Your company prepares and distributes frozen foods. The package claims a net weight of 24.5 ounces. A random sample of today's production was weighted, and the results were summarized as follows: average=24.41 ounces; standard deviation=0.11 ounce. Sample size=5 packages. Find the 95% confidence interval for the mean weight you would have found had you weighed all packages produced today. Interpret your results.

The 95% CI =  24.506 ≤ x≤ 24.314

The 95% CI is 0.9 less than the package claim at CI = 24.404 ≤ x≤ 24.596

Step-by-step explanation:

Here e have

μ = Population mean = 24.5 ounces

= Sample mean = 24.41 ounces

σ  = Standard deviation = 0.11 ounces

n = Sample size = 5 packages

At 95% confidence, we have

Where z = 1.96 and -z = -1.96

Therefore, the confidence interval is given by

CI =24.41 1.96 × =  24.506, 24.314 or

The 95% CI =  24.506 ≤ x≤ 24.314

When = μ, we have

which gives

CI = 24.404 ≤ x≤ 24.596

This shows that at 95 % confidence level, the observed mean of the packages produced on the day is approximately equivalent to the package claim or 0.09 less than package claim.

27

step-by-step explanation:

answer: solve 5\cdot 2^x=2405⋅2  

x

=2405, dot, 2, start superscript, x, end superscript, equals, 240.

solution

to solve for xxx, we must first isolate the exponential part. to do this, divide both sides by 555 as shown below. we do not multiply the 555 and the 222 as this goes against the order of operations!

\begin{aligned} 5\cdot 2^x& =240 2^x& =48 \\ \end{aligned}  

5⋅2  

x

2  

x

​    

=240

=48

​

step-by-step explanation: