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2 years ago

Pls help due in 10 minutes!!!!!!!!!!!!!!1

Mathematics

1 answer:

AlladinOne [14]2 years ago

5 0

Answer:

Step-by-step explanation:

Given expression:

$27 \cdot \left(\left(3^3\right)^{-1}\right)$

Following the order of operations, carry out the operations inside the parentheses first:

$\implies 27 \cdot \left(\left(3 \cdot 3\cdot 3\right)^{-1}\right)$

$\implies 27 \cdot \left(27^{-1}\right)$

$\textsf{Apply exponent rule} \quad a^{-n}=\dfrac{1}{a^n}:$

$\implies 27 \cdot \left(\dfrac{1}{27^1}\right)$

$\implies 27 \cdot \left(\dfrac{1}{27}\right)$

Convert 27 to a fraction:

$\implies \dfrac{27}{1} \cdot \dfrac{1}{27}$

Cross cancel the common factor:

$\implies \dfrac{\diagup\!\!\!\!\!27}{1} \cdot \dfrac{1}{\diagup\!\!\!\!\!27}$

$\implies \dfrac{1}{1}$

$\implies 1$

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In a test of weight loss programs, 40 randomly selected adults used the Atkins weight loss program. After 12 months, their mean

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Answer:

90% confidence interval for the mean weight loss for all such subjects is [0.82 lbs, 3.38 lbs].

Step-by-step explanation:

We are given that in a test of weight loss programs, 40 randomly selected adults used the Atkins weight loss program.

After 12 months, their mean weight loss was found to be 2.1 lbs., with a standard deviation of 4.8 lbs.

Firstly, the pivotal quantity for 90% confidence interval for the true mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean weight loss of 40 adults = 2.1 lbs

s = sample standard deviation = 4.8 lbs

n = sample of adults = 40

$\mu$ = population mean weight loss

Here for constructing 90% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.685 < $t_3_9$ < 1.685) = 0.90 {As the critical value of t at 39 degree

of freedom are -1.685 & 1.685 with P = 5%}

P(-1.685 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.685) = 0.90

P( $-1.685 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.685 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.685 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X +1.685 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.685 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.685 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $2.1-1.685 \times {\frac{4.8}{\sqrt{40} } }$ , $2.1+1.685 \times {\frac{4.8}{\sqrt{40} } }$ ]

= [0.82 , 3.38]

Therefore, 90% confidence interval for the true mean weight loss for all such subjects is [0.82 lbs, 3.38 lbs].

Interpretation of this confidence interval is that we are 90% confident that the mean weight loss for all such subjects will lie between 0.82 lbs and 3.38 lbs.

So, if our true mean weight will be between 0.82 lbs and 2.1 lbs then we can say that the Atkins program appear to be effective. But it is somewhat appear to be practical kind of aspect.

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