9 + 5 x (-3) - (6-2) / 4

1 answer:

6 0

P - Parentheses
E - Exponents
M - Multiplication ( Left to right )
D - Division ( Left to right )
A - Addition ( Left to right )
S - Subtraction ( Left to right )

PEMDAS

Final Answer : -7

Hope this helps! :)

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Answer: C. $(29,\ 37.8)$

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The confidence interval for difference of two population mean is given by :-

$\overline{x}_1-\overline{x}_2\pm z_{\alpha/2}\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}$

Given : Level of significance : $1-\alpha:0.99$

Then , significance level : $\alpha: 1-0.99=0.01$

Critical value : $z_{\alpha/2}=z_{0.005}=2.576$

$n_1=30\ ;\ n_2=50\\\\\overline{x}_1=46.7\ ;\ \overline{x}_2=13.3\\\\s_1=9.2\ ;\ s_2=1.9$

$46.7-13.3\pm(2.576)\sqrt{\dfrac{9.2^2}{30}+\dfrac{1.9^2}{50}}\approx33.4\pm4.38=(29.02\ ,37.78)\approx(29,\ 37.8)$

Hence, a 99% confidence interval for the difference between the mean wait times of everyone who rode both rides $(29,\ 37.8)$

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