Is 1/5 a solution of X+3/4-x=13/15

Mathematics

2 answers:

BartSMP [9]3 years ago

8 0

I do not think it is, because when you plug in 1/5 for X you do not get 13/15

WITCHER [35]3 years ago

3 0

(x + 3)/(4 - x) = 13/15
15(x + 3) = 13(4 - x)
15x + 45 = 52 - 13x
15x + 13x = 52 - 45
28x = 7
x = 7/28 = 1/4

Therefore, 1/5 is not a solution

You might be interested in

For every hour that Noah studies, his test score goes up 3 points. Explain which is the independent variable and which is the de

Arturiano [62]

The independent variable is hours and the score or points is the dependent variable .equation is s=3h

4 0

3 years ago

Read 2 more answers

A traffic light at a certain intersection is green 50% of the time, yellow 10% of the time, and red 40% of the time. A car appro

kati45 [8]

Answer:

$6.4\times 10^{-5} = 0.000064 = 0.0064\%$ .

Step-by-step explanation:

Probability that the car encounters a green light on the first day: $50 \% = 0.5$ .

To meet the question's conditions, the car needs to encounter another green light on the second day. Given that the colors of the light on each day are "independent," the chance that there's a green light followed by another green light will be

$(0.5) \times 0.5 = 0.25$ .

Condition is met on the first two days and green light on the third day: $(0.5 \times 0.5) \times 0.5 = 0.125$ .
Condition is met on the first three days and green light on the fourth day: $(0.5 \times 0.5 \times 0.5) \times 0.5$ .

To meet the condition on the fifth day, there needs to be a yellow light. The probability that the condition is met on the first four days and on the fifth day will be $(0.5 \times 0.5 \times 0.5 \times 0.5) \times 0.1 = 0.5^{4} \times 0.1$ .

To meet the condition on the sixth day, all prior days should meet the conditions. Besides, there needs to be a red light on the sixth day. $(0.5^{4} \times 0.1) \times 0.4$

Seventh day: $(0.5^{4} \times 0.1 \times 0.4 ) \times 0.4$
Eighth day: $(0.5^{4} \times 0.1 \times 0.4^2 ) \times 0.4$
Ninth day: $(0.5^{4} \times 0.1 \times 0.4^3 ) \times 0.4$
Tenth day: $(0.5^{4} \times 0.1 \times 0.4^4 ) \times 0.4 = 0.5^{4} \times 0.1 \times 0.4^{5}$

The question asks that the condition be met on all ten days. As a result, the probability of meeting the condition will be equal to the probability on the tenth day: $0.5^{4} \times 0.1 \times 0.4^{5} = 6.4\times 10^{-5} = 0.000064 = 0.0064\%$ .