore,
- 0.84 = (x - 19)/1
x = - 0.84 + 19 = 18.16
b) for the incubation times that make up the middle 97% of fertilized eggs, the probability is 97% that the incubation times lie below and above 19 days. Thus, we would determine 2 z values. From the normal distribution table, the two z values corresponding to 0.97 are
1.89 and - 1.89
For z = 1.89,
1.89 = (x - 19)/1
x = 1.89 + 19 = 20.89 days
For z = - 1.89,
- 1.89 = (x - 19)/1
x = - 1.89 + 19 = 17.11 days
the incubation times that make up the middle 97% of fertilized eggs are
17.11 days and 20.89 days
