Answer:
25 dollars
Step-by-step explanation:
1/4*60=15
1/3*60=20
20+15=35
60-35=25
Liza is driving to her sister's house 360 miles away. After 4 hours, Liza is 2/3 of the way there.
How many more hours does Liza have to drive?
<em><u>Answer:</u></em>
Liza need to drive 2 more hours to reach her sister house
<em><u>Solution:</u></em>
Given that Liza is driving to her sister's house 360 miles away
So total distance = 360 miles
After 4 hours, Liza is 2/3 of the way there.

So Liza has covered 240 miles in 4 hours
Let us find the speed
<em><u>The relation between distance and speed is given as:</u></em>


Thus speed = 60 miles per hour
Assuming speed is same throughout the journey, let us calculate the time taken to complete remaining distance
Remaining distance = 360 - 240 = 120 miles
Now distance = 120 miles and speed = 60 miles per hour

Thus Liza need to drive 2 more hours to reach her sister house
Answer:
yes
Step-by-step explanation:
Yes this triangle is right angled and can be justified using the Pythagorean theorem. The theorem states that:
. Substitute using the given values:
(9^2)+(12^2)=15^2
81 + 144 = 255
255 = 255.
It satisfies the theorem therefore it is a right angle triangle.
Answer:
The flower garden width would be 6 ft
Step-by-step explanation:
A = 12 ft^2 = x(x + 4) becomes:
x^2 + 4x - 12 = 0, or (by completing the square) x^2 + 4x + 4 - 4 - 12 = 0
Thus, we have (x + 2)^2 - 16 = 0, or (x + 2)^2 = 16
Taking the square root of both sides yields x + 2 = ±4.
In this context only a positive x value makes sense: x = 2 + 4 => x = 6.
The flower garden width would be 6 ft. The other dimension would be 6 - 4, or 2, ft.
if p is the probability that some bin ends up with 3 balls and q is the probability that every bin ends up with 4 balls. pq is 16.
First, let us label the bins with 1,2,3,4,5.
Applying multinomial distribution with parameters n=20 and p1=p2=p3=p4=p5=15 we find that probability that bin1 ends up with 3, bin2 with 5 and bin3, bin4 and bin5 with 4 balls equals:
5−2020!3!5!4!4!4!
But of course, there are more possibilities for the same division (3,5,4,4,4) and to get the probability that one of the bins contains 3, another 5, et cetera we must multiply with the number of quintuples that has one 3, one 5, and three 4's. This leads to the following:
p=20×5−2020!3!5!4!4!4!
In a similar way we find:
q=1×5−2020!4!4!4!4!4!
So:
pq=20×4!4!4!4!4!3!5!4!4!4!=20×45=16
thus, pq = 16.
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