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frozen [14]
3 years ago
13

A political polling organization wishes to select a smaller focus group from a group of 7 Republicans, 10 Democrats, and 2 Indep

endents. In how many ways can the group be chosen if: (a) it will consist of 1 Republican and 4 Democrats?(b) it will consist of 2 Republicans, 2 Democrats, and 3 Independents?
Mathematics
1 answer:
Alina [70]3 years ago
7 0

Answer:

a) 1470 ways

b) 2835 ways

Step-by-step explanation:

the question is not correct, in the correct question The number of independents should be 3 not 2

there are 7 republicans, 10 democrats and 3 independents.

C(n,r) is the number of different combinations of n distinct objects taken r at a time.

C(n,r) = \frac{n!}{(n-r)!r!}

a) it will consist of 1 Republican and 4 Democrats

The number of ways this can be chosen is = C(7,1) × C(10,4) = \frac{7!}{(7-1)!1!}  * \frac{10!}{(10-4)!4!} = 7 × 210 = 1470 ways

(b) it will consist of 2 Republicans, 2 Democrats, and 2 Independents (consist of 2 independents not 3)

The number of ways this can be chosen is = C(7,2) × C(10,2) × C(2,2)

 =\frac{7!}{(7-2)!2!}  * \frac{10!}{(10-2)!2!}* \frac{3!}{(3-2)!2!} = 21 × 45 × 3 = 2835 ways

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vampirchik [111]
\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h

Employ a standard trick used in proving the chain rule:

\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h

The limit of a product is the product of limits, i.e. we can write

\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)

The rightmost limit is an exercise in differentiating \sqrt x using the definition, which you probably already know is \dfrac1{2\sqrt x}.

For the leftmost limit, we make a substitution y=\sqrt x. Now, if we make a slight change to x by adding a small number h, this propagates a similar small change in y that we'll call h', so that we can set y+h'=\sqrt{x+h}. Then as h\to0, we see that it's also the case that h'\to0 (since we fix y=\sqrt x). So we can write the remaining limit as

\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}

which in turn is the derivative of \tan y, another limit you probably already know how to compute. We'd end up with \sec^2y, or \sec^2\sqrt x.

So we find that

\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}
7 0
3 years ago
Plzzzzz helps meee I don’t get this
Inessa05 [86]
1 way : 10d + 2d

another way: 5d + 5d + 2d
4 0
3 years ago
There is a bag filled with 4 blue, 3 red and 5 green marbles.
zhannawk [14.2K]

Answer:

You are selecting marbles with replacement. The marble selections (trials) are independent and the marble selection follows the binomial distribution.

The probability of selecting a red marble the first time is 1313.

(This is because 4 out of 12 marbles are red and412412 reduces to 1313.

The probability of selecting a red marble the second time is 1313.

The marble selections are independent and you can multiply the two probabilities to get the following:

probability of getting 2 reds = (13)2(13)2

=19=19.

So the probability of getting two reds is 1919.

7 0
2 years ago
Steph makes 90 % 90%90, percent of the free throws she attempts. She is going to shoot 3 33 free throws. Assume that the results
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Using the binomial distribution, it is found that there is a 0.027 = 2.7% probability that he makes exactly 1 of the 3 free throws.

For each free throw, there are only two possible outcomes, either he makes it, or he misses it. The results of free throws are independent from each other, hence, the binomial distribution is used to solve this question.

Binomial probability distribution

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

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In this problem:

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The probability that he makes exactly 1 is P(X = 1), hence:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 1) = C_{3,1}.(0.9)^{1}.(0.1)^{2} = 0.027

0.027 = 2.7% probability that he makes exactly 1 of the 3 free throws.

To learn more about the binomial distribution, you can take a look at brainly.com/question/24863377

3 0
2 years ago
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labwork [276]

Answer:

12,922

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