Question

1. if logan walks \\(\frac{7}{8}\\) mile in each \\(\frac{1}{3}\\) hour, how fast is he walking?

a. \\(\frac{7}{24}\\) miles per hour
b. \\(\frac{8}{21}\\) miles per hour
c. \\(2\frac{5}{8}\\) miles per hour
d. \\(3\frac{3}{7}\\) miles per hour

1. daniel has $25 to spend at the fair. the admission to the fair is $4, and the rides costs $1.50 each. daniel rides x rides at the fair. what inequality represents the situation?

a. \\(1.5x + 4 \leq 25\\)
b. \\(1.5x - 4 \leq 25\\)
c. \\(4x + 1.5 \leq 25\\)
d. \\(1.5x + 25 \leq 4\\)

1. a factory produces sunglasses at a rate of 500 pairs per day, every day. for the month of november, the factory cuts its daily by \\(\frac{1}{5}\\). how many pairs of sunglasses will the factory produce in the 30 days of november?

a. 3,000 pairs
b. 12,000 pairs
c. 15,000 pairs
d. 18,000 pair

1. the table shows the weekly change in the price of one gram of copper for four weeks.

cost of copper

week	change in price ($)
2	+ 2.15
3	-0.25
4	+2.50

Explanation:

Question 11

Step1: Recall speed formula

Speed \( v = \frac{\text{distance } d}{\text{time } t} \)
Given \( d = \frac{7}{8} \) mile, \( t = \frac{1}{3} \) hour.

Step2: Calculate speed

\( v=\frac{\frac{7}{8}}{\frac{1}{3}}=\frac{7}{8}\times3=\frac{21}{8}=2\frac{5}{8} \) miles per hour.

Total cost = admission cost + cost of rides.
Admission is $4, rides cost $1.50 each for \( x \) rides, so ride cost is \( 1.5x \).
Total cost must be ≤ $25 (what Daniel can spend).
Thus, \( 1.5x + 4 \leq 25 \).

Step1: Find new daily production

Original daily rate: 500 pairs.
Cut by \( \frac{1}{5} \), so new rate is \( 500\times(1 - \frac{1}{5}) = 500\times\frac{4}{5}=400 \) pairs per day.

Step2: Calculate November production

November has 30 days.
Total = \( 400\times30 = 12000 \) pairs.

Answer:

C. \( 2\frac{5}{8} \) miles per hour

Question 12