i thank it is 0.5 becose that is what retane out of this qestion so 0.5 is the anser
The length and the width of the rectangle are 20 and 14 inches respectively
<h3>How to find the length and width?</h3>
The given parameters are:
Length = 6 + Width
Perimeter = 68 inches
The perimeter of a rectangle is calculated as:
Perimeter = 2 * (Length + Width)
So, we have:
2 * (Length + Width) = 68
Divide both sides by 2
Length + Width = 34
Substitute Length = 6 + Width in Length + Width = 34
6 + Width + Width = 34
Evaluate the like terms
2 * Width = 28
Divide both sides by 2
Width = 14
Substitute Width = 14 in Length = 6 + Width
Length = 6 + 14
Evaluate
Length = 20
Hence, the length and the width of the rectangle are 20 and 14 inches respectively
Read more about perimeter at:
brainly.com/question/24571594
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Answer:
The confidence interval for the difference in proportions is
![-0.028\leq p_1-p_2 \leq 0.096](https://tex.z-dn.net/?f=-0.028%5Cleq%20p_1-p_2%20%5Cleq%200.096)
No. As the 95% CI include both negative and positive values, no proportion is significantly different from the other to conclude there is a difference between them.
Step-by-step explanation:
We have to construct a confidence interval for the difference of proportions.
The difference in the sample proportions is:
![p_1-p_2=x_1/n_1-x_2/n_2=(183/217)-(322/398)=0.843-0.809\\\\p_1-p_2=0.034](https://tex.z-dn.net/?f=p_1-p_2%3Dx_1%2Fn_1-x_2%2Fn_2%3D%28183%2F217%29-%28322%2F398%29%3D0.843-0.809%5C%5C%5C%5Cp_1-p_2%3D0.034)
The estimated standard error is:
![\sigma_{p_1-p_2}=\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2} } \\\\\sigma_{p_1-p_2}=\sqrt{\frac{0.843*0.157}{217}+\frac{0.809*0.191}{398} } \\\\\sigma_{p_1-p_2}=\sqrt{0.000609912+0.000388239}=\sqrt{0.000998151} \\\\ \sigma_{p_1-p_2}=0.0316](https://tex.z-dn.net/?f=%5Csigma_%7Bp_1-p_2%7D%3D%5Csqrt%7B%5Cfrac%7Bp_1%281-p_1%29%7D%7Bn_1%7D%2B%5Cfrac%7Bp_2%281-p_2%29%7D%7Bn_2%7D%20%7D%20%5C%5C%5C%5C%5Csigma_%7Bp_1-p_2%7D%3D%5Csqrt%7B%5Cfrac%7B0.843%2A0.157%7D%7B217%7D%2B%5Cfrac%7B0.809%2A0.191%7D%7B398%7D%20%7D%20%5C%5C%5C%5C%5Csigma_%7Bp_1-p_2%7D%3D%5Csqrt%7B0.000609912%2B0.000388239%7D%3D%5Csqrt%7B0.000998151%7D%20%5C%5C%5C%5C%20%5Csigma_%7Bp_1-p_2%7D%3D0.0316)
The z-value for a 95% confidence interval is z=1.96.
Then, the lower and upper bounds are:
![LL=(p_1-p_2)-z*\sigma_p=0.034-1.96*0.0316=0.034-0.062=-0.028\\\\\\UL=(p_1-p_2)+z*\sigma_p=0.034+1.96*0.0316=0.034+0.062=0.096](https://tex.z-dn.net/?f=LL%3D%28p_1-p_2%29-z%2A%5Csigma_p%3D0.034-1.96%2A0.0316%3D0.034-0.062%3D-0.028%5C%5C%5C%5C%5C%5CUL%3D%28p_1-p_2%29%2Bz%2A%5Csigma_p%3D0.034%2B1.96%2A0.0316%3D0.034%2B0.062%3D0.096)
The confidence interval for the difference in proportions is
![-0.028\leq p_1-p_2 \leq 0.096](https://tex.z-dn.net/?f=-0.028%5Cleq%20p_1-p_2%20%5Cleq%200.096)
<em>Can it be concluded that there is a difference in the proportion of drivers who wear a seat belt at all times based on age group?</em>
No. It can not be concluded that there is a difference in the proportion of drivers who wear a seat belt at all times based on age group, as the confidence interval include both positive and negative values.
This means that we are not confident that the actual difference of proportions is positive or negative. No proportion is significantly different from the other to conclude there is a difference.
Answer:
y=x+1
Step-by-step explanation:
The answer for this question is a) 2$
He bought 1500 shares (12000/8)
and sold all of them for 15000 which means that he sold each share for 10$ (15000/1500)
So he got got a profit of 2$ (10$-8$)