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umka21 [38]
2 years ago
14

Prior to hosting an international soccer match, the local soccer club needs to replace the artificial turf on their field with g

rass turf. The grass turf will cost $ 6.75 per square meter. If the field is 0.101 km by 0.065 km, how much will it cost the club to add the grass turf to their field
Mathematics
1 answer:
r-ruslan [8.4K]2 years ago
3 0

Answer:

$44,313.75

Step-by-step explanation:

First, we need to turn the km into meters. There are 1000 meters in a km meaning we need to multiply each of these values by 1000.

0.101 km * 1000= 101 meters

0.065 km * 1000 =  65 meters

Now we need to multiply these values together to find the area of the field

101m * 65m = 6,565m^{2}

Finally, we need to multiply this area by the cost per square meter

6,565m^{2} / 6.75 = $44,313.75

The total repair will cost $44,313.75

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855 cft

Step-by-step explanation:

(22/7)*4*4*17=855cft(rounded )

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1 year ago
Students at a private liberal arts college are classified as being freshmen, sophomores, juniors or seniors, and also according
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2 years ago
HELP!!!! 2 dot plots. Both number lines go from 0 to 10. Plot 1 is titled fifth grade. There are 2 dots above 1, 3 above 2, 1 ab
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Answer:

5th grade mean - 4.67

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7th-grade median - 3.5

7 0
1 year ago
Read 2 more answers
Ramina found the length of two pieces of ribbon to be 47.6 inches and 39.75 inches. Which is an effective estimation strategy fo
lukranit [14]

Answer:

<h3>Add 47.6 and 39.75, then round the answer</h3>

Step-by-step explanation:

If Ramina found the length of two pieces of ribbon to be 47.6 inches and 39.75 inches, the effective strategy of finding the sum of the two lengths is to:

1) First is to add the two values together

47.6 + 39.75

= (47+0.6)+(39+0.75)

= (47+39)+(0.6+0.75)

= 86 + 1.35

= 87.35

2) Round up the answer to nearest whole number.

87.35 ≈ 87 (note that we couldn't round up to 88 because the values after the decimal point wasn't up to 5)

Option C is correct

7 0
2 years ago
Read 3 more answers
The fish population in a pond with carrying capacity 1200 is modeled by the following logistic equation where N(t) denotes the n
MArishka [77]

9514 1404 393

Answer:

  (dN)/(dt) = (0.4)/(1200)N(1200-N) -50

  142 fish

Step-by-step explanation:

A) The differential equation is modified by adding a -50 fish per year constant term:

  (dN)/(dt) = (0.4)/(1200)N(1200-N) -50

__

B) The steady-state value of the fish population will be when N reaches the value that makes dN/dt = 0.

  (0.4/1200)(N)(1200-N) -50 = 0

  N(N-1200) = -(50)(1200)/0.4) . . . . rewrite so N^2 has a positive coefficient

  N^2 -1200N + 600^2 = -150,000 +600^2 . . . . complete the square

  (N -600)^2 = 210,000 . . . . . simplify

  N = 600 + √210,000 ≈ 1058

This steady-state number of fish is ...

  1200 - 1058 = 142 . . . . below the original carrying capacity

8 0
2 years ago
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