What is the value of the next term of the sequence?

The anand group is a shipping company that has just heard the bad news that in shipment of six televisions, two are defective. p

Marina86 [1]

2/3 or 4/6 chance of the tv to work 1/3 or 2/6 chance it won't work.

3 0

3 years ago

The amount of money that Li spends on a video game depends on how many game he purchases. if Li has at most $90 to spend on game

siniylev [52]

Answer:

Hence Domain set for function is {0,1,2,3,4}

Step-by-step explanation:

Given:

Total amount spend =90

22.5 $ each game cost

To Find:

What is domain for the function

Solution:

Now domain mean all possible value for input so that it gives real output as per condition mentioned

So Our condition is that li spends money on video game depends on how many game he purchases

so total money spent on game is output

i.e in order to get that much amount of games.

If he spends 22.5 $ for each game and Total amount spent is 90

So 90/22.5=4 Now amount of game he would purchase will be 4

i.e range for lil is 0 to 4 wiht 90 $ in hand

So Domain set will be ranging from{0,1,2,3,4}

7 0

3 years ago

How to solve for vertex form<br> y = -x^2 + 4x – 1

Nastasia [14]

Answer: y= -(x-2)^2+3

6 0

3 years ago

a hockey team ordered 12 wooden pedestals to mount their brass trophies on. The design pedestal is shown below. The teal is char

yKpoI14uk [10]

56 cause u need 12 divide by 0.08 12 times

8 0

3 years ago

Read 2 more answers

From a large number of actuarial exam scores, a random sample of scores is selected, and it is found that of these are passing s

Mnenie [13.5K]

<u>Supposing 60 out of 100 scores are passing scores</u>, the 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).

The lower limit is 0.5.
The upper limit is 0.7.

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $\frac{1+\alpha}{2}$ .

60 out of 100 scores are passing scores, hence $n = 100, \pi = \frac{60}{100} = 0.6$

95% confidence level

So $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so $z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6 - 1.96\sqrt{\frac{0.6(0.4)}{100}} = 0.5$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6 + 1.96\sqrt{\frac{0.6(0.4)}{100}} = 0.7$

The 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).

The lower limit is 0.5.
The upper limit is 0.7.

A similar problem is given at brainly.com/question/16807970

5 0

3 years ago