All categories

3 years ago

The sum of the digits of a two digit number is 14. The difference between the tens digit and the units digit is 2.

Mathematics

2 answers:

galben [10]3 years ago

7 0

Answer:

see explanation

Step-by-step explanation:

Sum of the 2 digits x and y is 14, thus

x + y = 14

The difference between the tens digit x and the ones digit y is 2, thus

x - y = 2

The system of equations representing the problem is

x + y = 14 → (1)

x - y = 2 → (2)

Add the2 equations term by term to eliminate y

2x = 16 ( divide both sides by 2 )

x = 8

Substitute x = 8 into (1)

8 + y = 14 ( subtract 8 from both sides )

y = 6

The 2 digit number is 86

Rama09 [41]3 years ago

6 0

Step-by-step explanation:

the system of equations representing the word problem is

x+y=14, y-x=2

let x be the tens digit and y be the ones digit

solve simultaneously the two equations,

x+y-x+y=14+2

2y=16

y=8

now put in y=8

8-x=2

x=8-2

x=6

so the digit is 68

You might be interested in

There are three local factories that produce radios. Each radio produced at factory A is defective withprobability .02, each one

diamong [38]

Answer:

The probability is 0.02667

Step-by-step explanation:

Let's call D1 the event that the first radio is defective and D2 the event that the second radio is defective.

So, if we select both radios any factory, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

P(D2/D1) = P(D2∩D1)/P(D1)

Taking into account that 0.02 is the probability that a radio produced at factory A is defective, P(D2/D1) for factory A is:

$P(D2/D1)_A=\frac{0.02*0.02}{0.02} =0.02$

At the same way, if both radios are from factory B, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

$P(D2/D1)_B=\frac{0.01*0.01}{0.01} =0.01$

Finally, if both radios are from factory C, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

$P(D2/D1)_C=\frac{0.05*0.05}{0.05} =0.05$

So, if the radios are equally likely to have been any factory, the probability to select both radios from any of the factories A, B or C are respectively:

P(A)=1/3

P(B)=1/3

P(C)=1/3

Then, the probability P(D2/D1) that the second radio is defective given that the first one is defective is:

$P(D2/D1)=P(A)P(D2/D1)_A+P(B)P(D2/D1)_B+P(C)P(D2/D1)_C$