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shusha [124]
3 years ago
7

Joel and Gunner both run the 800-meter race. The line graph below shows their times for the first five meets of the track season

. What was Gunner's range of times in the first five meets?
A) 1.0 Minute
B) 0.75 Minute
C) 0.5 Minute
D) 0.25 Minute

Part B

In how many meets did Joel have a faster time than Gunner?

A) 1
B) 2
C) 3
D) 4

Part C

If it were possible for Gunner to continue improving his time as he has been, what would be a good prediction for his time in the sixth meet.

A) 2.0 Minutes
B) 1.75 Minutes
C) 1.5 Minutes
D) 1.25 Minutes

Mathematics
2 answers:
alex41 [277]3 years ago
7 0
1. A 1.0 minute
Range can be calculated by subtracting 3.0 mins - 2.0 mins = 1.0 minute.

2. D. 4
Joel was faster than Gunner in meets 1, 2, 3 and 4.

3. B. 1.75 minutes
Gunner has consistent improvement at intervals of 0.25 minutes per meet. Subtracting 2.0 mins by 0.25 mins we get 1.75 minutes. 
Olenka [21]3 years ago
6 0

ANSWER:

d

Step-by-step explanation:

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3 years ago
City health officials will conduct a two-sample t-test for a difference in means to investigate whether
Anvisha [2.4K]

Answer:

The calculated value     t =   2.366 > 2.0301 at 0.05 level of significance

The null hypothesis is rejected

There is a difference between the means      

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given that the first sample size 'n₁' = 22

Given that the mean of the first sample x₁⁻ = 120

Given that the standard deviation of the sample (s₁) = 20

Given that the mean of the second sample x₂⁻ = 100

Given that the standard deviation of the second sample (S₂) = 30

<u><em>Step(ii):-</em></u>

T-test statistic

               t = \frac{x^{-} _{1} -x^{-} _{2} }{\sqrt{S^{2} (\frac{1}{n_{1} }+\frac{1}{n_{2} } ) } }

     where

             S^{2} = \frac{n_{1} S_{1} ^{2}+n_{2} S_{2} ^{2}   }{n_{1} +n_{2}-2 }

            S^{2} = \frac{22( 20)^{2}+ 15(30) ^{2}   }{22 +15-2 }

           S² =  637.1428

           S = √637.1428 = 25.24168  

The standard deviation of the sample 'S' = 25.24168   

<u><em>Step(iii):-</em></u>

<u><em>Null Hypothesis:H₀:</em></u>

There is no difference between the means

<u><em>Alternative Hypothesis:H₁:</em></u>

There is a difference between the means

         T-test statistic

               t = \frac{x^{-} _{1} -x^{-} _{2} }{\sqrt{S^{2} (\frac{1}{n_{1} }+\frac{1}{n_{2} } ) } }

              t = \frac{ 120 -100 }{\sqrt{637.14 (\frac{1}{22 }+\frac{1}{15 } ) } }

             t =   2.366

Degrees of freedom

           γ = n₁+n₂ -2 = 22+15-2 = 35

     t₀.₀₅ = 2.0301

The calculated value     t =   2.366 > 2.0301 at 0.05 level of significance

The null hypothesis is rejected

There is a difference between the means          

3 0
3 years ago
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