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aliina [53]
1 year ago
6

the chorus has 35 members. which expression represents the number of ways a group of 6 members can be chosen to do a special per

formance?
Mathematics
2 answers:
ch4aika [34]1 year ago
8 0

Answer:

^{35}C_6

Step-by-step explanation:

Given : The chorus has 35 members.

To Find:  which expression represents the number of ways a group of 6 members can be chosen to do a special performance?

Solution:

We will use combination to find the number of ways a group of 6 members can be chosen to do a special performance .

So, Formula : ^nC_r=\frac{n!}{r!(n-r)!}

Since we are given that total members are 35

So, n =35

No. of members in a group = 6

So, r =6

So,  ^{35}C_6=\frac{35!}{6!(35-6)!}

Thus expression represents the number of ways a group of 6 members can be chosen to do a special performance is  ^{35}C_6

zimovet [89]1 year ago
3 0
You have to choose 6 members from the group of 35 people. So there are \binom{35}{6} ways to do it.
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A square is constructed on side AD of quadrilateral ABCD such that FA lies on AB, as shown in the figure.
Elza [17]

Answer:  The co-ordinates of F are (4.4, -4.6) and the co-ordinates of D are (4.4, -1.4).

Step-by-step explanation:  Given that a square is constructed on side AD of quadrilateral ABCD such that FA lies on AB as shown in the figure. The co-ordinates of A are (6, -3) and the co-ordinates of B are (10, 1).

Also, AD : AB = 2 : 5 and the co-ordinates of the point E are (2.8, -3).

We are to select the correct co-ordinates of the points F and D.

Let, (a, b) are the co-ordinates of F and (c, d) are the co-ordinates of D.

Since ADEF is a square, so we have

AD = DE = EF = FA.

Given that

AD : AB = 2 : 5, so  FA : AB = 2 : 5.

That is, \left(\dfrac{c+4.4}{2},\dfrac{d-4.6}{2}\right)=\left(\dfrac{2.8+6}{2},\dfrac{-3-3}{2}\right)

We have, after applying the internal division formula that

\left(\dfrac{2\times 10+5\times a}{2+5},\dfrac{2\times 1+5\times b}{2+5}\right)=(6,-3)\\\\\\\Rightarrow \left(\dfrac{20+5a}{7},\dfrac{2+5b}{7}\right)=(6,-3)\\\\\\\Rightarrow \dfrac{20+5a}{7}=6,~~~~~\dfrac{2+5b}{7}=-3\\\\\\\Rightarrow 20+5a=42,~~~~\Rightarrow 2+5b=-21\\\\\\\Rightarrow 5a=22,~~~~~~~~~~\Rightarrow 5b=-23\\\\\\\Rightarrow a=4.4,~~~~~~~~~~~\Rightarrow b=-4.6.

So, the co-ordinates of F are (4.4, -4.6).

Now, since ADEF is a square, and the diagonals of a square bisect each other.

So, the mid-points of both the diagonals are same.

That is,

\textup{mid-point of DF}=\textup{mid-point of AE}\\\\\\\Rightarrow \left(\dfrac{c+4.4}{2},\dfrac{d-4.6}{2}\right)=\left(\dfrac{2.8+6}{2},\dfrac{-3-3}{2}\right)\\\\\\\Rightarrow \left(\dfrac{c+4.4}{2},\dfrac{d-4.6}{2}\right)=\left(\dfrac{8.8}{2},\dfrac{-6}{2}\right)\\\\\\\Rightarrow \dfrac{c+4.4}{2}=\dfrac{8.8}{2},~~~~~~\dfrac{d-4.6}{2}=-\dfrac{6}{2}\\\\\\\Rightarrow c+4.4=8.8,~~~~~\Rightarrow d-4.6=-6\\\\\Rightarrow c=4.4,~~~~~~~~~~~~\Rightarrow d=-1.4.

So, the co-ordinates of D are (4.4, -1.4).

Thus, the co-ordinates of F are (4.4, -4.6) and the co-ordinates of D are (4.4, -1.4).

7 0
1 year ago
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A random sample of 160 car purchases are selected and categorized by age. The results are listed below. The age distribution of
seraphim [82]

Answer:

The claim that all ages have purchase rates proportional to their driving rates is false.

Step-by-step explanation:

The complete question is:

A random sample of 160 car accidents are selected and categorized by the age of the driver determined to be at fault. The results are listed below. The age distribution of drivers for the given categories is 18% for the under 26 group, 39% for the 26-45 group, 31% for the 45-65 group, and 12% for the group over 65. Calculate the chi-square test statistic used to test the claim that all ages have crash rates proportional to their driving rates.

Age      >26     26-45       46-65      45<

Drivers 66    39            25          30

  A) 95.431

      B)101.324

      C)85.123

      D)75.101

Solution:

In this case we need to test whether there is sufficient evidence to warrant rejection of the claim that all ages have crash rates proportional to their driving rates.

A Chi-square test for goodness of fit will be used in this case.

The hypothesis can be defined as:

<em>H₀</em>: The observed frequencies are same as the expected frequencies.

<em>Hₐ</em>: The observed frequencies are not same as the expected frequencies.

The test statistic is given as follows:

\chi^{2}=\sum{\frac{(O-E)^{2}}{E}}

The values are computed in the table.

The test statistic value is 75.10.

The degrees of freedom of the test is:

n - 1 = 4 - 1 = 3

Compute the p-value of the test as follows:

p-value < 0.00001

*Use a Chi-square table.

p-value < 0.00001 < α = 0.05.

So, the null hypothesis will be rejected at any significance level.

Thus, there is sufficient evidence to warrant rejection of the claim that ages have crash rates proportional to their driving rates.

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2 years ago
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There are three outcomes of 4 out of eighteen outcomes, so the fraction of angle of spinner numbered 4 is \frac{3}{8}
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2 years ago
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What is the value of k such that x-5 is a factor of x3 – x2 + kx - 30?
Paraphin [41]

Answer:

k = - 14

Step-by-step explanation:

given that (x - 5) is a factor of the polynomial then x = 5 is a root and

x³ - x² + kx - 30 = 0 for x = 5, that is

5³ - 5² + 5k - 30 = 0

125 - 25 + 5k - 30 = 0

70 + 5k = 0 ( subtract 70 from both sides )

5k = - 70 ( divide both sides by 5 )

k = - 14


5 0
2 years ago
The lengths of the sides of triangle PQR are consecutive even integers. The perimeter of triangle PQR is 42 cm. What is the leng
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Answer:

16 cm

Step-by-step explanation:

Let the three consecutive even integers representing the sides of the triangle PQR be x, (x +2) and (x + 4).

Since, perimeter of triangle PQR = 42 cm

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1 year ago
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