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3 years ago

Use the debug ip rip command on r2 to determine the routes received in the rip updates from r3 and list them here.

1 answer:

7 0

10.0.0.0/8 is variably subnetted, 3 subnets, 2 masksC10.1.1.0/30 is directly connected, Serial0/0/0L10.1.1.1/32 is directly connected, Serial0/0/0R10.2.2.0/30 [120/1] via 10.1.1.2, 00:00:21, Serial0/0/0172.30.0.0/16 is variably subnetted, 2 subnets, 2 masksC172.30.10.0/24 is directly connected, GigabitEthernet0/1L172.30.10.1/32 is directly connected, GigabitEthernet0/1R3 only displays its own subnets for the 172.30.0.0 network. R3 does not have any routes for the172.30.0.0 subnets on R1.

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Find the missing value 8- = -5

podryga [215]

Answer:

x = 13

Step-by-step explanation:

call the missing number x

8 - x = - 5 Subtract 8 from both sides

8-8 - x = -5-8 Combine

-x = - 13 Multiply by -1

x = 13

6 0

2 years ago

Please I need some answers

Zarrin [17]

Answer:

Because each of the two numbers is a factor of the product, and since the

product has at least those two factors, it's a composite number by definition.

Step-by-step explanation:

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3 years ago

Y(0.5+2) please help me on this !!!

Alecsey [184]

Answer:

0.5y+2y

Step-by-step explanation:

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2 years ago

n HP laser printer is advertised to print text documents at a speed of 18 ppm (pages per minute). The manufacturer tells you tha

Dominik [7]

Answer:

0.321 is the probability that their mean printing speed of the sample is greater than 17.99 ppm.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 17.39 ppm

Standard Deviation, σ = 4.25 ppm

Sample size = 11

We are given that the distribution of printing speed is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

P(printing speed of the sample is greater than 17.99 ppm.)

P(x > 17.99)

$P( x > 17.99) = P( z > \displaystyle\frac{17.99-17.39}{\frac{4.25}{\sqrt{11}}}) = P(z > 0.468229)\\\\P( z > 0.468229) = 1 - P(z < 0.468229)$

Calculating the value from the standard normal table we have,

$1 - 0.679 = 0321= 32.1\%\\P( x > 17.99) = 32.1\%$

Thus, 0.321 is the probability that their mean printing speed of the sample is greater than 17.99 ppm.

8 0

3 years ago

A sandbox has an area of 10 2/3 square feet, and the length is 1/4 feet. What is the width of the sandbox?

AlekseyPX

Answer:

The width of the sandbox is $42\frac{2}{3} \ ft$ .

Step-by-step explanation:

Given,

Area of the sandbox = $10\frac{2}{3}=\frac{32}{3}\ ft^2$

Length = $\frac{1}{4}\ ft$

Solution,

Let the width of the sandbox be 'w'.

Since the sandbox is in shape of rectangle.

So we use the formula of area of rectangle.

$Area = Length\times Width$

On substituting the given values, we get;

$\frac{1}{4}\times w=\frac{32}{3}\\$

By cross multiplication method, we get;

$w=\frac{32\times4}{3\times 1} =\frac{128}{3} =42\frac{2}{3} \ ft$

Hence The width of the sandbox is $42\frac{2}{3} \ ft$ .

5 0

3 years ago