Answer:
Step-by-step explanation:
Answer:
The confidence interval for 90% confidence would be narrower than the 95% confidence
Step-by-step explanation:
From the question we are told that
The sample size is n = 41
For a 95% confidence the level of significance is
and
the critical value of
is 
For a 90% confidence the level of significance is
and
the critical value of
is 
So we see with decreasing confidence level the critical value decrease
Now the margin of error is mathematically represented as
given that other values are constant and only
is varying we have that

Hence for reducing confidence level the margin of error will be reducing
The confidence interval is mathematically represented as

Now looking at the above formula and information that we have deduced so far we can infer that as the confidence level reduces , the critical value reduces, the margin of error reduces and the confidence interval becomes narrower
Let's represent the two numbers by x and y. Then xy=60. The smaller number here is x=y-7.
Then (y-7)y=60, or y^2 - 7y - 60 = 0. Use the quadratic formula to (1) determine whether y has real values and (2) to determine those values if they are real:
discriminant = b^2 - 4ac; here the discriminant is (-7)^2 - 4(1)(-60) = 191. Because the discriminant is positive, this equation has two real, unequal roots, which are
-(-7) + sqrt(191)
y = -------------------------
-2(1)
and
-(-7) - sqrt(191)
y = ------------------------- = 3.41 (approximately)
-2(1)
Unfortunately, this doesn't make sense, since the LCM of two numbers is generally an integer.
Try thinking this way: If the LCM is 60, then xy = 60. What would happen if x=5 and y=12? Is xy = 60? Yes. Is 5 seven less than 12? Yes.
2 pounds ---- 100%
x pounds -----18%
x=(2*18)/100=36/100=0.36 pounds