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mr Goodwill [35]
3 years ago
13

Steven jogs around a circular field with a diameter of 70 yd. Find the area of the field.

Mathematics
2 answers:
sertanlavr [38]3 years ago
6 0

Answer:

3848.46 :)

Step-by-step explanation:

ValentinkaMS [17]3 years ago
5 0
Area: 3848.4510006475

1225π
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Does anyone know this?
Bumek [7]

Answer:

B

Step-by-step explanation:

Initial mean = 156, Changed mean = 163.92

So, mean increases. Only condition that satisfies this is B.

7 0
3 years ago
60.3 60.44 witch one is greater
Sindrei [870]
60.44 is greater than 60.3
-you can see this when you subtract 60.3 from 60.44 this would leave a 0.11 which is a positive number.
-thus meaning 60.44 is larger/greater.
4 0
2 years ago
Susie has a bag of marbles containing 3 Red, 7 Green, and 10 Blue marbles.
morpeh [17]

Answer:

1) a) 0.0945

b) 0.1062

2) a) 0.0138

b) 0.00081

3) a) 0.00094

b) 0.00083

4) 301.68 cents

5) 0.0056

Step-by-step explanation:

Since there are 3 Red, 7 Green, and 10 Blue marbles.

Total number of marbles N = 20

Probability (Red) = 3/20

Probability (Green) = 7/20

Probability (Blue) = 10/20

1) the probability of picking 5 marbles and getting at least one red marble

A) with replacement

P(at least 1 red of 5)= (3/20 * 17/20 * 17/20 * 17/20 * 17/20) + (3/20 * 3/20 * 17/20 * 17/20 * 17/20) + (3/20 * 3/20 * 3/20 * 17/20 * 17/20)

P(at least 1 red of 5) = 0.0783 +0.0138 + 0.0024 = 0.0945

B) without replacement

P(at least 1 red of 5) = ( 3/20 * 17/19* 16/18 * 15/17 * 14/16) + (3/20 * 2/19 * 17/18 * 16/17 * 15/16) + (3/20 * 2/19 * 1/18 * 17/17 * 16/16)

P(at least 1 red of 5) = 0.0921 + 0.01316 + 0.0009 = 0.1062

2) the probability of picking 6 marbles having 2 of each color

A) with replacement

P( 6, 2 of each) = 3/20 * 3/20 * 7/20 * 7/20 * 10/20 * 10/20

P( 6, 2 of each) = 0.0138

B) without replacement

P( 6, 2 of each) = 3/20 * 2/19 * 7/18 * 6/17 * 10/16 * 9/15

P( 6, 2 of each) = 0.00081

3) Pick 8 marbles: 4 green and 4 blue

A) with replacement

P(8, 4G and 4 B) = 7/20*7/20*7/20*7/20*10/20*10/20*10/20

P(8, 4G and 4 B) = 0.00094

B) without replacement

P(8, 4G and 4 B) = 7/20*6/19*5/18*4/17*10/16*9/15*8/14*7/13 = 0.00083

4) getting at least 6 marbles of the same color

Only Green and Blue marbles are more than 6

P(at least 6 marbles of the same color) = (7/20*6/19*5/18*4/17*3/16*2/15) + (10/20*9/19*8/18*7/17*6/16*5/15)

= 0.00018 + 0.00542

P(at least 6 marbles of the same color) = 0.0056

Cost of 6 .marbles= 6* 50 cents

C = 300 cents

Therefore, You will have to pay

(1 + 0.0056) 300 cent = 301.68 cents to be sure of getting at least 6 marbles of the same color

5) getting at least 6 marbles of the same color

Only Green and Blue marbles are more than 6

P(at least 6 marbles of the same color) = (7/20*6/19*5/18*4/17*3/16*2/15) + (10/20*9/19*8/18*7/17*6/16*5/15)

= 0.00018 + 0.00542

P(at least 6 marbles of the same color) = 0.0056

6 0
3 years ago
Plz show your work. Thx!! <br> 9m=6
grandymaker [24]
So first you would divide the 9 on both sides then 6 divided by 9 would be .6 repeating so m=.6 repeating.

5 0
3 years ago
2. A solar lease customer built up an excess of 6,500 kilowatt hours (kwh) during the summer using his solar
tia_tia [17]

9514 1404 393

Answer:

  a)  E = 6500 -50d

  b)  5000 kWh

  c)  the excess will last only 130 days, not enough for 5 months

Step-by-step explanation:

<u>Given</u>:

  starting excess (E): 6500 kWh

  usage: 50 kWh/day (d)

<u>Find</u>:

  a) E(d)

  b) E(30)

  c) E(150)

<u>Solution</u>:

a) The exces is linearly decreasing with the number of days, so we have ...

  E(d) = 6500 -50d

__

b) After 30 days, the excess remaining is ...

  E(30) = 6500 -50(30) = 5000 . . . . kWh after 30 days

__

c) After 150 days, the excess remaining would be ...

  E(150) = 6500 -50(150) = 6500 -7500 = -1000 . . . . 150 days is beyond the capacity of the system

The supply is not enough to last for 5 months.

3 0
2 years ago
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