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Irina18 [472]
2 years ago
12

Jacob is training for a marathon. His plan is to run the same distance for 3 days a week, then increase that distance by the sam

e amount each week of training. During week 6, Jacob runs 14 miles per day, which is 1.5 miles more per day than he ran during week 5. Which equation represents the daily running distance, in miles, as a function of time, t, in weeks?
a.f(t) = 0.5t + 7
b.f(t) = 0.5t + 11
c.f(t) = 1.5t + 5
d.f(t) = 1.5t + 12.5
Mathematics
1 answer:
Hunter-Best [27]2 years ago
3 0
Given:
Week 6 = 14 miles
Week 5 = 1.5 miles less than week 6.

1.5 miles is the common difference.

Week 6 = 14 miles
Week 5 = 12.5 miles
Week 4 = 11 miles
Week 3 = 9.5 miles
Week 2 = 8 miles
Week 1 = 6.5 miles

f(t) = first term + common difference(t-1)
f(t) = 6.5 miles + 1.5 miles (t-1)

f(1) = 6.5 + 1.5(1-1)
f(1) = 6.5 miles

f(2) = 6.5 + 1.5(2-1)
f(2) = 6.5 + 1.5
f(2) = 8 miles
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The mean weight of newborn infants at a community hospital is 6.6 pounds. A sample of seven infants is randomly selected and the
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Answer:

The correct option is  A

Step-by-step explanation:

From the question we are told that

    The  population is  \mu  =  6.6

     The level of significance is \alpha  =  5\%  = 0.05

      The sample data is  9.0, 7.3, 6.0, 8.8, 6.8, 8.4, and 6.6 pounds

The Null hypothesis is H_o  :  \mu =  6.6

 The Alternative hypothesis is  H_a :  \mu  > 6.6

The critical value of the level of significance obtained from the normal distribution table is

                       Z_{\alpha } = Z_{0.05 } = 1.645

Generally the sample mean is mathematically evaluated as

      \=x =  \frac{\sum x_i }{n}

substituting values

      \=x =  \frac{9.0 +  7.3 +  6.0+ 8.8+ 6.8+ 8.4+6.6 }{7}

      \=x =  7.5571

The standard deviation is mathematically evaluated as

           \sigma  =  \sqrt{\frac{\sum  [ x -  \= x ]}{n} }

substituting values

          \sigma  =  \sqrt{\frac{  [ 9.0-7.5571]^2 + [7.3 -7.5571]^2 + [6.0-7.5571]^2 + [8.8- 7.5571]^2 + [6.8- 7.5571]^2 + [8.4 - 7.5571]^2+ [6.6- 7.5571]^2 }{7} }\sigma =  1.1774

Generally the test statistic is mathematically evaluated as

            t  =  \frac{\= x - \mu }  { \frac{\sigma }{\sqrt{n} } }

substituting values

           t  =  \frac{7.5571  - 6.6  }  { \frac{1.1774 }{\sqrt{7} } }

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  What this implies is that there is no sufficient evidence to state that the sample data show as significant increase in the average birth rate

The conclusion is that the mean is  \mu = 6.6  \ lb

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