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Aloiza [94]
3 years ago
13

Our basketball team has 10 players. we need to divide into two teams of 5 for an intra-squad scrimmage. in how many ways can we

do this without restriction?
Mathematics
1 answer:
Alexeev081 [22]3 years ago
7 0

Answer:

1/2 or 10/5

Step-by-step explanation:

You might be interested in
The Wall Street Journal Corporate Perceptions Study 2011 surveyed readers and asked how each rated the Quality of Management and
kykrilka [37]

Answer:

The calculated χ² =  <u>17.0281</u>    falls in the critical region χ² ≥  9.49  so we reject the null hypothesis that  the quality of management and the reputation of the company are independent and conclude quality of management and the reputation of the company are dependent

The p-value is 0 .001909. The result is significant at p < 0.05

Part b:

40 > 8.5

35> 7.5

25> 4

Step-by-step explanation:

1) Let the null and alternative hypothesis as

H0: the quality of management and the reputation of the company are independent

against the claim

Ha: the quality of management and the reputation of the company are dependent

2) The significance level alpha is set at 0.05

3) The test statistic under H0 is

χ²= ∑ (o - e)²/ e where O is the observed and e is the expected frequency

which has an approximate chi square distribution with ( 3-1) (3-1)=  4 d.f

4) Computations:

Under H0 ,

Observed       Expected E              χ²= ∑(O-e)²/e

40                      35.00                          0.71

25                      24.50                         0.01

5                         10.50                         2.88  

35                      40.00                         0.62

35                      28.0                          1.75

10                       12.00                           0.33  

25                      25.00                             0.00

10                        17.50                              3.21

<u>15                       7.50                                 7.50  </u>

<u>∑                                                               17.0281</u>

     

     

Column Totals 100 70 30   200  (Grand Total)

5) The critical region is χ² ≥ χ² (0.05)2 = 9.49

6) Conclusion:

The calculated χ² =  <u>17.0281</u>    falls in the critical region χ² ≥  9.49  so we reject the null hypothesis that  the quality of management and the reputation of the company are independent and conclude quality of management and the reputation of the company are dependent.

7) The p-value is 0 .001909. The result is significant at p < 0.05

The p- values tells that the variables are dependent.

Part b:

If we take the excellent row total = 70 and compare it with the excellent column total= 100

If we take the good row total = 70 and compare it with the good column total= 80

If we take the fair row total = 50 and compare it with the fair column total= 30

The two attributes are said to be associated if

Thus we see that ( where (A)(B) are row and columns totals and AB are the cell contents)

AB> (A)(B)/N  

40 > 1700/200

40 > 8.5

35> 1500/200

35> 7.5

25> 800/200

25> 4

and so on.

Hence they are positively associated

8 0
3 years ago
Sofas being sold for $603 this is a 33% discount from the m what is the original price​
fgiga [73]
About $801.99 bc 603X.33 is 198.99 plus 603 is 801.99
6 0
3 years ago
15 + (5-x)/x=30 solve for x
elixir [45]

Answer:

5/16

Step-by-step explanation:

15+(5-x)/x=30

1) Subtract 15 from both sides:

(5-x)/x=15

2) Multiply both sides by x:

5-x=15x

3) Add x to both sides:

16x=5

4) Divide both sides by 16:

x=5/16

8 0
3 years ago
What would it mean if there was an outbreak of cholera? A. The disease has infected a few people spread out over a large distanc
Dmitry_Shevchenko [17]

Answer:

I think that it is B because its the only answer that makes sense

Step-by-step explanation:

5 0
4 years ago
An article reports that 1 in 500 people carry the defective gene that causes inherited colon cancer. In a sample of 2000 individ
Tema [17]

Answer:

a) P=0.558

b) P=0.021

Step-by-step explanation:

We can model this random variable as a Poisson distribution with parameter λ=1/500*2000=4.

The approximate distribution of the number who carry this gene in a sample of 2000 individuals is:

P(x=k)=\frac{\lambda^ke^{-\lambda}}{k!} =\frac{4^ke^{-4}}{k!}

a) We can calculate that the approximate probability that between 4 and 9 (inclusive) as:

P(4\leq x\leq 9)=\sum_{k=4}^9P(k)\\\\\\ P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\P(9)=4^{9} \cdot e^{-4}/9!=262144*0.0183/362880=0.013\\\\\\

P(4\leq x\leq 9)=\sum_{k=4}^9P(k)=0.195+0.156+0.104+0.060+0.030+0.013=0.558

b) The approximate probability that at least 9 carry the gene is:

P(x\geq9)=1-P(x\leq 8)\\\\\\

P(0)=4^{0} \cdot e^{-4}/0!=1*0.0183/1=0.018\\\\P(1)=4^{1} \cdot e^{-4}/1!=4*0.0183/1=0.073\\\\P(2)=4^{2} \cdot e^{-4}/2!=16*0.0183/2=0.147\\\\P(3)=4^{3} \cdot e^{-4}/3!=64*0.0183/6=0.195\\\\P(4)=4^{4} \cdot e^{-4}/4!=256*0.0183/24=0.195\\\\P(5)=4^{5} \cdot e^{-4}/5!=1024*0.0183/120=0.156\\\\P(6)=4^{6} \cdot e^{-4}/6!=4096*0.0183/720=0.104\\\\P(7)=4^{7} \cdot e^{-4}/7!=16384*0.0183/5040=0.06\\\\P(8)=4^{8} \cdot e^{-4}/8!=65536*0.0183/40320=0.03\\\\

P(x\geq9)=1-P(x\leq 8)\\\\P(x\geq9)=1-(0.018+0.073+0.147+0.195+0.195+0.156+0.104+0.060+0.030)\\\\P(x\geq9)=1-0.979=0.021

8 0
3 years ago
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