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timama [110]
3 years ago
15

In a room with 10 people, how many different handshakes are possible?

Mathematics
2 answers:
Ganezh [65]3 years ago
7 0
45 handshakes are possible
Tomtit [17]3 years ago
7 0

Answer:

45 handshake

Step-by-step explanation:

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What is the explicit rule for the pattern 16,11,6,1...
svlad2 [7]

Answer:

the explicit rule is the difference, which is -5.

Step-by-step explanation:

8 0
3 years ago
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Write the equation of the line that goes through (-2, 0) and (4,3).
Snezhnost [94]

Answer:

B

Step-by-step explanation:

I just put in in the desmos graphing calculator

8 0
3 years ago
Helppppppppppppppppppp
GaryK [48]

shovels=3

bucket =7

a_ 3:7

b- 7:3

8 0
3 years ago
The time needed for all college students to complete a certain paper-and-pencil maze follows a normal distribution with a popula
oee [108]

Answer:

The correct option is a

Step-by-step explanation:

From the question we are told that

   The population mean is  \mu = 30\ seconds

    The standard deviation is  \sigma = 3 \ seconds

     The sample size is  n  = 9

    The null hypothesis is H_o: \mu  = 30

     The alternative hypothesis is  H_a  :  \mu \ne 30

    The level of significance is  \alpha = 0.01

    The sample mean is  \= x = 31.2

Generally the test statistics is mathematically represented as

      z = \frac{ \= x - \mu }{ \frac{\sigma}{\sqrt{n} } }

=>   z = \frac{ 31.2 - 30}{ \frac{3}{\sqrt{9} } }

=>   z = 1.2

From the z table  the area under the normal curve to the right corresponding to  1.2  is    

         P(Z >  1.2) =  0.11507

Generally the p-value is mathematically represented as

         p-value = 2 * P(Z >  1.2)

=>       p-value = 2 *0.11507

=>       p-value = 0.230

From the value obtained w can see that

         p-value  > \alpha hence

The decision rule is  

Fail to reject the null hypothesis

The conclusion is there is not enough evidence to support the claim

6 0
3 years ago
Easy math, but not sure what I did wrong.
Ierofanga [76]

Answer:

8

Step-by-step explanation:

In his 5th year, he took 3 times as many exams as the first year.  So the number of exams taken in the 5th year must be a multiple of 3.

If a₁ = 1, then a₅ = 3.  However, this isn't possible because we need 4 integers between them, and a sum of 31.

If a₁ = 2, then a₅ = 6.  Same problem as before.

If a₁ = 3, then a₅ = 9.  This is a possible solution.

If a₁ = 4, then a₅ = 12.  If we assume a₂ = 5, a₃ = 6, and a₄ = 7, then the sum is 34, so this is not a possible solution.

Therefore, Alex took 3 exams in his first year and 9 exams in his fifth year.  So he took 19 exams total in his second, third, and fourth years.

3 < a₂ < a₃ < a₄ < 9

If a₂ = 4, then a₃ = 7 and a₄ = 8.

If a₂ = 5, then a₃ = 6 and a₄ = 8.

If a₂ = 6, then there's no solution.

So Alex must have taken 8 exams in his fourth year.

4 0
3 years ago
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