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2 years ago

Given circle P with arc AE=53, arc BA=68 and arc CB=72 match the following angles with their corresponding measurements

1 answer:

3 0

From the diagram;
1. Angle 2 = ADB+BDH
= arcAB/2 +90
= 34 +90
= 124°

2. Angle 4= 90°,
Reason ; the angle between a tangent and a radius is equal to 90. A tangent is a line that touches the circumference of a circle once even if prolonged.

3. Angle 5 = 90 -BDC (note the acr subtends twice the angle it subtends on the circumference to the center.
= 90-arc BC/2
= 90-36
= 54°

4. Angle 6 = BFD
= 180-ADB-FBD
= 180-AB/2-DE/2
But DE = 180 -121 = 59
Therefore, BFD = 180 -34-29.5
= 116.5°

5. Angle 1 = 180- BFD (angles on a straight line add up to 180°)
= 180- 116.5
= 63.5°

6. Angle 3 = 180 -(ADB+BFD)
= 180 -(34 +116.5)
= 180- 150.5
= 29.5°
similarly angle 3 = DE/2 = 59/2 = 29.5°

7. Angle 8= 90, because BD is diameter;
angles subtended by a diameter to the circumference is always a right angle (90°)

8.Angle 7 = BE
but BE= AB+AE
= 68+ 53
= 121°

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Harrison Water Sports has two retail outlets: Seattle and Portland. The Seattle store does 60 percent of the total sales in a ye

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Answer: P = 0.75

Step-by-step explanation:

Hi!

The sample space of this problems is the set of all the possible sales. It is divided in the disjoint sets:

$S_s = {\text{sales made in Seattle }}\\S_p={\text {sales made in Portland}}$

We have also the set of sales of boat accesories $S_b$ , the colored one in the image.

We are given the data:

$P(S_s) = 0.6\\P(S_b | S_s) = \frac{P(S_b\bigcap S_s)}{P(S_s)}=0.4\\P(S_b|S_p) =\frac{P(S_b\bigcap S_p)}{P(S_p)}=0.2$

From these relations you can compute the probabilities of the intersections colored in the image:

$pink\;set:\;P(S_b \bigcap S_s) =0.6*0.4=0.24\\blue\;set\;:P(S_b \bigcap S_p)=(1-0.6)*0.2 =0.08$

You are asked about the conditional probability:

$P(S_s|S_b) = \frac{P(S_s \bigcap S_b)}{P(S_b)}$

To calculate this, you need $P(S_b)$ . In the image you can see that the set $S_b$ is the union of the two disjoint pink and blue sets. Then:

$P(S_b)=P((S_b \bigcap S_s)\bigcup(S_b \bigcap S_p)) = 0.24 + 0.08 = 0.32$

Finally:

$P(S_s|S_b) = \frac{0.24}{0.32}=0.75$

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2 years ago

Cameron bought twelve pounds of candy corn for 79 cents a pound and eighteen pounds of M&Ms for $1.09 a pound, planning to m

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Emerson purchased a game that was on sale for 18% off. The sales tax in his county is 8%. Let y represent the original price of

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2 years ago

Read 2 more answers

In a certain community, eight percent of all adults over age 50 have diabetes. If a health service in this community correctly d

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Complete question is;

In a certain community, 8% of all people above 50 years of age have diabetes. A health service in this community correctly diagnoses 95% of all person with diabetes as having the disease, and incorrectly diagnoses 10% of all person without diabetes as having the disease. Find the probability that a person randomly selected from among all people of age above 50 and diagnosed by the health service as having diabetes actually has the disease.

Answer:

P(has diabetes | positive) = 0.442

Step-by-step explanation:

Probability of having diabetes and being positive is;

P(positive & has diabetes) = P(has diabetes) × P(positive | has diabetes)

We are told 8% or 0.08 have diabetes and there's a correct diagnosis of 95% of all the persons with diabetes having the disease.

Thus;

P(positive & has diabetes) = 0.08 × 0.95 = 0.076

P(negative & has diabetes) = P(has diabetes) × (1 –P(positive | has diabetes)) = 0.08 × (1 - 0.95)

P(negative & has diabetes) = 0.004

P(positive & no diabetes) = P(no diabetes) × P(positive | no diabetes)

We are told that there is an incorrect diagnoses of 10% of all persons without diabetes as having the disease

Thus;

P(positive & no diabetes) = 0.92 × 0.1 = 0.092

P(negative &no diabetes) =P(no diabetes) × (1 –P(positive | no diabetes)) = 0.92 × (1 - 0.1)

P(negative &no diabetes) = 0.828

Probability that a person selected having diabetes actually has the disease is;

P(has diabetes | positive) =P(positive & has diabetes) / P(positive)

P(positive) = 0.08 + P(positive & no diabetes)

P(positive) = 0.08 + 0.092 = 0.172

P(has diabetes | positive) = 0.076/0.172 = 0.442

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