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Rasek [7]
2 years ago
8

You are to drive to an interview in another town, at a distance of 270 km on an expressway. The interview is at 11:15 a.m. You p

lan to drive at 100 km/h, so you leave at 8:00 a.m. to allow some extra time. You drive at that speed for the first 110 km, but then construction work forces you to slow to 42.0 km/h for 43.0 km. What would be the least speed needed for the rest of the trip to arrive in time for the interview?
Mathematics
1 answer:
Yuri [45]2 years ago
6 0

Answer:

the least required speed  is 103.88 km/h

Step-by-step explanation:

Data provided in the question:

Total distance = 270 km

Total time to reach = Time between 8 a.m and 11:15 a.m i.e 3.25 hours

Now,

For the distance of 110 km speed was 100 km/h

therefore, the time taken to cover 110 km = \frac{\textup{Distance}}{\textup{speed}}

= \frac{\textup{110 km}}{\textup{100 km/h}}

= 1.1 hour

For another 43 km speed was 42 km/h

therefore, the time taken to cover 43 km =  \frac{\textup{Distance}}{\textup{speed}}

= \frac{\textup{43 km}}{\textup{42 km/h}}

= 1.0238 hours

Now,

The distance left to be covered = 270 - 110 - 43 = 117 km

Time left = 3.25 h - 1.1 h - 1.0238 h = 1.1262 h

Thus required speed = \frac{\textup{117 km}}{\textup{1.1262 h}}

= 103.88 km/h

Hence, the least required speed  is 103.88 km/h

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Write the equation of the given line in slope-intercept form.​
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Step-by-step explanation:

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7 0
3 years ago
sketch a unit circle with angle instandard positionquestions:a. For what values of 0 is the sine increasing? Decreasing?b. For w
liq [111]

We can draw a unit circle as:

The sine is the vertical leg (or vertical projection) of the triangle formed by the angle.

The cosine will be the horizontal leg (or horizontal projection) of this same triangle.

The hypotenuse of this triangle will be the radius if the unit circle (r = 1).

a) The sine will be increasing from θ = 0° until it reaches its maximum at 90°.

Then it will be decreasing for angles between 90° and 270°.

Between 270° and 0° it will be increasing again.

b) The cosine has a maximum value for an angle of 0°.

It will decrease until 180°.

Then it will be increasing from angles between 180° and 360°.

c) We can see in the unit circle that the sine is 0 when the angle is one the horizontal axis, like when θ = 0° or θ = 180°.

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a) Increasing: (0°, 90°) and (270°, 360°)

Decreasing: (90°, 270°)

b) Increasing: (180°, 360°)

Decreasing: (0°,180°).

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4 0
1 year ago
A student is getting ready to take an important oral examination and is concerned about the possibility of having an "on" day or
Tamiku [17]

Answer:

The students should request an examination with 5 examiners.

Step-by-step explanation:

Let <em>X</em> denote the event that the student has an “on” day, and let <em>Y</em> denote the

denote the event that he passes the examination. Then,

P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})

The events (Y|X) follows a Binomial distribution with probability of success 0.80 and the events (Y|X^{c}) follows a Binomial distribution with probability of success 0.40.

It is provided that the student believes that he is twice as likely to have an off day as he is to have an on day. Then,

P(X)=2\cdot P(X^{c})

Then,

P(X)+P(X^{c})=1

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Compute the probability that the students passes if request an examination with 3 examiners as follows:

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The probability that the students passes if request an examination with 3 examiners is 0.715.

Compute the probability that the students passes if request an examination with 5 examiners as follows:

P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})

        =[\sum\limits^{5}_{x=3}{{5\choose x}(0.80)^{x}(1-0.80)^{5-x}}]\times\frac{2}{3}+[\sum\limits^{5}_{x=3}{{5\choose x}(0.40)^{x}(1-0.40)^{5-x}}]\times\frac{1}{3}

       =0.734

The probability that the students passes if request an examination with 5 examiners is 0.734.

As the probability of passing is more in case of 5 examiners, the students should request an examination with 5 examiners.

8 0
2 years ago
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