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3 years ago

3 Elena and Jada are 32 miles apart on a path when they start

Mathematics

1 answer:

kumpel [21]3 years ago

8 0

Answer:

The hours it takes Elena to meet Jada = 4 hours

Step-by-step explanation:

The distance between Elena and Jada = 32 miles

The speed with which Elena runs = 6 miles per hour

The speed with which Jada walks = 2 miles per hour

Let t, represent the time it takes until Elena and Jada meets, we have;

Distance = Speed × Time

The total distance covered by both Elena and Jada before they meet = The distance between them = 32 miles

The distance covered by Elena before meeting Jada = Elena's speed × The time that elapsed for them to meet

∴ The distance covered by Elena before meeting Jada = 6 mph × t

The distance covered by Jada before meeting Elena = Jada's speed × The time that elapsed for them to meet

∴ The distance covered by Jada before meeting Elena = 2 mph × t

The total distance covered by both Elena and Jada before they meet = The distance covered by Elena before meeting Jada + The distance covered by Jada before meeting Elena = 32 miles

∴ The total distance covered by both Elena and Jada before they meet = 6 mph × t + 2 mph × t = 32 miles

t × (6 mph + 2 mph) = 32 miles

t × 8 mph = t × 8 miles/hour = 32 miles

t = 32 miles/(8 miles/hour) = 4 hours

The time it takes until Elena and Jada meets = t = 4 hours

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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: $0.5^{4} \times 0.1 \times 0.4^{5} = 6.4\times 10^{-5} = 0.000064 = 0.0064\%$ .