1/2in = 16 mi
2 3/4 in without whole numbers [(denominator ×whole) + numerator] / denominator
4×2 = 8 +3 = 11/4
Find common denominator for 1/2 and 11/4 (common D Is 4)
Multiply 1/2 by 2/2
2/4 in = 16 mi divide this by 2
2/4 ×1/2 = 1/4
16 /2 = 8
1/4 = 8
1/4 goes into 11/4 11 times
Multiply 8 times 11
88 miles (D)
End zone to end zone is = length so his scale is 6 in and in real life it is 360 ft
To find scale divide 360 by 6
1in = 60 ft (B)
160 ft wide
(divide 160 by 60 ft ) fraction
160/60 is a fraction now simplify
16/6
Common factors of both numbers is 2
Simplify by 2
8/3
Make into whole number fraction
3 goes into 8 two times (2×3=6)(2 becomes whole number) while there's is 2 left over( 8 -6=2) (becoming the numerator)
2 2/3 in (B)
120 ft long
80 ft wide
24 in is scale of "long"
Divide 120 by 24
120÷24=5
So 5in in scale equals 1ft in real life
To find width in scale divide 80 by 5
80/5= 16in (A)
If you have any trouble understanding feel free to let me know I will try to explain as best as I can :)
Yes it is correct, 7/12 is the equivalent to C. 7 divided by 12
86 feet
Assuming the court is rectangular in shape then
area = length × width
length = area / width = 
= 86 feet
Answer:
B. 1/3 or 33.3%
Step-by-step explanation:
<em>H-</em> 1 ,2 ,3, 4, 5, 6
<em>T-</em> 1, 2, 3, 4, 5, 6
Multiples of <em>3:</em> 3, 6, 9, 12, etc..
<em>2/6 simplified is 1/3 </em>
So, the probability that the penny will land on heads and the number cube will land on a multiple of 3, is 1/3 or 33.3%
Answer:
0.8413 is the required probability.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 4.00 centimeters
Standard Deviation, σ = 0.60 centimeters
Sample size, n = 16
We are given that the distribution of average diameter is a bell shaped distribution that is a normal distribution.
Formula:
Standard error due to sampling =
P(diameter of sample is more than 3.85 centimeter)
P(x > 3.85)
Calculation the value from standard normal z table, we have,
0.8413 is the probability that the the average diameter of sample of 16 sand dollars is more than 3.85 centimeters.
