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Basile [38]
3 years ago
5

Please help!!

Mathematics
1 answer:
Misha Larkins [42]3 years ago
5 0
Its 26.315% or rounded to 26.32% commission

Explanation; you take the total of the car, and then divide that by the exact amount Ken earned leaving you with the commission

18,250 ÷ 693.50 = 26.315789474 (normally using only the first two decimals when counting money, rounded to the nearest tenth) leaving you with 26.32
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Step-by-step explanation:

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2. Use the distributive property to rewrite the following expression.<br> 3(x+5)
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The answer is 3x+15

You multiply the x and 5 by 3 to get 3x and 15

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Which statement proves that parallelogram KLMN is a rhombus?
IrinaVladis [17]

Answer:

It is the last option.

Step-by-step explanation:

The diagonals of a rhombus (KM and NL ) are perpendicular. That is shown by the diagonals having slopes of 1 and -1.

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Suppose μ1 and μ2 are true mean stopping distances at 50 mph for cars of a certain type equipped with two different types of bra
Stella [2.4K]

Answer:

The 95% confidence interval would be given by -23.44 \leq \mu_1 -\mu_2 \leq -8.16  

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

\bar X_1 =114.1 represent the sample mean 1

\bar X_2 =129.9 represent the sample mean 2

n1=5 represent the sample 1 size  

n2=2 represent the sample 2 size  

s_1 =5.08 sample standard deviation for sample 1

s_2 =5.37 sample standard deviation for sample 2

\mu_1 -\mu_2 parameter of interest.

Confidence interval

The confidence interval for the difference of means is given by the following formula:  

(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}} (1)  

The point of estimate for \mu_1 -\mu_2 is just given by:

\bar X_1 -\bar X_2 =114.1-129.9=-15.8

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:  

df=n_1 +n_2 -1=5+5-2=8  

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,8)".And we see that t_{\alpha/2}=\pm 2.31  

The standard error is given by the following formula:

SE=\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}

And replacing we have:

SE=\sqrt{\frac{5.08^2}{5}+\frac{5.37^2}{5}}=3.306

Confidence interval

Now we have everything in order to replace into formula (1):  

-15.8-2.31\sqrt{\frac{5.08^2}{5}+\frac{5.37^2}{5}}=-23.437  

-15.8+2.31\sqrt{\frac{5.08^2}{5}+\frac{5.37^2}{5}}=-8.163  

So on this case the 95% confidence interval would be given by -23.44 \leq \mu_1 -\mu_2 \leq -8.16  

8 0
3 years ago
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