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jok3333 [9.3K]
3 years ago
15

Deer ticks can be carriers of either Lyme disease or human granulocytic ehrlichiosis (HGE). Based on a recent study, suppose tha

t of all ticks in a certain location carry Lyme disease, carry HGE, and of the ticks that carry at least one of these diseases in fact carry both of them. If a randomly selected tick is found to have carried HGE, what is the probability that the selected tick is also a carrier of Lyme disease?
Mathematics
1 answer:
Dmitry_Shevchenko [17]3 years ago
5 0

Answer:

There is a 100% probability that the selected tick is also a carrier of Lyme disease.

Step-by-step explanation:

The problem states that:

Of the ticks that carry at least one of these diseases in fact carry both of them.

So, if a tick carries one of these diseases, there is a 100% probability that is carries another.

What is the probability that the selected tick is also a carrier of Lyme disease?

There is a 100% probability that the selected tick is also a carrier of Lyme disease.

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Jim, Dan and David share some sweets in the ratio 3:1:5. Jim gets 21 sweets. How many more sweets does David get over Dan?
nadezda [96]

Answer:

David gets 28 more than Dan

Step-by-step explanation:

Jim: Dan : David: total

3   :    1    :  5       : 3+1+5 = 9

Jim gets 21

21/3 = 7

Multiply each number by 7

Jim: Dan : David: total

3*7 :  1*7  :  5 *7  :  9*7

21  :   7     :  35   :  63

David has 35 and Dan has 7

35-7 = 28

David gets 28 more than Dan

8 0
4 years ago
Write a polynomial that represents the area of the shaded region
marusya05 [52]
<h3>Answer: x^2-3x+36</h3>

=========================================

Explanation:

The larger rectangle has area of (x+1)(x+1) = x^2+2x+1 through the use of the FOIL rule or distribution

If you use distribution, then it might help to let y = x+1 so we'd have y(x+1) lead to xy+1y which becomes x(x+1)+1(x+1). From there it might be easier to see how to get x^2+2x+1 after everything distributes again and simplifies.

The smaller rectangle has area 5x-35 which is found by distributing 5(x-7)

To get the shaded area, we subtract the two rectangle areas found above

shaded area = (larger area) - (smaller area)

shaded area = (x^2+2x+1) - (5x - 35)

shaded area = x^2+2x+1 - 5x + 35

shaded area = x^2-3x+36

6 0
3 years ago
If your algorithm runs its critical section 1 + 2 + 3 + ... + (n-2) + (n-1) + n times, what is the asymptotic behavior of the al
Paraphin [41]
If 1 and 2 becomes multiplayed gave 3 and 3+3=4 (4-2)+(4-1)= 2+3=4 and 4 is n.
4 0
3 years ago
16,051 rounded to the nearest thousand
Kaylis [27]
It is 16,000. because 0 is lower than 5

3 0
3 years ago
Please help me with the photo below thanks
yanalaym [24]

Answer:

A) -12-4-5

B) X^2-X

Step-by-step explanation:

..

..

4 0
1 year ago
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