Answer:
Probability that at most 50 seals were observed during a randomly chosen survey is 0.0516.
Step-by-step explanation:
We are given that Scientists conducted aerial surveys of a seal sanctuary and recorded the number x of seals they observed during each survey.
The numbers of seals observed were normally distributed with a mean of 73 seals and a standard deviation of 14.1 seals.
Let X = <u><em>numbers of seals observed</em></u>
The z score probability distribution for normal distribution is given by;
Z =
~ N(0,1)
where,
= population mean numbers of seals = 73
= standard deviation = 14.1
Now, the probability that at most 50 seals were observed during a randomly chosen survey is given by = P(X
50 seals)
P(X
50) = P(
) = P(Z
-1.63) = 1 - P(Z < 1.63)
= 1 - 0.94845 = <u>0.0516</u>
The above probability is calculated by looking at the value of x = 1.63 in the z table which has an area of 0.94845.
If the question is 3^(2/3),
The answer is the cube root of 9.
Answer:
Reasonable estimation for constant of variation is 0.25 kWh per day.
Step-by-step explanation:
We are given the following information in the question:
- The graph represents the function where electricity usage.
- Electricity usage in kilowatts per hour of a clock radio varies directly with the number of days.
- The x-axis shows the number of days and usage in kilo-watt per hour is showed on the y-axis.
- Some coordinates of the graph are: (0,0), (2,0.5) and (6,1.5)
Formula for constant of variation:

Putting the values from the coordinates (2,0.5) and (6,1.5), we get:

Hence, reasonable estimation for constant of variation is 0.25 kWh per day.
Answer:
(x+3)(x+2)
Step-by-step explanation: