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olga2289 [7]
3 years ago
14

An athlete takes 33 minutes and 50 seconds to run 10 000 m.

Mathematics
1 answer:
lubasha [3.4K]3 years ago
3 0

Answer: 17.9 km/h

Step-by-step explanation:

Convert 33 minutes and 50 seconds to Hours:

= 33 ⁵⁰/₆₀ minutes

= 33.83 minutes

In hours that would be:

= 33.83 / 60

= 0.56 hours

10,000 meters in kilometers is:

= 10,000 / 1,000

= 10 kilometers

Speed = 10 / 0.56

= 17.9 km/h

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vazorg [7]
2)The goal is to isolate the variable and get is down to x<?. This is actually a two step inequality problem. 0.7x - 2 < 5.5       +2   +2 0.7x<7.5 This implies: x<10.714 The way to graph this is to place an open circle of the number 10.714 on the number line indicating this number is not included  as a solution to the inequality. Then draw the arrow going left from the open circle indicating all numbers from -∞ to 10.714.
8 0
3 years ago
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Cedric purchased a new refrigerator and stove at Home Depot for $3,729 on a deferred payment plan with no down
lesya [120]

Answer:

correct option is d. $242.81

Step-by-step explanation:

given data  

APR = 25.5% = \frac{25.5}{12}     = 2.125  

paid = $3,729

solution

we get here finance charge on the 1st month by multiplying 3,729 and now adding it to existing balance

so we get finance charge for the second and third months similarly as

APR ÷ 100 = \frac{2.125 }{100}  = 0.02125

so 1st  

= $3,729 × 0.02125  

= 79.25  

and  

$3,729  + $79.25 = $3808.24  

so for next  

= $3808.24 × 0.02125  

= 80.93

and  

$3808.24  + $80.93 = $3889.17  

so for next  

= $3889.17  × 0.02125  

= 82.64  

and  

$3889.17 + $ 82.64  = $3971.81

so  

finance charge =  3971.81 - 3729  

finance charge = 242.81  

so correct option is d. $242.81

8 0
3 years ago
On which interval is the average rate of change for bicycle production the greatest? A) 1950 to 1960 B) 1960 to 1970 C) 1970 to
Alborosie

Answer:

D) 1980 to 2000

Step-by-step explanation:

Finding the average rate of change in each interval to determine the greatest one.

Production per interval

1950-1960 = (21-11) \ million = 10 million

1960-1970 = (41-21)\ million = 20 million

1970-1980 = (71-41)\ million = 30 million

1980-2000 = (161-71)\ million= 90 million

Rate of change (1950-1960)= \frac{10}{11} \times 100= 90.90\%

Rate of change (1960-1970) = \frac{20}{21} \times 100= 95.23\%

Rate of change (1970-1980)=  \frac{30}{41} \times 100= 73.17\%

Rate of change (1980-2000)= \frac{90}{71} \times 100= 126.76\%

∴ rate of change between 1980 to 2000 is 126.78%

7 0
3 years ago
If there were 49 cars in a line that stretched 528 feet, what is the average car length? assume that the cars are lined up bumpe
lakkis [162]

If there were 49 cars in a line that stretched 528 feet, what is the average car length? assume that the cars are lined up bumper-to-bumper

Answer: We are given there are 49 cars

Also these 49 cars are in a line stretched 528 feet.

Now the average length of the car is:

Average length = \frac{(length-covered-by-49-cars)}{Number-of-cars}

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Therefore, the average length of car is 10.78 feet

                 

6 0
3 years ago
Solve this equation for A: A ÷ 2 = 4
Natali5045456 [20]
The answer to the equation is 8
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3 years ago
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