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3 years ago

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Mopeds (small motorcycles with an engine capacity below 50 cm3) are very popular in Europe because of their mobility, ease of op

eration, and low cost. Suppose the maximum speed of a moped is normally distributed with mean value 46.8 km/h and standard deviation 1.75 km/h. Consider randomly selecting a single such moped. (a) What is the probability that maximum speed is at most 49 km/h

1 answer:

Anika [276]3 years ago

8 0

Answer:

The probability that the maximum speed is at most 49 km/h is 0.8340.

Step-by-step explanation:

Let the random variable<em> </em><em>X</em> be defined as the maximum speed of a moped.

The random variable <em>X</em> is Normally distributed with mean, <em>μ</em> = 46.8 km/h and standard deviation, <em>σ</em> = 1.75 km/h.

To compute the probability of a Normally distributed random variable we first need to convert the raw score of the random variable to a standardized or <em>z</em>-score.

The formula to convert <em>X</em> into <em>z</em>-score is:

$z=\frac{X-\mu}{\sigma}$

Compute the probability that the maximum speed is at most 49 km/h as follows:

Apply continuity correction:

P (X ≤ 49) = P (X < 49 - 0.50)

= P (X < 48.50)

$=P(\frac{X-\mu}{\sigma}$

*Use a <em>z</em>-table for the probability.

Thus, the probability that the maximum speed is at most 49 km/h is 0.8340.

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Let
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tan x=opposite side angle x/adjacent side angle x
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3 years ago

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What are the coordinates of the orthocenter of △JKL with vertices at J(−4, −1) , K(−4, 8) , and L(2, 8) ?

Sati [7]

<u>Answer-</u>

<em>The coordinates of the orthocenter of △JKL is (-4, 8)</em>

<u>Solution-</u>

The orthocenter is the point where all three altitudes of the triangle intersect. An altitude is a line which passes through a vertex of the triangle and is perpendicular to the opposite side.

For a right angle triangle, the vertex at the right angle is the orthocentre of the triangle.

Here we are given the three vertices of the triangle are J(-4,-1), K(-4,8) and L(2,8)

If the triangle JKL satisfies Pythagoras Theorem, then triangle JKL will be a right angle triangle.

Applying distance formula we get,

$JK^2= (-4+4)^2+ (8+1)^2=0+81=81\\\\KL^2= (-4-2)^2+ (8-8)^2=36+0=36\\\\JL^2= (-4-2)^2+(8+1)^2=36+81=117$

As,

$\Rightarrow 117=81+36$

$\Rightarrow JL^2=JK^2+KL^2$

$\Rightarrow \text{JKL is a right angle triangle}$

$\Rightarrow \angle K=90^{\circ}$

Therefore, the vertex at K (-4, 8) is the orthocentre.

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2 years ago

655% of what number is 26,200?

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Answer:4000

Step-by-step explanation:

Let them number be a

655% of a=26200

655/100 x a=26200

655a/100=26200

Cross multiplying we get

655a=26200x100

655a=2620000

Divide both sides by 655

655a/655 = 2620000/655

a=4000

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3 years ago

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The total cost for buying new sneakers for the basketball team an be found by using the function y=25x + 65, is where x is the n

gtnhenbr [62]

Answer:

$Domain\ =\ \{x: 9,10,11....16\}$

<em></em> $Range\ =\ \{y:290, 315,340.....465\}$ <em></em>

Step-by-step explanation:

Given

$y = 25x + 65$

Required

Determine the range and domain when x = 9 to 16

The domain is the set of

$Domain\ =\ \{x: 9,10,11....16\}$

Solving for the range;

<em>When x = 9</em>

<em></em> $y = 25 * 9 + 65$ <em></em>

<em></em> $y = 225 + 65$ <em></em>

<em></em> $y = 290$ <em></em>

<em></em>

<em>When x = 10</em>

<em></em> $y = 25 * 10 + 65$ <em></em>

<em></em> $y = 250 + 65$ <em></em>

<em></em> $y = 315$ <em></em>

<em></em>

<em>When x = 11</em>

<em></em> $y = 25 * 11 + 65$ <em></em>

<em></em> $y = 275 + 65$ <em></em>

<em></em> $y = 340$ <em></em>

<em></em>

<em>You continue till you get to x = 16</em>

<em>---</em>

<em>--</em>

<em>-</em>

<em>When x = 16</em>

<em></em> $y = 25 * 16 + 65$ <em></em>

<em></em> $y = 400 + 65$ <em></em>

<em></em> $y = 465$ <em></em>

<em></em>

<em>Hence; the range is</em>

<em></em> $Range\ =\ \{y:290, 315,340.....465\}$ <em></em>

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2 years ago

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Step-by-step explanation:

The answer is definitely 2.

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