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aniked [119]
2 years ago
15

From home, Mary’s work is two thirds along the way to training. Training is 2.5km from work. Mary normally goes to work, then tr

aining and then home again. However, today she forgot her shoes. How far will Mary travel in total today if she has to go home before training to get her shoes?

Mathematics
1 answer:
polet [3.4K]2 years ago
3 0
First thing to do is to illustrate the problem, Since it was mentioned that work was along the way to training, the order is shown in the picture. Mary's home and workplace are nearer compared to her training center. It is also mentioned that the distance between work and home, denoted as x, is 2/3 of the total distance from home to training. The total distance is (x + 2.5). Thus,

x = 2/3(x+2.5)
x = 2/3 x + 5/3
1/3 x = 5/3
x = 5 km

Thus, the distance from home to work is 5 km. This means that Mary has to walk this distance twice to return home to get her shoes. Then, she will travel again the total distance of 5+2.5 = 7.5 km to get to her training center. So,

Total distance = 2(5km) + 7.5 km
Total distance = 17.5 km

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2 years ago
Question 13
kicyunya [14]

Answer:

The c intercept is 42

The t intercepts are: 6, -1 and 7

Step-by-step explanation:

Given

c(t) = (t - 6)(t +1)(t-7)

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Simply set t to 0

c(t) = (t - 6)(t +1)(t-7)

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Solving (b): The t intercept

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2 years ago
A study suggested that childrenbetween the ages of 6 and 11 in the US have anaverage weightof 74 lbs. with a standard deviation
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Answer:

The proportion of children in this age range between 70 lbs and 85 lbs is of 0.9306.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

A study suggested that children between the ages of 6 and 11 in the US have an average weightof 74 lbs, with a standard deviation of 2.7 lbs.

This means that \mu = 74, \sigma = 2.7

What proportion of childrenin this age range between 70 lbs and 85 lbs.

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X = 85

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Z = 4.07

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Hope this helps! :)

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