Part

Suppose that 35 people are divided in a random mannerinto two teams in such a way that one team contains10 people and the other

Alisiya [41]

To be precise, the size of your sample space is (2410)(2410). This number does go on the bottom of the fraction, and what goes on top is the size of the event. Break up the event into independent events 1. choose the 2 defective bulbs, and 2. choose the remaining 8 bulbs. I don't have much choice in event 1. There is only one way to choose both of the defective balls. In other words, (22)(22) (choosing 2 defective bulbs from a set of 2 defective bulbs). For event 2, there are 24−2=2224−2=22 nondefective bulbs, and I must choose 88 of them, so that's (228)(228). Finally, since events 1 and 2 are independent, we multiply the answers for the combined event: (22)(228)(22)(228)

P=(22)(228)(2410)P=(22)(228)(2410)

Or, since (22)=1(22)=1,

P=(228)(2410)P=(228)(2410)

Hope this helps!

7 0

3 years ago

A digital camcorder repair service has set a goal not to exceed an average of 5 working days from the time the unit is brought i

garri49 [273]

Answer:

The degrees of freedom first given by:

$df=n-1=12-1=11$

Then we can find the critical value taking in count the degrees of freedom and the alternative hypothesis and then we need to find a critical value who accumulates 0.05 of the area in the right tail and we got:

$t_{\alpha}= 1.796$

And for this case the rejection region would be:

b) Reject H0 if tcalc >1.7960

Step-by-step explanation:

Information given

5, 7, 4, 6, 7, 5, 5, 6, 4, 4, 7, 5.

System of hypothesis

We want to test if the true mean is higher than 5, the system of hypothesis are :

Null hypothesis: $\mu \leq 5$

Alternative hypothesis: $\mu > 5$

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

The degrees of freedom first given by:

$df=n-1=12-1=11$

Then we can find the critical value taking in count the degrees of freedom and the alternative hypothesis and then we need to find a critical value who accumulates 0.05 of the area in the right tail and we got:

$t_{\alpha}= 1.796$

And for this case the rejection region would be:

b) Reject H0 if tcalc >1.7960

6 0

3 years ago

painting measures 2.4 feet from left to right. It’s hung in the center of a rectangular wall measuring 20 feet from left to righ

Alenkasestr [34]

8.8, you do 20-2.4, then divide that sum

6 0

2 years ago

Read 2 more answers

How can you simplify 1+15