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Lapatulllka [165]
3 years ago
11

Novice drivers lack the experience necessary to reliably categorize and appropriately respond to ___________ hazards.

Computers and Technology
1 answer:
meriva3 years ago
7 0

Answer:

roadside

Explanation:

Novice drivers lack the experience necessary to reliably categorize and appropriately respond to roadside hazards. A roadside hazard refers to any roadside objects or features that have a diameter greater than 100mm that is on or near the roadway. These are objects or features that are likely to create a dangerous environment to the drivers.

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Tanzania [10]

Answer:

I'm not gonna give you the answer. I want you to think about this what is math, what is science, how does it help the world, and etc. These are some example questions to help.

Explanation:

5 0
3 years ago
In which of the following locations can you edit all of the Properties of a PowerPoint file?
Lapatulllka [165]
Access the File<span> menu, choose </span>Info Pane<span> to get to </span>Backstage view, you can see Properties on t<span>he area on the right side </span>of the current PowerPoint presentation.  <span>Within the </span>Properties<span> pane click the </span>Show All Properties<span>  option , T</span><span>his will displays properties such as </span>Size<span>, the number of </span>Slides<span>,  </span>Hidden Slides<span>, the number of </span>Multimedia Clips, etc.  Some of the entries are editable w<span>ithin the  </span>Properties pane, and some are not. Just move your mouse cursor over any detail of a property. The editable sections will change the cursor into edit mode.  
6 0
3 years ago
A datagram network allows routers to drop packets whenever they need to. The probability of a router discarding a packetis p. Co
tresset_1 [31]

Answer:

a.) k² - 3k + 3

b.) 1/(1 - k)²

c.) k^{2}  - 3k + 3 * \frac{1}{(1 - k)^{2} }\\\\= \frac{k^{2} - 3k + 3 }{(1-k)^{2} }

Explanation:

a.) A packet can make 1,2 or 3 hops

probability of 1 hop = k  ...(1)

probability of 2 hops = k(1-k)  ...(2)

probability of 3 hops = (1-k)²...(3)

Average number of probabilities = (1 x prob. of 1 hop) + (2 x prob. of 2 hops) + (3 x prob. of 3 hops)

                                                       = (1 × k) + (2 × k × (1 - k)) + (3 × (1-k)²)

                                                       = k + 2k - 2k² + 3(1 + k² - 2k)

∴mean number of hops                = k² - 3k + 3

b.) from (a) above, the mean number of hops when transmitting a packet is k² - 3k + 3

if k = 0 then number of hops is 3

if k = 1 then number of hops is (1 - 3 + 3) = 1

multiple transmissions can be needed if K is between 0 and 1

The probability of successful transmissions through the entire path is (1 - k)²

for one transmission, the probility of success is (1 - k)²

for two transmissions, the probility of success is 2(1 - k)²(1 - (1-k)²)

for three transmissions, the probility of success is 3(1 - k)²(1 - (1-k)²)² and so on

∴ for transmitting a single packet, it makes:

     ∞                             n-1

T = ∑ n(1 - k)²(1 - (1 - k)²)

    n-1

   = 1/(1 - k)²

c.) Mean number of required packet = ( mean number of hops when transmitting a packet × mean number of transmissions by a packet)

from (a) above, mean number of hops when transmitting a packet =  k² - 3k + 3

from (b) above, mean number of transmissions by a packet = 1/(1 - k)²

substituting: mean number of required packet =  k^{2}  - 3k + 3 * \frac{1}{(1 - k)^{2} }\\\\= \frac{k^{2} - 3k + 3 }{(1-k)^{2} }

6 0
3 years ago
Which one bc im struggling
Setler79 [48]

Answer:

cant really see it

Explanation:

3 0
2 years ago
What is another word for violation
serg [7]

Answer:

Infringement

Explanation:

The others are incorrect

4 0
3 years ago
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