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horsena [70]
2 years ago
7

46. An olympic athlete has just completed a four mile run. His average time was 6 minutes per mile. If he ran the first and last

mile in 5 minutes, and the other 2 were the same time as each other, what was the average time for the 2 miles in the middle
Mathematics
1 answer:
uranmaximum [27]2 years ago
4 0

Answer:

7 minutes

Step-by-step explanation:

average time = 6 minutes per mile

sum = average * count = 6 minutes / mile * 4 miles = 24 minutes

first + last mile = 5 + 5 = 10 minutes

first + second + third + last mile = sum = 24 minutes

second + third = sum - first - last = 24 - 5 - 5 = 24 - 10 = 14 minutes

average of 2 miles in the middle = sum / count = 14 minutes / 2 = 7 minutes

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Rachel made 28 hats. Rachel made 7 times as many hats as Greg. Let n be the number of hats that Greg made.
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28/7=n is what I believe would be the answer

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What is 12-9×3×4+3 in three different ways
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4 0
3 years ago
List all subsets of ta, b, c, d, e) containing a but not containing b
fomenos

Answer:

(a), (a,c), (a,d), (a,e), (a,c,d), (a,c,e), (a,d,e), (a,c,d,e)

Step-by-step explanation:

We are given the set (a,b,c,d,e).

Total number of subsets of the above set are 2^5 = 32

Subsets:

φ

(a,b,c,d,e)

(a), (b), (c), (d), (e)

(a,b), (a,c), (a,d), (a,e), (b,c), (b,d), (b,e), (c,d), (c,e), (d,e)

(a,b,c), (a,b,d), (a,b,e), (a,c,d), (a,c,e), (a,d,e), ( b,c,d), (b,c,e), (b,d,e), (c,d,e)

(a,b,c,d), (a,b,c,e), (a,b,d,e), (a,c,d,e), (b,c,d,e)

Subset having a but not b :

(a), (a,c), (a,d), (a,e), (a,c,d), (a,c,e), (a,d,e), (a,c,d,e)

7 0
3 years ago
Items for a fundraiser are packaged in small boxes shaped like rectangular prisms that are inches long, inches wide, and 8 inche
allochka39001 [22]

Answer:

- The number of small boxes that will fill the large box 1 = 64

- The number of small boxes that will fill the large box 2 = 56

Step-by-step explanation:

Complete Question

Items for a fundraiser are packaged in small boxes shaped like rectangular prisms that are 4.5 inches long, 4.5 inches wide, and 8 inches tall. To transport the items to an event, you want to know how many of the small boxes will fit in larger boxes. The larger boxes are available in two sizes. Large Box 1 is 24.25 inches long, 18 inches wide, and 24 inches tall. Large Box 2 is 20.5 inches long, 18.5 inches wide, and 24 inches tall. Both the small and large boxes must remain upright.

Solution

To know how many of the small boxes will fit in larger boxes, we need to obtain the volumes of the small box, large box 1 and large box 2.

Volume of a cuboid = L × W × H

For the small box,

Length = L = 4.5 inches

Width = W = 4.5 inches

Height = H = 8 inches

Volume of the small box = 4.5 × 4.5 × 8 = 162 in³

For large box 1,

Length = L = 24.25 inches

Width = W = 18 inches

Height = H = 24 inches

Volume of the large box 1 = 24.25 × 18 × 24 = 10,476 in³

For large box 2

Length = L = 20.5 inches

Width = W = 18.5 inches

Height = H = 24 inches

Volume of the large box 2 = 20.5 × 18.5 × 24 = 9,102 in³

The number of small boxes that'll fill the large box 1 = (10,476/162) = 64.667 = 64 small boxes (rounded down because the fraction cannot be forced into the large box 1.

The number of small boxes that will fill the large box 2 = (9,102/162) = 56.185 = 56 small boxes.

Hope this Helps!!!

7 0
3 years ago
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