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Burka [1]
3 years ago
10

A department store is having a “half price” sale. Which is equivalent to “half price”?

Mathematics
2 answers:
Alex73 [517]3 years ago
7 0

Answer:

50% off

Step-by-step explanation:

50 is half of 100

Delvig [45]3 years ago
5 0

Answer:

50% off

Step-by-step explanation:

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(5)(4)(3)(2)(1)= <u>120. </u>your answer will be 120

Step-by-step explanation:

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for a class project, a political science student at a large university wants to estimate the percent of students that are regist
nekit [7.7K]

Answer:

90% confidence interval for the true percent of students that are registered voters is [0.56 , 0.64].

Step-by-step explanation:

We are given that for a class project, a political science student at a large university wants to estimate the percent of students that are registered voters.

He surveys 500 students and finds that 300 are registered voters.

Firstly, the pivotal quantity for 90% confidence interval for the true proportion is given by;

                         P.Q. = \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of students who are registered voters = \frac{300}{500} = 0.60

           n = sample of students = 500

           p =  true percent of students

<em>Here for constructing 90% confidence interval we have used One-sample z proportion test statistics.</em>

So, 90% confidence interval for the true proportion, p is ;

P(-1.6449 < N(0,1) < 1.6449) = 0.90  {As the critical value of z at 5%

                                            level of significance are -1.6449 & 1.6449}  

P(-1.6449 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 1.6449) = 0.90

P( -1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.90

P( \hat p-1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.90

<u>90% confidence interval for p</u> =[\hat p-1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }]

 = [ 0.60-1.6449 \times {\sqrt{\frac{0.60(1-0.60)}{500} } } , 0.60+1.6449 \times {\sqrt{\frac{0.60(1-0.60)}{500} } } ]

 = [0.56 , 0.64]

Therefore, 90% confidence interval for the true percent of students that are registered voters is [0.56 , 0.64].

Interpretation of the above confidence interval is that we are 90% confident that the true percent of students that are registered voters will lie between 0.56 and 0.64.

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3 years ago
Solve using standard algorithm 12.2 * 4.
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Answer and Step-by-step explanation:

12.2

*   4

____

<em><u>48.8</u></em>

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4 times 2 is 8.

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