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garri49 [273]
3 years ago
8

Can somebody give me a Quick answer. First person gets Brainliest.

Mathematics
2 answers:
Evgen [1.6K]3 years ago
4 0

Answer:D

Step-by-step explanation:

Doss [256]3 years ago
3 0

Answer:

D. -4½

Step-by-step explanation:

Remember PEMDAS:

Parentheses

Exponents

Multiply and Divide (in any order)

Add and Subtract (in any order)

\begin{array}{rl}& \dfrac{3}{4}\left(\dfrac{2}{3} + \dfrac{4}{3} \right ) - 9 \div\dfrac{1}{2} \times\dfrac{1}{3} \\\\\text{Do the addition inside parentheses} &\dfrac{3}{4}\left(\dfrac{6}{3} \right ) - 9 \div\dfrac{1}{2} \times\dfrac{1}{3} \\ \\\text{Simplify the new fraction} &\dfrac{3}{4}(2) - 9 \div\dfrac{1}{2} \times\dfrac{1}{3} \\ \\\text{Multiply} &\dfrac{3}{2} - 9 \div\dfrac{1}{2} \times\dfrac{1}{3} \\\\\end{array}

\begin{array}{rl}\text{Invert the fraction and change divide to multiply} &\dfrac{3}{2} - 9 \times 2 \times\dfrac{1}{3} \\\\\text{Cancel the threes} &\dfrac{3}{2} - 3 \times 2\\\\\text{Multiply} &\dfrac{3}{2} - 6 \\\\\text{Find lowest common denominator } &\dfrac{3}{2} - \dfrac{12}{2} \\\\\text{Add the numerators} &\dfrac{-9}{2} \\\\\text{Convert to a mixed number} &=\mathbf{-4\frac{1}{2}} \\\\\end{array}

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Assuming that the population is normally​ distributed, construct a 9090​% confidence interval for the population mean for each o
jasenka [17]

Given:

Set A: 1 4 4 4 5 5 5 8

Mean: 4.5

Standard dev: 1.9

 

Set B:

Mean: 4.5

Standard dev: 2.45

 

% =  90 

Set A:

Standard Error, SE = s/ √n =    1.9/√8 = 0.67  

Degrees of freedom = n - 1 =   8 -1 =  7   

t- score =  1.89457861

<span> <span><span> <span>   </span> </span> </span></span>

Width of the confidence interval = t * SE =     1.89457861* 0.67 = 1.272685913

Lower Limit of the confidence interval = x-bar - width =      4.5 - 1.272685913 = 3.23

Upper Limit of the confidence interval = x-bar + width =      4.5 + 1.272685913 = 5.77

The 90% confidence interval is [3.23, 5.77]

 

Set B:

Standard Error, SE = s/ √n =    2.45/√8 = 0.87  

Degrees of freedom = n - 1 =   8 -1 = 7   

t- Score =  1.89457861

<span> <span><span> <span>   </span> </span> </span></span>

Width of the confidence interval = t * SE =     1.89457861* 0.87 = 1.641094994

Lower Limit of the confidence interval = x-bar - width =      4.5 - 1.641094994 = 2.86

Upper Limit of the confidence interval = x-bar + width =      4.5 + 1.641094994 = 6.14

The 90% confidence interval is [2.86, 6.14]

 

<span>We can obviously see that sample B has more variation in the scores than sample A. The fact that the standard deviation is 2.45 for B and 1.9 for A). Therefore, they yield dissimilar confidence intervals even though they have the same mean and range.</span>

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Step-by-step explanation:

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