Answer:
They buy 6 hotdogs and 5 popcorn
Step-by-step explanation:
Assume that they buy x hotdogs and y popcorn
∵ They buy a total of 11 hotdogs and popcorn
∵ The number of hotdogs is x and the number of popcorn is y
∴ x + y = 11 ⇒ (1)
∵ Hot dogs cost $2.50 each
∵ Popcorn costs $1.00 each
∵ They spend $20 on hot dogs and popcorn
→ Multiply x by 2.5 and y by 1, add the products and equate the sum by 20
∴ 2.5(x) + 1(y) = 20
∴ 2.5x + y = 20 ⇒ (2)
Now we have a system of equations to solve it
→ Subtract equation (1) from equation (2)
∵ (2.5x - x) + (y - y) = (20 - 11)
∴ 1.5x + 0 = 9
∴ 1.5x = 9
→ Divide both sides by 1.5 to find x
∴ x = 6
→ Substitute the value of x in equation (1) to find y
∵ 6 + y = 11
→ Subtract 6 from both sides
∴ 6 - 6 + y = 11 - 6
∴ y = 5
∴ They buy 6 hotdogs and 5 popcorn
Answer:
Wassup Obito
Step-by-step explanation:
I would have done it if you had asked for only 1 question but you have asked someone to do more than 10 q
You must be mad to think someone will do the entire thing
Mark me brainliest
Answer:
24
Step-by-step explanation:
30 secs = 0.5 min
thus 84 / 3.5 min = 168 / 7 = 24
Stand by !
The next thing you will read in this block
will be two monomials whose greatest
common factor is 4m :
68 x n p m q r y s
and
124 a b g c f m t z
____________________________
And now . . .
Here are two more:
8 m x
and
12 k m
(a) Average time to get to school
Average time (minutes) = Summation of the two means = mean time to walk to bus stop + mean time for the bust to get to school = 8+20 = 28 minutes
(b) Standard deviation of the whole trip to school
Standard deviation for the whole trip = Sqrt (Summation of variances)
Variance = Standard deviation ^2
Therefore,
Standard deviation for the whole trip = Sqrt (2^2+4^2) = Sqrt (20) = 4.47 minutes
(c) Probability that it will take more than 30 minutes to get to school
P(x>30) = 1-P(x=30)
Z(x=30) = (mean-30)/SD = (28-30)/4.47 ≈ -0.45
Now, P(x=30) = P(Z=-0.45) = 0.3264
Therefore,
P(X>30) = 1-P(X=30) = 1-0.3264 = 0.6736 = 67.36%
With actual average time to walk to the bus stop being 10 minutes;
(d) Average time to get to school
Actual average time to get to school = 10+20 = 30 minutes
(e) Standard deviation to get to school
Actual standard deviation = Previous standard deviation = 4.47 minutes. This is due to the fact that there are no changes with individual standard deviations.
(f) Probability that it will take more than 30 minutes to get to school
Z(x=30) = (mean - 30)/Sd = (30-30)/4.47 = 0/4.47 = 0
From Z table, P(x=30) = 0.5
And therefore, P(x>30) = 1- P(X=30) = 1- P(Z=0.0) = 1-0.5 = 0.5 = 50%
