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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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Trava [24]

Law of Cosines says


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3 years ago
The 4th and 8th term of a G.P. are 24 and 8/27 respectively. find the 1st term and common ratio
beks73 [17]

Answer:

see explanation

Step-by-step explanation:

The n th term of a geometric progression is

• a_{n} = a₁ r^{n-1}

where a₁ is the first term and r the common ratio

given a₄ = 24, then

a₁r^{3} = 24 → (1)

Given a₈ = \frac{8}{27}, then

a₁r^{7} = \frac{8}{27} → (2)

Divide (2) by (1)

r^{4} = \frac{\frac{8}{27} }{24} = \frac{1}{81}

Hence r = \sqrt[4]{\frac{1}{81} } = \frac{1}{3}

Substitute this value into (1)

a₁ × (\frac{1}{3} )³ = 24

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4 0
3 years ago
There are x number of students at helms. If the number of students increases by 7.8% each year, how many students will be there
vodomira [7]

There will be 1.078x students next year and equation is number of students in next year = x + 7.8% of x

<h3><u>Solution:</u></h3>

Given, There are "x" number of students at helms.  

The number of students increases by 7.8% each year which means if there "x" number of students in present year, then the number of students in next year will be x + 7.8% of x

Number of students in next year = number of students in present year + increased number of students.

\begin{array}{l}{\text { Number of students in next year }=x+7.8 \% \text { of } x} \\\\ {\text { Number of students in next year }=x\left(1+\frac{7.8}{100}\right)} \\\\ {\text { Number of students in next year }=x(1+0.078)=1.078 x}\end{array}

Thus there will be 1.078x students in next year

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3 years ago
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