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makvit [3.9K]
3 years ago
12

Using the diagram below, what is the horizontal distance between the control tower and the plane?

Mathematics
1 answer:
Natali [406]3 years ago
6 0

Answer:

x = 664.093 m

Step-by-step explanation:

tan∅ = opposite over adjacent

Since we are trying to find the horizontal distance, that means we are dealing with the bottom length and not the hypotenuse (the diagonal length):

Our theta (∅) = 52°

Our opposite = 850 m

Our adjacent = x m

tan52° = 850/x

xtan52° = 850

x = 850/tan52°

x = 664.093 m

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seraphim [82]

Answer:

2 3/4 dozen cookies per hour

Step-by-step explanation:

To find his unit rate per hour, divide 5 1/2 by 2:

5.5/2

= 2.75 or 2 3/4 dozen cookies

So, his unit rate in dozen cookies per hour is 2 3/4

6 0
3 years ago
Given: Isosceles TriangleDEF. The measure of vertex Angle D is four times the measure of each base angle E and F.
777dan777 [17]

Answer:

D = 120° , E = 30° , F = 30°

Step-by-step explanation:

Let D = 4x

Let E&F = x

4x + x + × = 180°[angles in triangle is 180°]

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(Correct me if i am wrong)

4 0
3 years ago
The university is on a bearing of 050 degrees from the stadium and 300 degrees from the hospital. Mark the position of the unive
Licemer1 [7]

Answer:

Check Explanation

Step-by-step explanation:

The map isn't available, but, one can draw this bearings with the descriptions given

Angles for bearings are usually drawn, reading from the north direction.

A sketch of the described bearings and locations of the stadium, university and the hospital is presented in the attached image to this question.

For the sketch on the map.

On The map, draw a four-cardinal points' cross at the stadium draw a line from the bearing 050° from the stadium's four cardinal points' cross. Leave the line as it is.

Then, draw another four cardinal points' cross at the hospital and draw a line from the bearing 300° from the hospital's four cardinal points' cross.

Wherever the line from the hospital meets the line from the stadium, is where the University is located.

Hope this Helps!!!!

6 0
3 years ago
I need help on this question
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Answer:

d

Step-by-step explanation:

8 0
3 years ago
Suppose a particular type of cancer has a 0.9% incidence rate. Let D be the event that a person has this type of cancer, therefo
natita [175]

Answer:

There is a 12.13% probability that the person actually does have cancer.

Step-by-step explanation:

We have these following probabilities.

A 0.9% probability of a person having cancer

A 99.1% probability of a person not having cancer.

If a person has cancer, she has a 91% probability of being diagnosticated.

If a person does not have cancer, she has a 6% probability of being diagnosticated.

The question can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

P = \frac{P(B).P(A/B)}{P(A)}

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

In this problem we have the following question

What is the probability that the person has cancer, given that she was diagnosticated?

So

P(B) is the probability of the person having cancer, so P(B) = 0.009

P(A/B) is the probability that the person being diagnosticated, given that she has cancer. So P(A/B) = 0.91

P(A) is the probability of the person being diagnosticated. If she has cancer, there is a 91% probability that she was diagnosticard. There is also a 6% probability of a person without cancer being diagnosticated. So

P(A) = 0.009*0.91 + 0.06*0.991 = 0.06765

What is the probability that the person actually does have cancer?

P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.91*0.009}{0.0675} = 0.1213

There is a 12.13% probability that the person actually does have cancer.

3 0
3 years ago
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