$\overline{x}\pm z_c\dfrac{\sigma}{\sqrt{n}}$ , where $\overline{x}$ is the sample mean , $\sigma$ is the population standard deviation , n is the sample size and $z_c$ is the two-tailed test value for z.

Let x represents the time taken to mail products for all orders received at the office of this company.

As per given , we have

Confidence level : 95%

n= 62

sample mean : $\overline{x}=65$ hours

Population standard deviation : $\sigma=18$ hours

z-value for 93% confidence interval: $z_c=1.8119$ [using z-value table]

Now, 93% confidence the mean time taken to mail products for all orders received at the office of this company :-

$65\pm (1.8119)\dfrac{18}{\sqrt{62}}\\\\ 65\pm4.142\\\\=(65-4.142,\ 65+4.142)\\\\= (60.858,\ 69.142)$

Thus , 93% confidence the mean time taken to mail products for all orders received at the office of this company : (60.858, 69.142)

8 0

4 years ago

Sooooooooo<br> technically i've neevr lost a fight with a tiger

Aloiza [94]

Answer:

oh thats cool

Step-by-step explanation:

3 0

3 years ago

Choose the inverse of y = x2 - 10x.

PilotLPTM [1.2K]

=Y2-10Y

We move all terms to the left:

-(Y2-10Y)=0

We add all the numbers together, and all the variables

-(+Y^2-10Y)=0

We get rid of parentheses

-Y^2+10Y=0

We add all the numbers together, and all the variables

-1Y^2+10Y=0

a = -1; b = 10; c = 0;

Δ = b2-4ac

Δ = 102-4·(-1)·0

Δ = 100

The delta value is higher than zero, so the equation has two solutions

We use following formulas to calculate our solutions:

Y1=−b−Δ√2aY2=−b+Δ√2a

Δ‾‾√=100‾‾‾‾√=10

Y1=−b−Δ√2a=−(10)−102∗−1=−20−2=+10

Y2=−b+Δ√2a=−(10)+102∗−1=0−2=0

3 0

3 years ago