Tim wants to collect 400 stickers. If he makes pages of ten how will he know when he has 400?

Karen and Sara invent a game. They roll a number cube twice and read the two digits shown as a two-digit number. So, if Karen ge

Daniel [21]

1) 11 12 13 14 15 16 21 22 23 24 25 26 31 32 33 34 35 36 41 42 43 44 45 46 51 52 53 54 55 56 61 62 63 64 65 66
2) least number is 11
3) greatest number is 66
4) Yes they are equally likely because it’s a dice so it a 16.6 chance of u rolling a number

6 0

2 years ago

Does anyone know this answer

aniked [119]

Answer:

C = -30

Step-by-step explanation:

C = 5/9 (F - 32)

C = 5/9 ( - 22 - 32)

C = 5/9 ( -54)

C = -30

3 0

2 years ago

Anyone listens to Texako, Summrs, Ect ?

LenKa [72]

Answer:

No, but i would like to know who they are

Step-by-step explanation:

4 0

1 year ago

Nora and Eli are making homemade spring rolls for a party. Nora can make 8 spring rolls in 10 minutes. Eli can make 10 spring ro

NemiM [27]

Answer:

Could you possible elaborate on the question more?

6 0

3 years ago

Suppose that in a random sample of 300 employed Americans, there are 57 individuals who say that they would fire their boss if t

otez555 [7]

Answer:

The 95% confidence interval for the population proportion is (0.1456, 0.2344). This means that we are 95% sure that the true proportion of employed American who say that they would fire their boss if they could is between 0.1456 and 0.2344.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

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

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$n = 300, \pi = \frac{57}{300} = 0.19$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.19 - 1.96\sqrt{\frac{0.19*0.81}{300}} = 0.1456$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.19 + 1.96\sqrt{\frac{0.19*0.81}{300}} = 0.2344$

The 95% confidence interval for the population proportion is (0.1456, 0.2344). This means that we are 95% sure that the true proportion of employed American who say that they would fire their boss if they could is between 0.1456 and 0.2344.

4 0

3 years ago