Answer:
Let's define the variables:
A = price of one adult ticket.
S = price of one student ticket.
We know that:
"On the first day of ticket sales the school sold 1 adult ticket and 6 student tickets for a total of $69."
1*A + 6*S = $69
"The school took in $150 on the second day by selling 7 adult tickets and student tickets"
7*A + 7*S = $150
Then we have a system of equations:
A + 6*S = $69
7*A + 7*S = $150.
To solve this, we should start by isolating one variable in one of the equations, let's isolate A in the first equation:
A = $69 - 6*S
Now let's replace this in the other equation:
7*($69 - 6*S) + 7*S = $150
Now we can solve this for S.
$483 - 42*S + 7*S = $150
$483 - 35*S = $150
$483 - $150 = 35*S
$333 = 35*S
$333/35 = S
$9.51 = S
That we could round to $9.50
That is the price of one student ticket.
Answer:
The answer is 11.1
Step-by-step explanation:
First you need to figure out what all of the possible combinations of the two dice being thrown then divide it by the number of faces on both dice and you get your probability.
Answer:v= x a h add then multiply
Step-by-step explanation:
Well 19 would be 399 and for 20. the discount is 12.87 and the price of it after the discount is 26.13
Answer:
Step-by-step explanation:
Given that for a sample of eight bears, researchers measured the distances around the bears' chests and weighed the bears.
r = linear correlation coeff = 0.952
H_0: r =0\\
H_1 :r\neq 0
(Two tailed test)
r difference = 0.952
n=8
Std error = \sqrt{\frac{1-r^2}{n-2} } =0.12496
Test statistic t = 0.952/0.12496 = 7.618
Alpha = 0.05
df = 6
p value = 0.000267
This implies H0 is rejected.
There exists a linear relation between the variables and r cannot be 0
0.952^2 = 0.906=90.6% of variation in weight can be explained by the linear relationship between weight and chest size
