Suppose that the amount spent per person per month on health insurance in the u. s. is normally distributed with an unknown mean and standard deviation. the monthly health insurance costs of 17 randomly sampled individuals in the u. s. are used to estimate the mean of the population. what t-score should be used to find the 90% confidence interval for the population mean?

1.745884

Step-by-step explanation:

Given : Level of significance :

Then , significance level :

Sample size :

Then , the degree of freedom for t-distribution:

The t-score for a level of confidence is given by the formula below :-

, where df is the degree of freedom and is the significance level.

With the help of the normal t-distribution table, we have

Thus, the t-score should be used to find the 90% confidence interval for the population mean =1.745884

given that triangle abc is transformed two times to get triangle

a"b"c"

first abc is transformed in to a'b'c' as follows:

the coordinates of a are (1,-1) and transformed into a'(-1,1)

similarly b (4,-2) became b'(-4,2) and

c(7,-2) became c'(-7,2)

i.e. (x,y) becomes (-x,-y)

this is nothing but reflection about a point here origin.

thus first transformation is reflection on the origin.

next is exactly shifting 3 units down

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