Can you help me with this , I am having a very hard time ?

Mathematics

1 answer:

nlexa [21]3 years ago

3 0

QS is 126 (63 * 2), since T is the midpoint of QS, and therefore QT and TS are of equal lengths.

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The student body President of a high school claims to know the names of at least 1000 of the 1800 students at the school. To tes

hoa [83]

Answer:

The 99% confidence interval for p in this case is (0.3317, 0.5883).

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 $1 - \frac{\alpha}{2}$ .

Randomly selects 100 students from the school and asks the President to name each one. The President is able to correctly name 46 of the students.

This means that:

$n = 100, \pi = \frac{46}{100} = 0.46$

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 limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.46 - 2.575\sqrt{\frac{0.46*0.54}{100}} = 0.3317$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.46 + 2.575\sqrt{\frac{0.46*0.54}{100}} = 0.5883$

The 99% confidence interval for p in this case is (0.3317, 0.5883).

4 0

3 years ago

A 2 liter bottle of coke for $1.39 or a 12 pack of 12 ounce cams of pepsi for $3.49? which is the best deal

pav-90 [236]

A 2-liter bottle of Coke for $1.39 is the best deal. No need to divide.

5 0

3 years ago

Read 2 more answers

Please help me with this #5

shepuryov [24]

$\text{Given that,}\\\\a_1 = 10~ \text{ and}~ a_n = a_{n-1} -4~\text{for}~ n\geq 2\\\\a_2 = a_{2-1} -4 = a_1 -4 = 10 -4 = 6\\\\a_3 = a_{3-1} - 4=a_2 -4 = 6 -4 =2\\\\a_4 = a_{4-1} -4 = a_3 -4 = 2 -4 = -2\\\\\text{The first four terms are:}~ 10,~ 6,~ 2~, -2$