Answer:
3.41 feet
Step-by-step explanation:
Area = Length × Breath
Area of the rectangular lawn = 100 × 50
= 5000 feet²
The sidewalk must occupy an area no more than 10% of the total lawn area.
So, the area of the sidewalk would be not more than = 10% × 5000
= 0.10 × 5000
= 500 feet²
Let the width of the sidewalk = x feet
area of the side walk = (L×W of the long way) + ((L-x)×W of the short way)
(100 × x) + ((50 - x) × x) < 500
100x + (50-x)(x) < 500
-x² + 150x < 500
-x² + 150x = 500
-x² + 150x - 500 = 0
By using quadratic formula
or
x = 3.41089 ≈ 3.41 feet or x = 146.58
Therefore, width of the sidewalk would be 3.41 feet.
Answer:
<em>The probability that the second ball is red is 71%</em>
Step-by-step explanation:
<u>Probabilities</u>
We know there are 5 red balls and 2 green balls. Let's analyze what can happen when two balls are drawn in sequence (no reposition).
The first ball can be red (R) or green (G). The probability that it's red is computed by
The probability is's green is computed by
If we have drawn a red ball, there are only 4 of them out of 6 in the urn, so the probability to draw a second red ball is
If we have drawn a green ball, there are still 5 red balls out of 6 in the urn, so the probability to draw a red ball now is
The total probability of the second ball being red is
The probability that the second ball is red is 71%
6x - 3.5 = -15.5
Add 3.5 to both sides
6x = -12
Divide by 6
x = -2
Carly has the correct response
Answer:
1) Event A = 2/3
Event B = 1/2
2) 1/2
Step-by-step explanation:
1)
Event A :
No. we need on dice = 4
Total numbers on dice = 6
Hence sample space of the event = 6
P( getting 4) = 4/6 = 2/3
Event B :
A coin has a head & a tail.
Hence sample space of the event = 2
But as we need tail only ,
P ( getting Tail ) = 1/2 [ if only tossed once ]
2)
Total numbers on a die = 6
Total no. of odd numbers on die = 3 (∵ 1 , 3 & 5 are odd )
Sample space of this event = 6
P (getting an odd number) = 3/6 = 1/2