Data on caloric intake or teens for two types of diet are given below.

Typical Diet Caloric intake Sample mean St. Dev.
No fast food 2310 2295 2280 2340 2235 2265 2315 2291.429 34.8466
food Fast 2579 2160 2165 2580 2558 2591 2614 2518 2583.125 33.0646

Give a 95% confidence interval for the difference of the two means, (Uf - Un), where is the mean calorie intake for teens who typically eat fast food and is that for teens who do not typically eat fast food assuming two normal populations, independent random samples, and equal variances for the two populations

Step-by-step explanation:

For "no fast food,

n1 = 9

Mean = (2310 + 2295 + 2280 + 2340 + 2235 + 2265 + 2315 + 2291.429 + 34.8466)/9

Mean, m1 = 2041

Standard deviation, s1 = √summation(x - u)²/n

summation(x - u)² =

(2310 - 2041)^2 + (2295 - 2041)^2 + (2280 - 2041)^2 + (2340 - 2041)^2 + (2235 - 2041)^2 + (2265 - 2041)^2 + (2315 - 2041)^2 + (2291.429 - 2041)^2 + (34.8466 - 2041)^2

= 4533653.14837256

s = √4533653.14837256/9

s = 709.75

For " fast food",

n2 = 10

Mean = (2579 + 2160 + 2165 + 2580 + 2558 + 2591 + 2614 + 2518 2583.125 + 33.0646)/10

Mean,m2 = 2238

summation(x - u)² =

(2579 - 2238)^2 + (2160 - 2238)^2 + (2165 - 2238)^2 + (2580 - 2238)^2 + (2558 - 2238)^2 + (2591 - 2238)^2 + (2614 - 2238)^2 + (2518 - 2238)^2 + (2583.125 - 2238)^2 + (33.0646 - 2238)^2

= 5672294.38379816

s2 = √5672294.38379816/10

s2 = 753.15

For a confidence interval of 95%, z = 1.96

The formula for confidence interval is

m1 - m2 ± z × √(s1²/n1 + s2²/n2)

= 2041 - 2238 ± 1.96 × √(709.75²/9 + 753.15²/10)

= - 197 ± 1.96 × √(55971.6736 + 56723.4923)

= - 197 ± 1.96 × 335.7

= - 197 ± 657.972

The lower end of the interval is

- 197 - 657.972 = - 854.972

The upper end of the interval is

- 197 + 657.972 = 460.972

a) 0

step-by-step explanation:

i^413 x i^0 = 0 as any number multiplied by a zero is

happy to

pls mark as brainliest.

step-by-step explanation:

points a and b are equidistant from the desired line, so the line you find by using them will go through point r.

using points other than a and b will give the desired line only if they are equidistant from r. if they are randomly located, the perpendicular line you create will not necessarily go through point r, hence will not be a solution to the problem.