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3 years ago

Given: AD || BC and AD ≅ BC Prove: AB || CD

Mathematics

1 answer:

Svet_ta [14]3 years ago

3 0

Answer:

Statement 1: it is a parallelogram

Reason 1: if one pair of sides of a quadrilateral are parallel and congruent sides, then it is a parallelogram.

Statement 2: AB is parallel to CD

Reason 2: in a parallelogram opposite sides are parallel.

QED

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Let f(x) = Ax + B and g(x) = Cx + D where A, B, C, and D are non-zero constants. Use upper case letters for A, B, C, D and lower

Elodia [21]

Answer:

Let f(x) = Ax + B and g(x) = Cx + D where A, B, C, and D are non-zero constants. Use upper case letters for A, B, C, D and lower case letter for x in your answers below. (a) Find f(g(x)). f(g(x)) = A(Cx + D) + B (b) Find the slope of the function you found in part (a). slope = (c) Find the vertical intercept of the function you found in part (a). Do not write your answer as a point

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Anne bought 6 cans of paint and 2/3 of a pint of special paint additive formulated to reduce

iogann1982 [59]

Answer:

1/9 pints

Step-by-step explanation:

[Given] 2/3 / 6

["Keep, change, flip"] 2/3 * 1/6

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[Simplify] 1/9

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2 years ago

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Two gears are connected and are rotating simultaneously. The smaller gear has a radius of 4 inches, and the larger gear has a ra

pychu [463]

Draw a diagram to illustrate the problem as shown below.

When the smaller gear rotates through a revolution, it sweeps an arc length of
2π(4) = 8π inches.

Part 1
The same arc length is swept by the larger gear. The central angle of the larger gear, x, is
7x = 8π
x = (8π)/7 radians = (8π)/7 * (180/π) = 205.7°

Answer: 205.7° (nearest tenth)

Part 2
When the larger gear makes one rotation, it sweeps an arc length of
2π(7) = 14π inches.
If the central angle for the smaller gear is y radians, then
4y = 14π
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Answer:
The smaller gear makes 1.75 rotations

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3 years ago

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.9 minutes and a standard deviation of 2.9

Eva8 [605]

Answer:

a) 0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) 0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes

c) 0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 8.9 minutes and a standard deviation of 2.9 minutes.

This means that $\mu = 8.9, \sigma = 2.9$

Sample of 37:

This means that $n = 37, s = \frac{2.9}{\sqrt{37}}$

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

320/37 = 8.64865

Sample mean below 8.64865, which is the p-value of Z when X = 8.64865. So

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

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{8.64865 - 8.9}{\frac{2.9}{\sqrt{37}}}$

$Z = -0.53$

$Z = -0.53$ has a p-value of 0.2981

0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

275/37 = 7.4324

Sample mean above 7.4324, which is 1 subtracted by the p-value of Z when X = 7.4324. So

$Z = \frac{X - \mu}{s}$

$Z = \frac{7.4324 - 8.9}{\frac{2.9}{\sqrt{37}}}$

$Z = -3.08$

$Z = -3.08$ has a p-value of 0.001

1 - 0.001 = 0.999

0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Sample mean between 7.4324 minutes and 8.64865 minutes, which is the p-value of Z when X = 8.64865 subtracted by the p-value of Z when X = 7.4324. So

0.2981 - 0.0010 = 0.2971

0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

7 0

2 years ago