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mrs_skeptik [129]
3 years ago
14

6. Suppose that a fair coin is tossed 2 times, and the result of each toss (H or T) is recorded.

Mathematics
1 answer:
nekit [7.7K]3 years ago
5 0

Answer:

a) S= {HH, HT, TH, TT}

b) P(X=0) = (2C0) (0.5)^0 (1-0.5)^{2-0}= 0.25

P(X=1) = (2C1) (0.5)^1 (1-0.5)^{2-1}= 0.5

P(X=2) = (2C2) (0.5)^2 (1-0.5)^{2-2}= 0.25

And we have the following table:

X     |     0   |     1   |      2

P(X) |  0.25 |  0.5 |  0.25

Step-by-step explanation:

Let's define first some notation

H= represent a head for the coin tossed

T= represent tails for the coin tossed

We are going to toss a coin 2 times so then the size of the sample size is 2^2 = 4

a. What is the sample space for this chance experiment?

The sampling space on this case is given by:

S= {HH, HT, TH, TT}

b. For this chance experiment, give the probability distribution for the random variable of the total number of heads observed.

The possible values for the number of heads are X=0,1,2. If we assume a fair coin then the probability of obtain heads is the same probability of obtain tails and we can find the distribution like this:

P(X=0) = (2C0) (0.5)^0 (1-0.5)^{2-0}= 0.25

P(X=1) = (2C1) (0.5)^1 (1-0.5)^{2-1}= 0.5

P(X=2) = (2C2) (0.5)^2 (1-0.5)^{2-2}= 0.25

And we have the following table:

X     |     0   |     1   |      2

P(X) |  0.25 |  0.5 |  0.25

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IceJOKER [234]

Answer:

x > 4

Step-by-step explanation:

4x + 3 > 19

4x + 3 - 3 > 19 - 3

4x > 16

4x/4 > 16/4

x > 4

Hope this helps.

6 0
3 years ago
the change in water vapor in a cloud is modeled by a polynomial function, c (x). describe how to find the x - intercepts of c (x
lukranit [14]

The change in the water vapors is modeled by the polynomial function c(x). In order to find the x-intercepts of a polynomial we set it equal to zero and solve for the values of x. The resulting values of x are the x-intercepts of the polynomial.

Once we have the x-intercepts we know the points where the graph crosses the x-axes. From the degree of the polynomial we can visualize the end behavior of the graph and using the values of maxima and minima a rough sketch can be plotted.

Let the polynomial function be c(x) = x ² -7x + 10

To find the x-intercepts we set the polynomial equal to zero and solve for x as shown below:

x ² -7x + 10 = 0

Factorizing the middle term, we get:

x ² - 2x - 5x + 10 = 0

x(x - 2) - 5(x - 2) =0

(x - 2)(x - 5)=0

x - 2 = 0 ⇒ x=2

x - 5 = 0 ⇒ x=5

Thus the x-intercept of our polynomial are 2 and 5. Since the polynomial is of degree 2 and has positive leading coefficient, its shape will be a parabola opening in upward direction. The graph will have a minimum point but no maximum if the domain is not specified. The minimum points occurs at the midpoint of the two x-intercepts. So the minimum point will occur at x=3.5. Using x=3.5 the value of the minimum point can be found. Using all this data a rough sketch of the polynomial can be constructed. The figure attached below shows the graph of our polynomial.

8 0
3 years ago
Read 2 more answers
Hi guys this is my question
sasho [114]

Answers -

statement B, D, E are correct .

Solution -

As figure B is the scaled copy of figure A, so it is smaller than B and its perimeter is also less than that of A.

On the other hand , as B is the smaller replica of A, so it has same number of edges ,angels and sides. Though the length of the sides are not same, the angels are same. so the perimeters of both figures are different but sum of all the angels are same.

5 0
3 years ago
A baseball player got 102 hits in the last 300 times at bat. Explain how you would find the percent of times at bat.​
slava [35]

Answer:

34%

Step-by-step explanation:

Divide the number of hits, 102, by the times at bat, 300. Then move the decimal 2 places to the right and add the percent sign to get 34%.

6 0
3 years ago
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Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i
jenyasd209 [6]

Answer:

If the limit that you want to find is \lim_{x\to \infty}\dfrac{P(x)}{Q(x)} then you can use the following proof.

Step-by-step explanation:

Let P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0} and Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0} be the given polinomials. Then

\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}

Observe that

\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}

and

\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}

Then

\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}

3 0
3 years ago
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