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Tresset [83]
1 year ago
11

What is the total number of different routes that a fire

Mathematics
1 answer:
Pavlova-9 [17]1 year ago
4 0

The total number of different routes that a fire truck can travel the m-distance from F to Z = 4 (Option C).

<h3 /><h3>What is a graph?</h3>

A graph refers to a diagram that shows how a variable varies in relation to one or more other variables, such as a collection of points, lines, line segments, curves, or regions.

Now,

Firstly, refer to the graph attached.

  • The graph can be used to manually calculate the number of routes that lead from F to Z.
  • F is four m-distances from Z.
  • Now, manually calculate the m-distance of the path from F to Z using the 4 boxes that represent it. We discover 6 distinct pathways, each with an m-distance of 4.

Hence, The total number of different routes that a fire truck can travel the m-distance from F to Z = 4 (Option C).

To learn more about graphs, refer to the link: brainly.com/question/4025726

#SPJ4

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Solve 3(a + 3) – 6 = 21.
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Answer:

a=6

Step-by-step explanation:

to find the value of a you need to simplify the equation first. so...

3(a+3)-6=21 (you remove the bracket first)

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3a+3=21 (you collect the like terms then)

3a=21-3

3a=18 (then you both divide both sides by 3 to find the value of a)

a=18/3

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to check your answer substitute 3 instead of a

3(a+3)-6=21

3(6+3)-6=21

3(9)-6=21 (according to BODMAS since multiplication comes first you multiply 3 with 9 before subtracting it from 6.)

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Correct Ans:
Option A. 0.0100

Solution:
We are to find the probability that the class average for 10 selected classes is greater than 90. This involves the utilization of standard normal distribution.

First step will be to convert the given score into z score for given mean, standard deviation and sample size and then use that z score to find the said probability. So converting the value to z score:

z-score= \frac{90-83.6}{ \frac{8.7}{ \sqrt{10} } } \\  \\ &#10;z-score =2.326

So, 90 converted to  z score for given data is 2.326. Now using the z-table we are to find the probability of z score to be greater than 2.326. The probability comes out to be 0.01.

Therefore, there is a 0.01 probability of the class average to be greater than 90 for the 10 classes.
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