Construct a 90% confidence interval of the population proportion using the given information.

X= 175

n= 250

Tbe confidence interval marked the range of values for which the true population mean is estimated to be given a certain level of confidence. Hence, the confidence interval is (0.6523, 0.7477)

Since the sample size is large enough, we use the the Z distributuon :

Confidence interval is defined thus :

Margin of Error :

Hence,

Lower confidence boundary = 0.7 - 0.0290 = 0.6523

Upper confidence boundary = 0.7 + 0.0290 = 0.7477

Therefore, the confidence interval is (0.6523, 0.7477)

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The 90% confidence interval of the population proportion is (0.6523, 0.7477).

Step-by-step explanation:

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

In which

z is the zscore that has a pvalue of .

For this problem, we have that:

90% confidence level

So , z is the value of Z that has a pvalue of , so .

The lower limit of this interval is:

The upper limit of this interval is:

The 90% confidence interval of the population proportion is (0.6523, 0.7477).

answer: 2 and 1

step-by-step explanation: got this question

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

step-by-step explanation:

given   that triangle lno is congruent to triangle lnm.

we have to fill up the blanks in the given question.

according to above order of congruency   kept in mind we can say that

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

ln ≅ ln due to common in both.

hence, by sas congruency,

the side angle side rule   states that if two sides and the included angle of one triangle are congruent to another triangle, then these two triangles are congruent.