The answer to your question is: Yes, someone undoubtedly can.
Although you haven't asked to be told or shown how to solve it, I'm here
already, so I may as well stick around and go through it with you.
The sheet is telling you to find the solutions to two equations, AND THEN
DO SOMETHING WITH THE TWO SOLUTIONS. But you've cut off the
instructions in the pictures, so all we have are the two equations, and
you'll have to figure out what to do with their solutions.
<u>First equation:</u>
(2/5) x - 6 = -2
Add 6 to each side:
(2/5) x = 4
Multiply each side by 5:
2x = 20
Divide each side by 2 :
<u>x = 10</u>
<u>Second equation:</u>
-3y + 1/4 = 13/4
Subtract 1/4 from each side:
-3y = 12/4
Multiply each side by 4 :
-12 y = 12
Divide each side by -12 :
<u> y = -1</u>
the first option is the answer
Step-by-step explanation:
4c + 6a <= 120 [building hours]
4c + 4a <= 100 [testing hours]
4 * 20 + 6 * 6 <= 120
80 + 36 <= 120
116 <= 120
true, they have enough building hours
4 * 20 + 4 * 6 <= 100
80 + 24 <= 100
104 <= 100
false, they won't have enough testing hours
the first option is the answer
301 meters
Since 2cm=5m you would have to divide 120.4 by 2 to get 60.2 and times it by 5 to get 301
Answer:
21 child tickets
Step-by-step explanation:
For this problem, you'd use a system of equations.
First, define your variables.
x = # of child tickets sold
y = # of adult tickets sold
There were 4 times as many adult tickets (y) as child tickets (x) sold, so:
4 x = y
4 x - y = 0
93 (4 x - y = 0)
372 x - 93 y = 0
The total revenue was $909.30, adult tickets were $9.30 each, child tickets were $6.10 each, so:
6.10 x + 9.30 y = 909.30
10 (6.10 x + 9.30 y = 909.30)
61 x + 93 y = 9093
372 x - 93 y = 0
433 x = 9093
433 x / 433 = 9093 / 433
x = 21
The solution for this problem is:
We know the problem has the following given:
Sample size of 200
X = 182
And the probability of .9005; computation: 1 - .0995 = .9005
So in order to get the probability:
P (x >= 182) = 1 – 0.707134 = .292866 is the probability
that when 200 reservations
are recognized, there are more passengers showing up than there
are seats vacant.
The other solution is:
p (>= 182) = p(183) +
P(184) + P(185) + ... + P(199) + P(200) = 0.292866