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timofeeve [1]
4 years ago
10

The horse racing record for a 1.50 mi track is shared by two horses. Fiddle isle, who ran the race in 143 s on march 21, 1970, a

nd john henry, who ran the same distance in an equal time on march 16, 1980. What were the horses' average speeds in:
a.Mi/s
Mathematics
1 answer:
egoroff_w [7]4 years ago
5 0

we know that

the average speed is equal to

average\ speed =\frac{distance}{time}

in this problem we have

distance=1.50\ miles\\ time=143\ seconds

substitute the values in the formula

average\ speed =\frac{1.5}{143}

average\ speed =0.01\frac{miles}{seconds}

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Step-by-step explanation:

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3 0
4 years ago
When it is operating properly, a chemical plant has a daily production rate that is normally distributed with a mean of 885 tons
-BARSIC- [3]

Answer:

<em>The test statistic Z = 1.844 < 1.96 at 0.05 level of significance</em>

<em>Null hypothesis is accepted </em>

<em>Yes he is right</em>

<em>The manager claims that at least 95 % probability that the plant is operating properly</em>

Step-by-step explanation:

<u>Explanation</u>:-

Given data Population mean

                                           μ     = 885 tons /day

Given random sample size

                                          n = 60

mean of the sample

                                        x⁻  = 875 tons/day

The standard deviation of the Population

                                      σ = 42 tons/day

<em><u>Null hypothesis</u></em><em>:- H₀: </em>The manager claims that at least 95 % probability that the plant is operating properly

<u><em>Alternative Hypothesis :H₁</em></u>: The manager do not claims that at least 95 % probability that the plant is operating properly

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

The test statistic

 Z = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }

Z = \frac{875 -885}{\frac{42}{\sqrt{60} } }

Z = \frac{-10}{5.422} = -1.844

|Z| = |-1.844| = 1.844

<em>The tabulated value</em>

<em>                          </em>Z_{\frac{0.05}{2} } = Z_{0.025} = 1.96<em></em>

<em>The calculated value Z = 1.844 < 1.96 at 0.05 level of significance</em>

<em>Null hypothesis is accepted </em>

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

<em>The manager claims that at least 95 % probability that the plant is operating properly</em>

<em></em>

<em></em>

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