We wish to give a 90% confidence interval for the mean value of a normally distributed random variable. we obtain a simple random sample of 9 elements from the population on which the variable is defined. the sample has a mean value of 10.2 and a sample standard deviation 1.5. find the 90% confidence interval for the mean value.

Step-by-step explanation:

Given : Sample size : n= 9

Degree of freedom = df =n-1 =8

Sample mean :

sample standard deviation :

Significance level ;

Since population standard deviation is not given , so we use t- test.

Using t-distribution table , we have

Critical value =

Confidence interval for the population mean :

90% confidence interval for the mean value will be :

Hence, the 90% confidence interval for the mean value=

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:

(8+64)^2= 5184 < answer

step-by-step explanation:

2^3 = 8

4^3= 64

(8+64)^2 = 5184