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Anastasy [175]
3 years ago
10

Ashley, Frank, and Chang sent a total of 73 text messages over their cell phones during the weekend. Frank sent 7 more messages

than Ashley. Chang sent 2 times as many messages as Frank. How many messages did they each send?
Mathematics
1 answer:
jenyasd209 [6]3 years ago
6 0

Answer:

Ashley=13 messages

Frank=20 messages

Chang=2x+14=40 messages

Step-by-step explanation:

Number of texts sent by each person;

Ashley=x

Frank=x+7

Chang=2(x+7)=2x+14

Total number of text messages=Number sent by Ashley+Number sent by Frank+Number sent by Chang

where;

Total number of text messages=73

Replacing;

73= x+(x+7)+(2x+14)

x+x+2x+7+14=73

4x+21=73

4x=73-21

4x=52

x=52/4

x=13

Ashley=13 messages

Frank=x+7=13+7=20 messages

Chang=2x+14=(2×13)+14=40 messages

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Find the hcf*lcm for the number 105 and 125
Naily [24]

Answer:

15, 840.

Step-by-step explanation:

-HCM(GCM)

105's factor are 1, 3, 5, 7, <em>15</em>, 21, 35, 105.

125's factor are 1, 2, 3, 4, 5, 6, 8, 10, 12, <em>15</em>, 20, 24, 30, 40, 60, 120.

so, the answer is 15!

6 0
3 years ago
Write an equation<br> write an solution<br> and do a check step
melomori [17]

what is one easy addition equation?

1 + 1 = 2

how did you get this ?

you add 1 and 1 together and you will get your solution.

7 0
3 years ago
Find one counter example to show that the conjecture is false. angle 1 and angle 2 are​ supplementary, so one of the angles is a
Mrac [35]

Answer:

B. m ∠ 1 = 90° and m ∠ 2 = 90°

Step-by-step explanation:

For most situations, the conjecture would probably be true, but there is one exception that makes this statement false.

When two right angles are supplementary, none of them is acute.

For an angle to be acute it needs to be lesser than 90°, and for a pair of angles to be supplementary they should add up to exactly 180°.

With a pair of right angles (90° each), their sum adds up to 180° but neither of them are acute.

Therefore, the answer is B. m ∠ 1 = 90° and m ∠ 2 = 90°

4 0
4 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
A random sample of bolts is taken from inventory, and their lengths are measured. The average length in the sample is 5.3 inches
LekaFEV [45]

Answer:

L_x=5.3 inches

Step-by-step explanation:

Average length \=x =5.3 inches

Standard deviation \sigma=0.2 inches

Sample size n=50

Generally The point estimate for the mean length of all bolts in inventory is

L_x= \=x

L_x=5.3 inches

3 0
3 years ago
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