The figure shown is made up of a triangle and a square. The perimeter of the figure is 56 inches. Complete
1 answer:
Image of figure is missing so i have attached it.
Answer:
The total length of three sides of the square = 36 in
Side length of square = 12 in
A_triangle = 48 in²
Area of square = 144 in²
Total area of figure = 192 in²
Step-by-step explanation:
Total perimeter = 56 inches
Let the side of the square be s.
Thus, perimeter of the 3 sides of the square = 3s
Therefore, adding all the external sides of the figure, we have;
3s + 10 + 10 = 56
3s + 20 = 56
3s = 56 - 20
3s = 36
s = 36/3
s = 12
Area of triangle is given by the formula;
A_triangle = ½bh
Where;
b is base
h is height
Thus;
A_triangle = ½ × 12 × 8
A_triangle = 48 in²
Area of square = 12 × 12 = 144 in²
Total area of figure = 144 + 48 = 192 in²
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Answer:
a) 85.31%
b) 56 minutes
c) 17 days
d) 0.1721
Step-by-step explanation:
In order to make the calculations easier, let's <em>standardize the curve</em> by doing the change
where is the average trip-time and
the standard deviation and
P(X≤ t) is the probability that the trip takes less than t minutes.
a)
Here we are looking for the value of P(X>20).
For X=20 we have Z = (20-24)/3.8 = -1.05
So, we want the area under the standard normal curve for Z > -1.05
that we can compute either using a table or a computer and we find this area equals 0.8531
So, he arrives late to work 85.31% of the times.
(See picture 1 attached)
b)
In this case we are looking for a value t of time such that
P(X≥ t) = 20% = 0.2
So, we are seeking a value Z such that the area under the normal curve to the left of Z equals 0.2
By using a table or a computer we find Z = 0.842
(See picture 2 attached)
By inserting this value in the equation on standardization
So 20% of the longest trips takes 55.996 ≅ 56 minutes
c)
Since the average of days he arrives late is 85.31%, it is expected that in 20 work days he arrives late 85.31% of 20, which equals 17 days.
d)
Since the probability that he does not arrive late is 1-0.8531 = 0.1469 and the probability of arriving early is independent of the previous trip,
the probability of arriving early n days in a row is
0.1496*0.1496*...*0.1496 n times and
the probability of being early most than 10 trips in 20 days is
