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soldier1979 [14.2K]
3 years ago
9

11. Two cyclists start biking from a trail’s start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 h

ours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?
Mathematics
1 answer:
Margarita [4]3 years ago
8 0

Answer:

<em>4.5 hours</em>

Step-by-step explanation:

distance = velocity × time

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Answer:

x = 1\2 and y = 0

Step-by-step explanation:

8x + 3y = 4 * 5\3

14x - 5y = 7 * 1

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82\3x = 41\3

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10

Step-by-step explanation:

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Let us suppose we have data on the absorbency of paper towels that were produced by two different manufacturing processes. From
maksim [4K]

Answer:

The 95% CI for the difference of means is:

-155.45 \leq \mu_1-\mu_2 \leq -44.55

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>"Find a 95% confidence interval on the difference of the towels mean absorbency produced by the two processes. Assumed that the standard deviations are estimated from the data. Round to two decimals places."</em>

Process 1:

- Sample size: 10

- Mean: 200

- S.D.: 15

Process 2:

- Sample size:  4

- Mean: 300

- S.D.: 50

The difference of the sample means is:

M_d=M_1-M_2=200-300=-100

The standard deviation can be estimated as:

\sigma_d=\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}\\\\\sigma_d=\sqrt{\frac{15^2}{10}+\frac{50^2}{4}} =\sqrt{22.5+625}=\sqrt{647.5}=25.45

The degrees of freedom are:

df=n_1+n_2-2=10+4-2=12

The t-value for a 95% confidence interval and 12 degrees of freedom is t=±2.179.

Then, the confidence interval can be written as:

M_d-t\cdot \sigma_d\leq \mu_1-\mu_2 \leq M_d+t\cdot \sigma_d\\\\-100-2.179*25.45\leq \mu_1-\mu_2 \leq -100+2.179*25.45\\\\-100-55.45 \leq \mu_1-\mu_2 \leq -100+55.45\\\\ -155.45 \leq \mu_1-\mu_2 \leq -44.55

8 0
3 years ago
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