Answer:
7.5 cups of raisins, 10 cups of peanuts and 5 cups of chocolate chips.
Step-by-step explanation:
We are given that,
The amount of ingredients needed for the trail mix are,
Raisins = cup
Peanuts = 1 cup
Chocolate chips = cup
Since, Kiran needs to make 10 cups of the trail mix.
The amount of ingredients needed will be,
Raisins = = = 7.5 cups
Peanuts = 1 × 10 = 10 cups
Chocolate chips = = 5 cups
Hence, the ingredients needed are 7.5 cups of raisins, 10 cups of peanuts and 5 cups of chocolate chips.
I found the answer to this by first dividing 34 by 4 to get 8.5. We already know what 4 divided by 4 is. For every one pound of almonds, it costs $8.50. Since we know that we need to find out how much it costs for 5 pounds of almonds, we simply multiply 8.5 by 5 to get 42.5. It costs $42.50 to buy 5 pounds of almonds.
688
x 29
6192
1376 x
19952
So you start by taking 9 from 29 and multiplying it by 688. when you get the answer put it below the line. then multiply the 2 from the 29 and whne you get the answer put a zero on the end and then add the answer you got from 9x688 (6192) and 2x688 (13760) and then thats youre answer. hope it helps!
We assume the lunch prices we observe are drawn from a normal distribution with true mean
and standard deviation 0.68 in dollars.
We average
samples to get
.
The standard deviation of the average (an experiment where we collect 45 samples and average them) is the square root of n times smaller than than the standard deviation of the individual samples. We'll write

Our goal is to come up with a confidence interval (a,b) that we can be 90% sure contains
.
Our interval takes the form of
as
is our best guess at the middle of the interval. We have to find the z that gives us 90% of the area of the bell in the "middle".
Since we're given the standard deviation of the true distribution we don't need a t distribution or anything like that. n=45 is big enough (more than 30 or so) that we can substitute the normal distribution for the t distribution anyway.
Usually the questioner is nice enough to ask for a 95% confidence interval, which by the 68-95-99.7 rule is plus or minus two sigma. Here it's a bit less; we have to look it up.
With the right table or computer we find z that corresponds to a probability p=.90 the integral of the unit normal from -z to z. Unfortunately these tables come in various flavors and we have to convert the probability to suit. Sometimes that's a one sided probability from zero to z. That would be an area aka probability of 0.45 from 0 to z (the "body") or a probability of 0.05 from z to infinity (the "tail"). Often the table is the integral of the bell from -infinity to positive z, so we'd have to find p=0.95 in that table. We know that the answer would be z=2 if our original p had been 95% so we expect a number a bit less than 2, a smaller number of standard deviations to include a bit less of the probability.
We find z=1.65 in the typical table has p=.95 from -infinity to z. So our 90% confidence interval is

in other words a margin of error of
dollars
That's around plus or minus 17 cents.