Ask question

Ask question

All categories

3 years ago

5

The probability a person is infected by a certain cold virus is 0.2. If a random sample of 12 people is taken, what is the proba

bility that at least one person will be affected by the cold virus?

1 answer:

mezya [45]3 years ago

3 0

Answer:

Step-by-step explanation:

It is given that The probability a person is infected by a certain cold virus is 0.2 that is p=0.2, then q=1-0.2=0.8.

Now, we have to find the probability that at least one person will be affected by the cold virus, that is P(x>1).

P(x>1)=1-(P(x=0)+P(x=1))

= $1-(nc_{x}p^xq^{n-x}+nc_{x}p^xq^{n-x})$

= $1-(12c_{0}p^0q^{12}+12c_{1}p^1q^{11})$

= $1-((0.8)^{12} +12(0.2)(0.8)^{11})$

= $1-(0.8+12(0.2))(0.8)^{11}$

= $1-(0.8+2.4)(0.0858)$

= $1-(3.2)(0.0858)$

= $1-0.274$

= $0.726$

Thus, probability that at least one person will be affected by the cold virus will be=0.726.

You might be interested in

Am i correct? please be honest

nadya68 [22]

Answer:

Yep. Those are 5 (-4)'s.

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

A flash disk has the capacity of storing 1 megabyte of data. How many kilobytes of data can be stored in the same flash disk

Luden [163]

For this sort of problem, you need to be familiar with prefixes.

So:

megabyte = $1 *10^6$ bytes

kilobyte = $1*10^3$ bytes

and a regular byte would just be $1$

Now, you will need to do a conversion:

$1megabyte*\frac{1*10^6bytes}{1megabyte}*\frac{1*10^{-3}kilobyte}{1byte}$

I'm going to explain this just in case, but we convert megabytes to regular bytes in the first half by using the information I gave you above. (In one megabyte, there are $1*10^6$ bytes!)

In the second fraction thought, remember one kilobyte is $1*10^3$ bytes. However, you usually only see one of each thing on the bottom of fractions, so you need to add a negative sign to the 3. (technically you could just do $\frac{1kilobyte}{1*10^3bytes}$ , but I think it is more correct to do write it out how I did).

*** They do equal the same thing though! Do whichever way is easier for you!

Now, your answer should be:

$10^3$ kilobytes (1000 kilobytes)

Hope this helped!

6 0

3 years ago

The expressions below represent either exponential growth or exponential decay. Check all the expressions that represent exponen

asambeis [7]

Answer:

Answers B, E and F (1, 5 and 6).

Step-by-step explanation:

In an exponential equation, (y=ab^x), the "b" value must be a decimal less than 1 (but greater than 0) for it to be considered an exponential decay function. Only equations B, E and F have a "b" value that fits this criteria.

6 0

3 years ago

Help me with this, Its in the doc below

zvonat [6]

Answer:

https://faq.brainly.com/hc/en-us/articles/360014661139.

Step-by-step explanation:

5 0

3 years ago

A teenager who is 5 feet tall throws an object into the air. The quadratic function LaTeX: f\left(x\right)=-16x^2+64x+5f ( x ) =

tia_tia [17]

Answer:

At approximately x = 0.08 and x = 3.92.

Step-by-step explanation:

The height of the ball is modeled by the function:

$f(x)=-16x^2+64x+5$

Where f(x) is the height after x seconds.

We want to determine the time(s) when the ball is 10 feet in the air.

Therefore, we will set the function equal to 10 and solve for x:

$10=-16x^2+64x+5$

Subtracting 10 from both sides:

$-16x^2+64x-5=0$

For simplicity, divide both sides by -1:

$16x^2-64x+5=0$

We will use the quadratic formula. In this case a = 16, b = -64, and c = 5. Therefore:

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Substitute:

$\displaystyle x=\frac{-(-64)\pm\sqrt{(-64)^2-4(16)(5)}}{2(16)}$

Evaluate:

$\displaystyle x=\frac{64\pm\sqrt{3776}}{32}$

Simplify the square root:

$\sqrt{3776}=\sqrt{64\cdot 59}=8\sqrt{59}$

Therefore:

$\displaystyle x=\frac{64\pm8\sqrt{59}}{32}$

Simplify:

$\displaystyle x=\frac{8\pm\sqrt{59}}{4}$

Approximate:

$\displaystyle x=\frac{8+\sqrt{59}}{4}\approx 3.92\text{ and } x=\frac{8-\sqrt{59}}{4}\approx0.08$

Therefore, the ball will reach a height of 10 feet at approximately x = 0.08 and x = 3.92.

7 0

3 years ago