Using the second equation, let’s find x
x + 7y = 8
x = 8 - 7y
Substitute x in the other equation
3/8(8 - 7y) + 1/3y = 17/24
3 - 21/8y + 1/3y = 17/24
-55/24y = -55/24
y = 1
Plug in y in any equation
x + 7(1) = 8
x + 7 = 8
x = 1
x = 1 and y = 1
Answer:
$8.50
Step-by-step explanation:
ok so if you divide 3.40 by 5 you get .68 and you multiply that to get 8.16 then you divide .68 to get the cost of half a pound and you get .34 so you add that to 8.16 and get 8.5.
your answer is $8.50
hope this helps!
Which data set has an outlier? 25, 36, 44, 51, 62, 77 3, 3, 3, 7, 9, 9, 10, 14 8, 17, 18, 20, 20, 21, 23, 26, 31, 39 63, 65, 66,
umka21 [38]
It's hard to tell where one set ends and the next starts. I think it's
A. 25, 36, 44, 51, 62, 77
B. 3, 3, 3, 7, 9, 9, 10, 14
C. 8, 17, 18, 20, 20, 21, 23, 26, 31, 39
D. 63, 65, 66, 69, 71, 78, 80, 81, 82, 82
Let's go through them.
A. 25, 36, 44, 51, 62, 77
That looks OK, standard deviation around 20, mean around 50, points with 2 standard deviations of the mean.
B. 3, 3, 3, 7, 9, 9, 10, 14
Average around 7, sigma around 4, within 2 sigma, seems ok.
C. 8, 17, 18, 20, 20, 21, 23, 26, 31, 39
Average around 20, sigma around 8, that 39 is hanging out there past two sigma. Let's reserve judgement and compare to the next one.
D. 63, 65, 66, 69, 71, 78, 80, 81, 82, 82
Average around 74, sigma 8, seems very tight.
I guess we conclude C has the outlier 39. That one doesn't seem like much of an outlier to me; I was looking for a lone point hanging out at five or six sigma.
Answer:
(a) Stops, turns around and starts travelling back towards home.
(b) 8 km/hour.
Step-by-step explanation:
(b) Jo travels 4 kilometres in 1/2 hour = 9 km/hour.
