How do you do this? Please help!

Simplify: 9(a + b) + 4(3a + 2b)<br> A) 13a 13b <br> B) 21a + 17b <br> C) 21a + 2b <br> D) 38ab

Aneli [31]

The answer is
B) 21a + 17b

8 0

4 years ago

enrollment in the ski/snowboard club increased by 30% this year. there are now 182 students in the club. how many students were

eduard

Enrolement increased by 30% that means last year=100%
incease of 30%=30+100=130% now

now=182

percent means parts out of 100 so
130%=130/100=13/10

182=13/10 of x (x represents the number last year)
of means multipy so
182=13/10 times x
multiply both sides by 10/13 to clear fraction
182 times 10/13=x=140

140 students last year

8 0

3 years ago

3

erastovalidia [21]

ummm i think is bussin dems toes, but im not sure

7 0

3 years ago

While researching the cost of school lunches per week across the state, you use a sample size of 45 weekly lunch prices. The sta

Drupady [299]

We assume the lunch prices we observe are drawn from a normal distribution with true mean $\mu$ and standard deviation 0.68 in dollars.

We average $n=45$ samples to get $\bar{x}$ .

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

$\sigma = 0.68 / \sqrt{45} = 0.101$

Our goal is to come up with a confidence interval (a,b) that we can be 90% sure contains $\mu$ .

Our interval takes the form of $( \bar{x} - z \sigma, \bar{x} + z \sigma )$ as $\bar{x}$ 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

$( \bar{x} - 1.65 (.101), \bar{x} + 1.65 (.101) )$