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boyakko [2]
2 years ago
6

John’s motorcycle average speed on motorcycle is 50 km/h. If he drives it for 3 hours how far does he travel? ​

Mathematics
2 answers:
marysya [2.9K]2 years ago
6 0
  • Speed=50km/h
  • Time=3h

Distance.

  • Avg. Speed×Time
  • 50(3)
  • 150km
professor190 [17]2 years ago
4 0

\qquad\qquad\huge\underline{{\sf Answer}}

Here we go ~

Average speed of John's motorcycle is 50 km/h and he drives for 3 hours.

We have been asked to find the distance travelled by John in this tike interval.

So, let's apply the formula :

\qquad \sf distance = avg. \: speed \times time

\qquad \sf  \dashrightarrow  \: d = 50 \times 3

\qquad \sf  \dashrightarrow  \: d =150 \: km

So, he travelled 150 km in 3 hours ~

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A teacher categorized students by whether they regularly completed their homework (or not) and whether they passed the course (o
ella [17]

Probabilities are used to determine the chances of an event.

Let P represents the event that a student passes the course, and C represents the event that a student completes the assignment

<u />

<u>(a) The probability a student passed the course given that they completed their homework?</u>

From the table, we have:

\mathbf{n(P\ n\ C) = 25}

\mathbf{n(C) = 30}

So, the required probability is:

\mathbf{Pr = \frac{n(P\ n\ C)}{n(C)}}

This gives

\mathbf{Pr = \frac{25}{30}}

\mathbf{Pr = 0.8333}

Express as percentage

\mathbf{Pr = 83.33\%}

Hence, the probability a student passed the course given that they completed their homework is 83.33%

<u>(b) The probability a student passed the course given that they didn't complete their homework</u>

From the table, we have:

\mathbf{n(P\ n\ C'= 7}

\mathbf{n(C') = 25}

So, the required probability is:

\mathbf{Pr = \frac{n(P\ n\ C')}{n(C')}}

This gives

\mathbf{Pr = \frac{7}{25}}

\mathbf{Pr = 0.28}

Express as percentage

\mathbf{Pr = 28\%}

Hence, the probability a student passed the course given that they didn't complete their homework is 28%

<u>(c) The probability a student completed their homework given that passed the course?</u>

From the table, we have:

\mathbf{n(P\ n\ C) = 25}

\mathbf{n(P) = 32}

So, the required probability is:

\mathbf{Pr = \frac{n(P\ n\ C)}{n(P)}}

This gives

\mathbf{Pr = \frac{25}{32}}

\mathbf{Pr = 0.78125}

Express as percentage

\mathbf{Pr = 78.125\%}

Approximate

\mathbf{Pr = 78.13\%}

Hence, the probability a student completed their homework given that they passed the course is 78.13%

<u>(d) The probability a student didn't complete their homework given that they didn't pass the course </u>

From the table, we have:

\mathbf{n(P'\ n\ C'= 18}

\mathbf{n(P) = 23}

So, the required probability is:

\mathbf{Pr = \frac{n(P'\ n\ C')}{n(P')}}

This gives

\mathbf{Pr = \frac{18}{23}}

\mathbf{Pr = 0.7826}

Express as percentage

\mathbf{Pr = 78.26\%}

Hence, the probability a student didn't complete their homework given that they didn't pass the course is 78.26%

Read more about probabilities at:

brainly.com/question/11234923

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