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sergejj [24]
3 years ago
14

Ashley runs around the following track 100 m • 60 m 그 How many times must she run around the track in order to run a total dista

nce of 4 km? Show your work Hint: 1 km = 1000 m​ pls help

Mathematics
1 answer:
In-s [12.5K]3 years ago
6 0

9514 1404 393

Answer:

  10.3 laps

Step-by-step explanation:

On each lap, Ashley runs two lengths of 100 m and the circumference of a 60 m circle. That circumference is ...

  C = πd

  C = π(60 m) ≈ 188.496 m

So, the length of one lap is ...

  2·100 m + 188.496 m = 388.496 m

The number of times around for 4000 m will be ...

  (4000 m)/(388.496 m/lap) = 10.30 laps

Ashley must run 10.3 times around in order to run a total distance of 4 km.

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In a litter of 7 kittens, each kitten weighs less than 3.5 ounces. Find all the possible values of the combined weights of the k
Dima020 [189]

Let's use w to symbolize the weight of kittens. Given by the statement "each kitten weighs less than 3.5 ounces" we know that

1*w < 3.5

We can multiply both sides of the inequality by 7 to determine the total weights

7*(1*w) < 7*(3.5)

7*w < 24.5

Since there are 7 kittens, the combined weight of the kittens is 7w, therefore the above expression could be read as "The combined weight of the kittens is less than 24.5 ounces"

6 0
3 years ago
In analyzing the city and highway fuel efficiencies of many cars and trucks, the mean difference in fuel efficiencies for the 60
Olegator [25]

Answer:

7.38-1.96\frac{2.51}{\sqrt{601}}=7.18    

7.38+1.96\frac{2.51}{\sqrt{601}}=7.58    

For this case the confidence interval is given by :

7.18 \leq \mu \leq 7.58

Step-by-step explanation:

Data provided

\bar X=7.38 represent the sample mean for the fuel efficiencies

\mu population mean

s=2.51 represent the sample standard deviation

n=601 represent the sample size  

Confidence interval

The formula for the confidence interval of the true mean is given by:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

The degrees of freedom for this case is given by:

df=n-1=601-1=600

The Confidence level for this case is 0.95 or 95%, and the significance level \alpha=0.05 and \alpha/2 =0.025 and the critical value is given by t_{\alpha/2}=1.96

Replcing into the formula for the confidence interval is given by:

7.38-1.96\frac{2.51}{\sqrt{601}}=7.18    

7.38+1.96\frac{2.51}{\sqrt{601}}=7.58    

For this case the confidence interval is given by :

7.18 \leq \mu \leq 7.58

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3 years ago
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