Answer:
Step-by-step explanation:
x+y=10
x=10-y
x=10-5
x=5
3y= -3x+30
3(10-y)= -3(10-y)+30
30-3y= -30+3y+30
30+30-30=6y
30=6y
y=5
3a) 3:1 or 3/1
3b) 3:4 or 3/4
4) 125 pages a day
5) 10 skittle per ounce in a bag
6) Market: $23.60/5= $4.72*7= $33.04, Orchard: $32.76/7= $4.68*5= $23.40, Which? Orchard because it is cheaper to buy 7 there than to buy 7 at market place because it is $00.30 cents cheaper at Orchard
7) Because money is always on top of time, amount, weight, etc.
Step-by-step explanation:
3x(x2 + 2x – 6) + 4(x2
– 6x + 2)
Answer: 0.1824
Step-by-step explanation:
Given : The mileage per day is distributed normally with
Mean : 
Standard deviation : 
Let X be the random variable that represents the distance traveled by truck in one day .
Now, calculate the z-score :-

For x= 132 miles per day.

For x= 159 miles per day.

Now by using standard normal distribution table, the probability that a truck drives between 132 and 159 miles in a day will be :-

Hence, the probability that a truck drives between 132 and 159 miles in a day =0.1824
Answer:
1.3x
Step-by-step explanation:
X=1
1+0.3=1.3 (don't forget to add 'X')
So, the answer is 1.3x
hope this helps!