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Lesechka [4]
3 years ago
7

If the Internet consisted of four computers, there would be six possible connections. If it consisted of five computers, there w

ould be ten possible connections. How many connections are possible with ten computers?
Mathematics
1 answer:
Luba_88 [7]3 years ago
3 0

Answer:

<em>45 possible connections</em>

<em />

Step-by-step explanation:

The general equation for finding the possible number of connections in a network is given as

\frac{n*(n - 1)}{2}

where n is the number of computers on the network.

for 4 computers, we'll have

\frac{4*(4 - 1)}{2} = \frac{4*3}{2} = 6

for 5 computers, we'll have

\frac{5*(5 - 1)}{2} = \frac{5*4}{2} = 10.

therefore, for 10 computers, we will have

\frac{10*(10 - 1)}{2} = \frac{10*9}{2} = <em>45 possible connections</em>

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What number pattern is generated by the following
jenyasd209 [6]

Answer:

nth term = 1 + n(13/4)

Step-by-step explanation:

3n + n/4 + 1

n = 1, 3(1) + (1)/4 + 1 = 4 + ¼ = 17/4

n = 2, 3(2) + (2)/4 + 1 = 7 + ½ = 30/4

n = 3, 3(3) + (3)/4 + 1 = 10 + ¾ = 43/4

n = 4, 3(4) + (4)/4 + 1 = 14 = 56/4

The difference between consecutive terms is 13/4

At n = 1, 17/4 = 1 + 1(13/4)

At n = 2, 30/4 = 1 + 2(13/4)

nth term = 1 + n(13/4)

3 0
3 years ago
(ar^b) ^4 = 16r^20 where a and b are positive integers work our a and b​
ivolga24 [154]

Answer:

a = 2

b = 5

Step-by-step explanation:

Given :

(ar^b)^4 = 16r^20 ; a and b are positive integers :

Opening the bracket :

a^4r^4b = 16r^20

a^4 = 16 - - - - - (1)

r^4b = r^20 - - - (2)

a^4 = 16

Take the 4th root of both sides :

(a^4)^(1/4) = 16^1/4

a = 2

From (2)

r^4b = r^20

4b = 20

Divide both sides by 4

4b/4 = 20/4

b = 5

Hence ;

a = 2

b = 5

6 0
2 years ago
Picture of raadius is 10 cm​
ss7ja [257]

Step-by-step explanation:

And,what is the question so that i can help u??

6 0
3 years ago
Order the angles measures from least to greatest
enot [183]

Answer:

M < P < N

Step-by-step explanation:

Greater angles oppose sides of greater length.

6 0
2 years ago
A company with a large fleet of cars wants to study the gasoline usage. They check the gasoline usage for 50 company trips chose
Nookie1986 [14]

Answer:

The 95% confidence interval is given by (25.71536 ;28.32464)

And if we need to round we can use the following excel code:

round(lower,2)

[1] 25.72

round(upper,2)

[1] 28.32

And the interval would be (25.72; 28.32)  

Step-by-step explanation:

Notation and definitions  

n=50 represent the sample size  

\bar X= 27.2 represent the sample mean  

s=5.83 represent the sample standard deviation  

m represent the margin of error  

Confidence =88% or 0.88

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Calculate the critical value tc  

In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 88% of confidence, our significance level would be given by \alpha=1-0.88=0.12 and \alpha/2 =0.06. The degrees of freedom are given by:  

df=n-1=50-1=49  

We can find the critical values in R using the following formulas:  

qt(0.06,49)

[1] -1.582366

qt(1-0.06,49)

[1] 1.582366

The critical value tc=\pm 1.582366  

Calculate the margin of error (m)  

The margin of error for the sample mean is given by this formula:  

m=t_c \frac{s}{\sqrt{n}}  

m=1.582366 \frac{5.83}{\sqrt{50}}=14.613  

With R we can do this:

m=1.582366*(5.83/sqrt(50))

m

[1] 1.304639

Calculate the confidence interval  

The interval for the mean is given by this formula:  

\bar X \pm t_{c} \frac{s}{\sqrt{n}}  

And calculating the limits we got:  

27.02 - 1.582366 \frac{5.83}{\sqrt{50}}=25.715  

27.02 + 1.582366 \frac{5.83}{\sqrt{50}}=28.325

Using R the code is:

lower=27.02-m;lower

[1] 25.71536

upper=27.02+m;upper

[1] 28.32464

The 95% confidence interval is given by (25.71536 ;28.32464)  

And if we need to round we can use the following excel code:

round(lower,2)

[1] 25.72

round(upper,2)

[1] 28.32

And the interval would be (25.72; 28.32)  

6 0
3 years ago
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