Answer: The answers are
Step-by-step explanation: Given in the question that Lana completed a 10-mile race in 90 minutes by running and walking both. She usually runs at a pace of 8.5 minutes per mile and walks at a pace of 12 minutes per mile. We need to find the number of minutes for which she ran and walk throughout the race.
Let, Lana runs for 'x' miles and walks for 'y' miles. Then, we have
Multiplying the first equation by 8.5 and subtracting from the second equation, we have
Therefore,
Thus, time for which Lana ran is
And, the time for which Lana walks is
Thus, the answers are
3400
Step-by-step explanation:
70×60=4200
10+20=30
70-30=40
40×20=800
4200-800=3400
Answer:
72mm^2 (72 mm squared)
Step-by-step explanation:
so i would work out the little triangle which is 8x6/2=<em><u>24m</u></em>m^2
then do 24x8/2 which is <u><em>96m</em></u>m^2
then take away 24 from 96 which gives the answer of <u><em>72mm^2</em></u>
Let N be the number of blue counters. This implies that 3N counters are red, and there are 4N counters in total.
Assuming that you don't reinsert the first counter, for the first pick, you have N blue counters over 4N total counters, so you'll pick a blue counter with probability
For the second pick, you're left with N-1 blue counters over 4N-1 counters, so you'll pick a blue counter with probability
The probability of picking two blue counters with two picks is the product of the two probabilities:
And we want this to equal 1/20, so we have
We can expand the left hand side and solve for N:
So, there are 4 blue counters and 12 red counters, for a total of 16 counters in the bag.
We can indeed verify that the probabilities of picking the two blue counters is
Answer:
(a) C = +9
(b) C = -9
Step-by-step explanation:
Given:
The equation to solve is given as:
In order to solve this for 'C', we have to isolate 'C' on the left side of the equation.
Adding -15 on both sides, we get:
Now, taking square root on both the sides, we get:
Therefore, there are two values of 'C'.
Therefore, options (a) and (b) are correct.