Answer:
I think its 1/3/ and 4. But I dont know tho. just make sure with other.
Step-by-step explanation:
The correct answer is: [B]: "40 yd² " .
_____________________________________________________
First, find the area of the triangle:
The formula of the area of a triangle, "A":
A = (1/2) * b * h ;
in which: " A = area (in units 'squared') ; in our case, " yd² " ;
" b = base length" = 6 yd.
" h = perpendicular height" = "(4 yd + 4 yd)" = 8 yd.
___________________________________________________
→ A = (1/2) * b * h = (1/2) * (6 yd) * (8 yd) = (1/2) * (6) * (8) * (yd²) ;
= " 24 yd² " .
___________________________________________________
Now, find the area, "A", of the square:
The formula for the area, "A" of a square:
A = s² ;
in which: "A = area (in "units squared") ; in our case, " yd² " ;
"s = side length (since a 'square' has all FOUR (4) equal side lengths);
A = s² = (4 yd)² = 4² * yd² = "16 yd² "
_________________________________________________
Now, we add the areas of BOTH the triangle AND the square:
_________________________________________________
→ " 24 yd² + 16 yd² " ;
to get: " 40 yd² " ; which is: Answer choice: [B]: " 40 yd² " .
_________________________________________________
3(12)/10(12) + 7(10)/12(10) = 36/120 + 70/120
==> 18/60 + 35/60 = 53/60
Answer:
P(a junior or a senior)=1
Step-by-step explanation:
The formula of the probability is given by:

Where P(A) is the probability of occurring an event A, n(A) is the number of favorable outcomes and N is the total number of outcomes.
In this case, N is the total number of the students of statistics class.
N=18+10=28
The probability of the union of two mutually exclusive events is given by:

Therefore:
P(a junior or a senior) =P(a junior)+P(a senior)
Because a student is a junior or a senior, not both.
n(a junior)=18
n(a senior)=10
P(a junior)=18/28
P(a senior) = 10/28
P(a junior or a senior) = 18/28 + 10/28
Solving the sum of the fractions:
P(a junior or a senior) = 28/28 = 1