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goblinko [34]
3 years ago
11

There are 240 coins in a jar, all dimes and quarters. The value of the coins in the jar is $27.60. There are nine times as many

dimes as there are quarters. How many dimes (d ) are in the jar?
Mathematics
1 answer:
sammy [17]3 years ago
4 0
I believe it is 28 dimes
.25 × 30 = 7.50
.10 × 190 = 19
19 + 7.50 = 26.50
Then, you add 4 more quarters which is 27.50 and the 28 dime makes it 27.60
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F plus 4 divided by h= 6
Sergeeva-Olga [200]
F is 8 h is 2 hope that helpssss
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an NFL team kicker misses 2 out of every 11 field goals if he missed 8 filed goals throughout the season how many did he attempt
Aleks [24]
For this equation, you want to do it in fractions/ratios to properly solve it. You would have his average misses out of every field goal and his real missed attempts over total. It would look like this

\frac{2}{11}  =   \frac{8}{x}

You want to solve for x since x is the total amount of field goals that he attempted. You can do this by doing cross multiplication:

(2)(x) = (8)(11)

From here you can get:

2x = 88

Divide each side by 2 to isolate x and you get:

x=  44

So he made a total of 44 field goals.
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3 years ago
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The graph of an absolute value function opens up and has a vertex of (0, -3).
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Absolute value function domain is all real numbers

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The deepest part of the Mariana trench (the deepest trench in the world’s oceans) is at an elevation of -11.033 kilometers. The
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8 0
3 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
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