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2 months ago

Q13. A stone is dropped from a height of 5 km. The distance it falls through varies directly with the

Mathematics

2 answers:

Leona [12.6K]2 months ago

8 0

I’m not really certain, but I think C could be the answer; my apologies if that's inaccurate.

zzz [12.3K]2 months ago

3 0

Answer:

During the 5th second, the stone travels 36 meters. The correct answer is A.

Step-by-step explanation:

Proportions

Two quantities are considered proportional if one can be derived from the other by multiplying by a constant. If we denote these variables as y and x, it can be expressed as:

y=k.x

There are other forms of proportions where the relationship is non-linear. For example, if y varies with the square of x, we can represent this as:

$y=k.x^2$

According to the problem's parameters, the distance a stone drops from a height of 5 km is directly proportional to the square of the time taken during the descent. If d denotes the distance and t indicates time, we have:

$d=k.t^2$

To determine the constant k, we apply the provided information: the stone falls d=66 meters in t=4 seconds. By substituting:

$64=k.4^2=16k$

Solving this yields:

$k=64/16=4$

Using the value of k in the equation completes the model.

$d=4.t^2$

Next, we calculate the distance after t=5 seconds:

$d=4\cdot 5^2=4\cdot 25=100$

After 5 seconds, the stone has traveled 100 meters. However, we need the distance for just the 5th second, specifically between 4 and 5 seconds.

We already know the distance at t=4 seconds is 64 meters and at t=5 seconds it's 100 meters:

distance in the 5th second = 100 m - 64 m = 36 m

The stone covers 36 m in the 5th second. Select option A.

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Read 2 more answers

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(B) 0.1875

(C) Cannot be determined

Step-by-step explanation:

The time it takes for a student to finish a statistics quiz is uniformly distributed between 32 and 64 minutes.

Let's denote X as the duration needed for the student to complete the statistics quiz

Thus, X ~ U(32, 64)

The probability density function (PDF) for a uniform distribution is expressed as;

f(X) = $\frac{1}{b-a}$ , a < X < b where a = 32 and b = 64

The cumulative distribution function (CDF) is given by P(X <= x) = $\frac{x-a}{b-a}$

(A) The probability of a student taking longer than 59 minutes to complete the quiz = P(X > 59)

P(X > 59) = 1 - P(X <= 59) = 1 - $\frac{x-a}{b-a}$ = 1 - $\frac{59-32}{64-32}$ = $1-\frac{27}{32}$ = 0.15625

(B) The probability that a student completes the quiz between 37 and 43 minutes = P(37 <= X <= 43) = P(X <= 43) - P(X < 37)

P(X <= 43) = $\frac{43-32}{64-32}$ = $\frac{11}{32}$ = 0.34375

P(X < 37) = $\frac{37-32}{64-32}$ = $\frac{5}{32}$ = 0.15625

P(37 <= X <= 43) = 0.34375 - 0.15625 = 0.1875

= P(X = 44.74)

This probability cannot be calculated as it is a continuous distribution, which doesn't provide probabilities for specific points.

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