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vichka [17]
3 years ago
12

Denise has 6 hours to spend training for an upcoming race. She completes her training by running full speed the distance of the

race and walking back the same distance to cool down. If she runs at a speed of 7mph and walks back at a speed of 3mph, how long should she plan to spend walking back?
Mathematics
1 answer:
VMariaS [17]3 years ago
5 0
  To solve this problem,you need to use the formula d = rd (distance = rates x time)She runs at a speed of 7 mph and walks at a speed of 3 mph. Her distance running is d = 7trwhere tr is the time she spends running Her distance walking isd = 3twwhere tw is the time she spends walking The distances are the same so7tr = 3tw We also know that the total time is 4 hourstr + tw = 4tr = 4-tw Substitute this value of tr in the first equation7tr = 3tw7(4-tw) = 3tw28-7tw = 3tw28 = 10tw2.8 = tw Denise will spend 2.8 hours (2 hours, 48 minutes) walking back and 1.2 hours (1 hour, 12 minutes running.

Hope I helped :)
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A national grocery store chain wants to test the difference in the average weight of turkeys sold in Detroit and the average wei
Arte-miy333 [17]

Answer:

<em>Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom</em>

<em>The null hypothesis is accepted . </em>

<em>Assume the population variances are approximately the same</em>

<u><em>Step-by-step explanation:</em></u>

<u>Explanation</u>:-

Given data a random sample of 20 turkeys sold at the chain's stores in Detroit yielded a sample mean of 17.53 pounds, with a sample standard deviation of 3.2 pounds

<em>The first sample size  'n₁'= 20</em>

<em>mean of the first sample 'x₁⁻'= 17.53 pounds</em>

<em>standard deviation of first sample  S₁ = 3.2 pounds</em>

Given data a random sample of 24 turkeys sold at the chain's stores in Charlotte yielded a sample mean of 14.89 pounds, with a sample standard deviation of 2.7 pounds

<em>The second sample size  n₂ = 24</em>

<em>mean of the second sample  "x₂⁻"= 14.89 pounds</em>

<em>standard deviation of second sample  S₂ =  2.7 poun</em>ds

<u><em>Null hypothesis</em></u><u>:-</u><u><em>H₀</em></u><em>: The Population Variance are approximately same</em>

<u><em>Alternatively hypothesis</em></u><em>: H₁:The Population Variance are approximately same</em>

<em>Level of significance ∝ =0.05</em>

<em>Degrees of freedom ν = n₁ +n₂ -2 =20+24-2 = 42</em>

<em>Test statistic :-</em>

<em>    </em>t = \frac{x^{-} _{1} -  x_{2} }{\sqrt{S^2(\frac{1}{n_{1} } }+\frac{1}{n_{2} }  }

<em>    where         </em>S^{2}   = \frac{n_{1} S_{1} ^{2}+n_{2}S_{2} ^{2}   }{n_{1} +n_{2} -2}

                      S^{2} = \frac{20X(3.2)^2+24X(2.7)^2}{20+24-2}

<em>              substitute values and we get  S² =  40.988</em>

<em>     </em>t= \frac{17.53-14.89 }{\sqrt{40.988(\frac{1}{20} }+\frac{1}{24}  )}<em></em>

<em>  </em>   t =  1.3622

  Calculated value t = 1.3622

Tabulated value 't' =  2.081

Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom

<u><em>Conclusion</em></u>:-

<em>The null hypothesis is accepted </em>

<em>Assume the population variances are approximately the same.</em>

<em>      </em>

<em>                        </em>

<em>                    </em>

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Where P(Px | Py) is the center of the circle.

If you insert one point of the circle, the equation must be true.

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