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kipiarov [429]
4 years ago
10

Two cyclists leave a city at the same time, one going east and the other going west. The westbound cyclist bikes at 3 mph faster

than the eastbound cyclist. If after 6 hours they are 162 miles apart, how fast is each cyclist riding?
If you can solve this, can you please show your work and explain what I need to plug in where, and what equation I need to use? This is so confusing.
Mathematics
1 answer:
NikAS [45]4 years ago
7 0
We will say that the eastbound cyclist is going x mph. This means that the westbound cyclist is going x+3 mph.
They are both cycling for 6 hours and end up 162 miles apart.
Distance=rate*time
6x+6(x+3)=162
6x+6x+18+162
12x=144
x=12
So, the eastbound cyclist is going 12 mph.
Remember the westbound cyclist is going 3 mph faster, so s/he is going 15 mph.
Hope this helps!!
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The price of a tv was decreased by 20% to £1440. What was the price before the decrease?
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Dmitry_Shevchenko [17]

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s defined to be the dollar value of loans defaulted divided by the total dollar value of all loans made. Banking officials claim
Amiraneli [1.4K]

Answer:

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the mean bad debt ratio for Ohio banks is different than the Midwestern average (3.5%).

Test statistic t = 1.431

Critical values tc = ±2.447

P-value = 0.203

Step-by-step explanation:

This is a hypothesis t-test for the population mean.

The claim is that the mean bad debt ratio for Ohio banks is different than the Midwestern average (3.5%).

Then, the null and alternative hypothesis are:

H_0: \mu=3.5\\\\H_a:\mu\neq 3.5

The significance level is 0.05.

The sample has a size n=7.

We calculate  the sample mean and standard deviation as:

M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{7}(7+4+6+3+5+4+2)\\\\\\M=\dfrac{31}{7}\\\\\\M=4.43\\\\\\

s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{6}((7-4.43)^2+(4-4.43)^2+(6-4.43)^2+. . . +(2-4.43)^2)}\\\\\\s=\sqrt{\dfrac{17.71}{6}}\\\\\\s=\sqrt{2.95}=1.72\\\\\\

The sample mean is M=4.43.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=1.72.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{1.72}{\sqrt{7}}=0.65

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{4.43-3.5}{0.65}=\dfrac{0.93}{0.65}=1.431

The degrees of freedom for this sample size are:

df=n-1=7-1=6

This test is a two-tailed test, with 6 degrees of freedom and t=1.431, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=2\cdot P(t>1.431)=0.203

As the P-value (0.203) is bigger than the significance level (0.05), the effect is not significant.

If we use the critical value approach, for this level of confidence, the critical values are tc = ±2.447. The test statistic is within the bounds of the critical values and falls within the acceptance region.

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the mean bad debt ratio for Ohio banks is different than the Midwestern average (3.5%).

3 0
3 years ago
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