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2 years ago

There is something wrong with this definition for a pair of vertical angles: “If AB and

Mathematics

1 answer:

Oduvanchick [21]2 years ago

8 0

Answer:

If AB and CD intersect at point P such that P is between A and B and is also between C and D, then ∠APC and ∠BPD are a pair of vertical angles.

Step-by-step explanation:

The diagram is a sketch of a counterexample. Notice that ∠APC and ∠BPD are actually the same angles and are not vertical angles.

If P is between A and B and is also between C and D, then ∠APC and ∠BPD are vertical angles. Then the statement can be completed as:

If AB and CD intersect at point P such that P is between A and B and is also between C and D, then ∠APC and ∠BPD are a pair of vertical angles.

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The mean cost of domestic airfares in the United States rose to an all-time high of $385 per ticket (Bureau of Transportation St

saveliy_v [14]

Answer:

a)0.067

b)0.111

c)0.612

d)$687.28

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = $385

Standard Deviation, σ = $110

We are given that the distribution of domestic airfares is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(domestic airfare is $550 or more)

P(x > 550)

$P( x > 550) = P( z > \displaystyle\frac{550 - 385}{110}) = P(z > 1.5)$

$= 1 - P(z \leq 1.5)$

Calculation the value from standard normal z table, we have,

$P(x > 550) = 1 - 0.933 = 0.067= 6.7\%$

b) P(domestic airfare is $250 or less)

$P(x \leq 250) = P(z \leq \displaystyle\frac{250-385}{110}) = P(z \leq -1.22)$

Calculating the value from the standard normal table we have,

$P( x \leq 250) = 0.111 = 11.1\%$

c))P(domestic airfare is between $300 and $500)

$P(300 \leq x \leq 500) = P(\displaystyle\frac{300 - 385}{110} \leq z \leq \displaystyle\frac{500-385}{110}) = P(-0.77 \leq z \leq 1.04)\\\\= P(z \leq 1.04) - P(z < -0.77)\\= 0.851 - 0.239 = 0.612 = 61.2\%$

$P(300 \leq x \leq 500) = 61.2\%$

d) P(X=x) = 0.03

We have to find the value of x such that the probability is 0.03.

P(X > x)

$P( X > x) = P( z > \displaystyle\frac{x - 385}{110})=0.03$

$= 1 -P( z \leq \displaystyle\frac{x - 385}{110})=0.03$

$=P( z \leq \displaystyle\frac{x - 385}{110})=0.997$

Calculation the value from standard normal z table, we have,

$\displaystyle\frac{x - 385}{110} = 2.748\\x = 687.28$

Hence, the domestic fares must be $687.28 or greater for them to lie in the highest 3%.

7 0

3 years ago

Answer with memes best meme gets brainliest!

alukav5142 [94]

Answer:

for the lazy anti social people

5 0

3 years ago

Read 2 more answers

Kristian ordered 3 DVDS by mail each dvd cost the same amount with a $6 shipping charge the total cost was $42 how much did each

Natasha2012 [34]

Answer:

$12

Step-by-step explanation:

42 - 6 = 36

36/3

7 0

2 years ago

At the school football game, there are 42 adults

galina1969 [7]

Answer:

54 students

Step-by-step explanation:

42 / 7 = 6

9 x 6 = 54

42 : 54 = 7 : 9

4 0

2 years ago

someone please explain to me how to find the area PLEASE! Find the area and perimeter of quadrilateral ABCD below. Explain your

Lunna [17]

Answer:

Part A) The area of the figure is $24\ units^{2}$

Part B) The perimeter of the figure is $20\ units$

Step-by-step explanation:

step 1

Find the area of the figure

we know that

The area of the figure is equal to the area of triangle ABD plus the area of triangle BCD

The area of triangle is equal to

$A=\frac{1}{2}bh$

<u>Area of triangle ABD</u>

Observing the graph

$b=BD=(-2+8)=6\ units$

$h=(9-5)=4\ units$

substitute

$A=\frac{1}{2}(6)(4)=12\ units^{2}$

<u>Area of triangle BCD</u>

Observing the graph

$b=BD=(-2+8)=6\ units$

$h=(5-1)=4\ units$

substitute

$A=\frac{1}{2}(6)(4)=12\ units^{2}$

The area of the figure is

$12\ units^{2}+12\ units^{2}=24\ units^{2}$

step 2

Find the perimeter of the figure

we know that

The perimeter of the figure is equal to

$P=AB+BC+CD+AD$

we have

$A(-5,1),B(-8,5),C(-5,9),D(-2,5)$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Find the distance AB

$d=\sqrt{(5-1)^{2}+(-8+5)^{2}}=5\ units$

Find the distance BC

$d=\sqrt{(9-5)^{2}+(-5+8)^{2}}=5\ units$

Find the distance CD

$d=\sqrt{(5-9)^{2}+(-2+5)^{2}}=5\ units$

Find the distance AD

$d=\sqrt{(5-1)^{2}+(-2+5)^{2}}=5\ units$

substitute the values

$P=5+5+5+5=20\ units$

7 0

3 years ago