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.
10500-4500=6000
6000/250=24
So the answer is 24 months, or 2 years.
6p=13.50
there are 6 notebooks and it costs $13.50 altogether. P represents the cost of one notebook, therefore the equation would be 6p=13.5. It can be rearranged to p=13.5/6
Answer:
92%
Step-by-step explanation:
The percentage of apartments rented for less than $600 is given by the sum of all of the apartments in the 300 to 400, 400 to 500 and 500 to 600 intervals divided by the total number of apartments sampled (250).
92% of the apartments rented for less than $600.
The first thing we are going to assume for this case is that the tree and the post are located in the same place.
From that place, both cast a shadow in the same direction.
We then have two similar triangles.
Therefore, we have the following relationship:
From here, we clear the value of x.
We have then:
Rewriting:
Answer:
the tree is 30 feet tall
