A group of 59 randomly selected students have a mean score of 29.5 with a standard

deviation of 5.2 on a placement test. What is the 90% confidence interval for the mean

score, ", of all students taking the test?

And we can conclude that the true mean for the scores are given by this interval (28.37; 30.63) at 90% of confidence

Step-by-step explanation:

Information given

represent the sample mean

population mean  

s= 5.2 represent the sample standard deviation

n=59 represent the sample size  

Confidence interval

The confidence interval for the true mean is given by the following formula:

  (1)

The degrees of freedom are given by:

The Confidence interval is 0.90 or 90%, the significance is and , and the critical value is

Now we have everything in order to replace into formula (1):

And we can conclude that the true mean for the scores are given by this interval (28.37; 30.63) at 90% of confidence

3

step-by-step explanation:

in standard form:

a= dictionaries

b= maps

you have to buy 3 dictionaries, so put a 3 next to the a (dictionaries) so that you can multiply them by each other.

a3+b= 60 (the amount of money you have)

we already know the cost of a=15 so replace a with $15

($15 x 3)+b=60

multiply 15 and 3 to get $45 and replace it with $45

45+b=60

and then replace b with 15 to get the total:

45+15=60

that means you have $15 left after buying 3 dictionaries.

15 divided by the cost of a map ($5) equals 3.

so you can buy 3 maps.

sorry if it was confusing!

it is 10.6 according to my calculations