The formula for distance is equal to:
d = v * t
where d is distance, v is velocity or speed, and t is
time
Since the distance travelled by the two airplane is
similar, therefore we can create the initial equation:
v1 * t1 = v2 * t2
We know that v1 = 496, and v2 = 558 so:
496 t1 = 558 t2
or
x = 558 t2 / 496
We also know that airplane 1 travelled 30 minutes (0.5
hours) earlier than airplane 2, therefore:
x = t2 + 0.5
Hence,
496 (t2 + 0.5) = 558 t2
496 t2 + 248 = 558 t2
t2 = 4 hours
x = t2 + 0.5 = 4 + 0.5
x = 4.5 hours
So the equation is:
x = 558 t2 / 496
And the first plane travelled:
x = 4.5 hours
Answer:
what do u mean
Step-by-step explanation:
Answer:
Step-by-step explanation:
10x^2(3x^2 - 5x) Remove the brackets (Distributive Property)
10*3*x^2 * x^2 - 5x * 10x^2
30 x^(2 + 2) - 5*10x^(2+1)
30x^4 - 50x^3
A) x=2; i’m guessing if you meant by 4x+1=x+7
Since 5 winning numbers are draw and there are exactly 2 winning numbers, the other 3 numbers chosen have to be incorrect.
The 2 numbers picked right, there are 5C2=10 different possibilities.
The other 3 numbers are just picked from the rest of the 32 numbers. Getting there are 32C3=4960 different possibilities.
For each set of 2 correct winning numbers, you could have the 4960 different losing numbers to match up to make a unique set. This meant that there are 4690*10=46900 different total possibilities.
Now the total different outcomes of how you can choose the numbers are 37C5=435897 outcomes.
Now the way to find probabilities is want/total
The want is 46900 and the total is 435897
Doing the division you get the number rounded to the nearest thousandths as 0.107 or in percent form as
10.759% chance of picking exactly 2 winning numbers.
This seems like a competition problem of some sort therefore I assume that you already know what combinations in form nCk and permutation in form nPk means.