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fredd [130]
2 years ago
10

at one of george washington's parties, each man shook hands with everyone except his spouse, and no handshakes took place betwee

n women. if 1313 married couples attended, how many handshakes were there among these 2626 people?
Mathematics
1 answer:
Keith_Richards [23]2 years ago
7 0

The total number of handshakes was 234.

  • The first man shook hands with 12 other men.
  • The second man shakes hands with 11 other men, as he had already shaken hands with the first man.
  • Like-wise, the last man does not need to shake hands again as every man has already shaken hands with him in their respective turns.
  • The number of handshakes among men is 0 + 1 + 2 + … + 12.
  • The sum of "n" natural numbers = (n)(n+1)/2
  • The number of handshakes among men is 12(13)/2.
  • The number of handshakes among men is 78.
  • Each man shakes hands with every woman except his spouse.
  • Each man shakes hands with 12 other women.
  • The number of handshakes between men and women is 13*12.
  • The number of handshakes between men and women is 156.
  • The total number of handshakes is 78 + 156.
  • The total number of handshakes is 234.

To learn more about natural numbers, visit :

brainly.com/question/17429689

#SPJ4

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The science of classifying and naming organisms based on their different characteristics is called .
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Given the following information about a hypothesis test of the difference between two means based on independent random samples,
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Answer:

S^2_p =\frac{(13-1)(5)^2 +(10 -1)(3)^2}{13 +10 -2}=18.143

And the deviation would be just the square root of the variance:

S_p=4.259

Then the statistic is given by:

t=\frac{(12 -9)-(0)}{4.259\sqrt{\frac{1}{13}+\frac{1}{10}}}=1.674

And the correct option would be:

t = 1.674

Step-by-step explanation:

Data given:

n_1 =13 represent the sample size for group 1

n_2 =10 represent the sample size for group 2

\bar X_1 =12 represent the sample mean for the group 1

\bar X_2 =9 represent the sample mean for the group 2

s_1=5 represent the sample standard deviation for group 1

s_2=3 represent the sample standard deviation for group 2

We are assuming two independent samples from two normal distributions with equal variances we are assuming that  

\sigma^2_1 =\sigma^2_2 =\sigma^2

And the statistic is given by this formula:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

Where t follows a t distribution with n_1+n_2 -2 degrees of freedom and the pooled variance S^2_p is given by this formula:

\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

The system of hypothesis on this case are:

Null hypothesis: \mu_1 \leq \mu_2

Alternative hypothesis: \mu_1 > \mu_2

The pooled variance is given by:

S^2_p =\frac{(13-1)(5)^2 +(10 -1)(3)^2}{13 +10 -2}=18.143

And the deviation would be just the square root of the variance:

S_p=4.259

Then the statistic is given by:

t=\frac{(12 -9)-(0)}{4.259\sqrt{\frac{1}{13}+\frac{1}{10}}}=1.674

And the correct option would be:

t = 1.674

4 0
3 years ago
Show how u got the answer 309×42
luda_lava [24]
                3            
                1
                390
              x  42
               -------
                1282
              1564
          ----------------
                       
the answer would be 16380
Hope this helped :)
4 0
4 years ago
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