If there is 99 apples and i take one of them how many apples are there now
2 answers:
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Using the normal distribution, it is found that there are 68 students with scores between 72 and 82.
<h3>Normal Probability Distribution</h3>The z-score of a measure X of a normally distributed variable with mean and standard deviation
is given by:
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
In this problem, the mean and the standard deviation are given, respectively, by:
The proportion of students with scores between 72 and 82 is the <u>p-value of Z when X = 82 subtracted by the p-value of Z when X = 72</u>.
X = 82:
Z = 1
Z = 1 has a p-value of 0.84.
X = 72:
Z = 0
Z = 0 has a p-value of 0.5.
0.84 - 0.5 = 0.34.
Out of 200 students, the number is given by:
0.34 x 200 = 68 students with scores between 72 and 82.
More can be learned about the normal distribution at brainly.com/question/24663213
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Answer:
A) Weighted graph is attached
B) Shortest routes are;
1. A → C → B → D → A
2. A → D → B → C → A
Step-by-step explanation:
A) We are told their home is in City A. So that's where any journey will begin from.
Furthermore we are told that;
City A to City B = 296 miles
City A to City C = 206 miles
City A to City D = 79 miles
City B to City C = 497 miles
City B to City D = 241 miles
City C to City D = 281 miles.
I have attached an image of the weighted graph showing the distances on the appropriate edges.
B) We want to find the shortest route using Brute force method. The brute force method is by solving a particular problem by checking all the possible cases/routes to get the desired result we are looking for.
In this case, the desired result is the shortest route for the family to complete their vacation. So, i have attached a diagram showing the different routes via brute force method.
From the brute force method, the shortest length route is 1023 miles and this routes are from Cities;
1. A → C → B → D → A
2. A → D → B → C → A
