Please a,b,c ko answer plz plz plz

%error=100(estimated-actual)/actual

%error=100(96-(96-16))/(96-16)

%error=100(16/80)

%error=1600/80

%error=20%

She overestimated how many bracelets she would sell by 20%

8 0

4 years ago

Read 2 more answers

A chess player won 35 of the games he played. If his ratio of wins to loses was 5:6, how many games did he play total?

kobusy [5.1K]

42 because he won 35 so 5 times 7 equals 35 so you would multiply the other side by seven and 6 times 7 is 42

4 0

3 years ago

1. A poll conducted in 2018 found that 54% of US adult Twitter users get at least some news on Twitter. The standard error for t

Harrizon [31]

Answer:

The 99% confidence interval for the fraction of US adult Twitter users who get some news on Twitter is (0.4872, 0.6018). It means that we are 99% sure that the true proportion of US adult Twitter users who get some news on Twitter is between 0.4872 and 0.6018.

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.

$\pi \pm zs$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ , and s is the standard error.

54% of US adult Twitter users get at least some news on Twitter.

This means that $\pi = 0.54$

The standard error for this estimate was 2.4%

This means that $s = 0.024$

99% confidence level

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$

The lower bound is:

$\pi - zs = 0.54 - 2.575*0.024 = 0.4782$

The upper bound is:

$\pi + zs = 0.54 + 2.575*0.024 = 0.6018$

The 99% confidence interval for the fraction of US adult Twitter users who get some news on Twitter is (0.4872, 0.6018). It means that we are 99% sure that the true proportion of US adult Twitter users who get some news on Twitter is between 0.4872 and 0.6018.

7 0

3 years ago

Please a,b,c ko answer plz plz plz ​

Please a,b,c ko answer plz plz plz