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Dovator [93]
3 years ago
14

PLEASE HELP!!!!!!!dgbdgdbhdndcn

Mathematics
1 answer:
bogdanovich [222]3 years ago
6 0
Problem 1)

AC is only perpendicular to EF if angle ADE is 90 degrees

(angle ADE) + (angle DAE) + (angle AED) = 180
(angle ADE) + (44) + (48) = 180
(angle ADE) + 92 = 180
(angle ADE) + 92 - 92 = 180 - 92
angle ADE  = 88

Since angle ADE is actually 88 degrees, we do NOT have a right angle so we do NOT have a right triangle

Triangle AED is acute (all 3 angles are less than 90 degrees)

So because angle ADE is NOT 90 degrees, this means AC is NOT perpendicular to EF

-------------------------------------------------------------

Problem 2)

a) The center is (2,-3) 

The center is (h,k) and we can see that h = 2 and k = -3. It might help to write (x-2)^2+(y+3)^2 = 9 into (x-2)^2+(y-(-3))^2 = 3^3 then compare it to (x-h)^2 + (y-k)^2 = r^2

---------------------

b) The radius is 3 and the diameter is 6

From part a), we have (x-2)^2+(y-(-3))^2 = 3^3 matching (x-h)^2 + (y-k)^2 = r^2

where
h = 2
k = -3
r = 3

so, radius = r = 3
diameter = d = 2*r = 2*3 = 6

---------------------

c) The graph is shown in the image attachment. It is a circle with center point C = (2,-3) and radius r = 3.

Some points on the circle are

A = (2, 0)
B = (5, -3)
D = (2, -6)
E = (-1, -3)

Note how the distance from the center C to some point on the circle, say point B, is 3 units. In other words segment BC = 3.

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Wouldnt this be something like x and y squared
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(select ALL of the correct answers)
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Answer:

the answer is C,2^6×3^3

Step-by-step explanation:

6^3×2^6/2^3

=6^3×2^6-^3

=6^3×2^3

=2^6×2^3

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If 1 mile equals 5280 feet and an escalator moves at a rate of 2 feet per second. At what rate does the escalator move in miles
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Ratio=2 ft/s
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Answer: 1.36 miles/h
6 0
2 years ago
Read 2 more answers
Suppose that one person in 10,000 people has a rare genetic disease. There is an excellent test for the disease; 98.8% of the pe
nirvana33 [79]

Answer:

A)The probability that someone who tests positive has the disease is 0.9995

B)The probability that someone who tests negative does not have the disease is 0.99999

Step-by-step explanation:

Let D be the event that a person has a disease

Let D^c be the event that a person don't have a disease

Let A be the event that a person is tested positive for that disease.

P(D|A) = Probability that someone has a disease given that he tests positive.

We are given that There is an excellent test for the disease; 98.8% of the people with the disease test positive

So, P(A|D)=probability that a person is tested positive given he has a disease = 0.988

We are also given that  one person in 10,000 people has a rare genetic disease.

So,P(D)=\frac{1}{10000}

Only 0.4% of the people who don't have it test positive.

P(A|D^c) = probability that a person is tested positive given he don't have a disease = 0.004

P(D^c)=1-\frac{1}{10000}

Formula:P(D|A)=\frac{P(A|D)P(D)}{P(A|D)P(D^c)+P(A|D^c)P(D^c)}

P(D|A)=\frac{0.988 \times \frac{1}{10000}}{0.988 \times (1-\frac{1}{10000}))+0.004 \times (1-\frac{1}{10000})}

P(D|A)=\frac{2470}{2471}=0.9995

P(D|A)=0.9995

A)The probability that someone who tests positive has the disease is 0.9995

(B)

P(D^c|A^c)=probability that someone does not have disease given that he tests negative

P(A^c|D^c)=probability that a person tests negative given that he does not have disease =1-0.004

=0.996

P(A^c|D)=probability that a person tests negative given that he has a disease =1-0.988=0.012

Formula: P(D^c|A^c)=\frac{P(A^c|D^c)P(D^c)}{P(A^c|D^c)P(D^c)+P(A^c|D)P(D)}

P(D^c|A^c)=\frac{0.996 \times (1-\frac{1}{10000})}{0.996 \times (1-\frac{1}{10000})+0.012 \times \frac{1}{1000}}

P(D^c|A^c)=0.99999

B)The probability that someone who tests negative does not have the disease is 0.99999

8 0
3 years ago
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Nataly_w [17]

Answer:

500/6 min if I'm correct

Step-by-step explanation:

8 0
2 years ago
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