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.
Answer
Given,
Mass of bag = 6 Kg
Area covered by the bag = 2.5 x 5 = 12.5 m²
a) Bags required to cover area = 7.5 x 10.2
= 76.5 m²
Number of bags, N = 
N = 6.12 = 7 (approx.)
7 Bags of 6 Kg fertilizers will be needed to cover the area.
b) Cost of 6 Kg fertilizers = $15.50
Cost of 7 bags of 6 Kg fertilizers = 7 x $15.50
= $108.50.
It is (13 + 34), (43 + 4). Hope this helps!!! It couldn’t be the other ones because they don’t include all of the numerals.