Plz help taking a quiz???

Need help with question 13B i can not figure it out please help.

Vsevolod [243]

Short Answer 10
The tough part is the number of drinks.
Let the number of them be x
4.5 x + 3 + (10/100) * (4.5x + 3) + (7/100) * (4.5 x + 3) ≤ 60

The tip is not included in the tax, and the tax does not include the tip. You can shorten this up quite a bit.
4.5 x + 3 + (10/100) * (4.5x + 3) + (7/100) * (4.5 x + 3) ≤ 60
(1 + 10/100 + 7 /100) (4.5x + 3) ≤ 60 Combine the tax, tip, scone and Lattes

(1.17) * (4.5x + 3) ≤ 60 Divide by 1.17
4.5x + 3 ≤ 60 / 1.17
4.5x + 3 ≤ 51.28 Subtract 3
4.5x ≤ 48.28 Divide by 4.5
x ≤ 48.28 / 4.5
x ≤ 10.72

Discussion.
There's a problem. He can't buy 0.72 of a Latte. He can only buy 10 of them. In this case we must round down so what happens if he buys 10 lattes? Does the price the taxes and the tip allow him to get another Latte? The answer should be no because then you would be 11 lattes and the tip and taxes would push you over 60 dollars. But be careful, sometimes things like this can alter you answer..

Answer: 10 <<<<

7 0

3 years ago

A rectangle pyramid is sliced parallel to its base as shown in the figure. What is the shape of the resulting two-dimensional cr

Aleks [24]

C hope it helps sorry if I'm wrong

8 0

3 years ago

Read 2 more answers

A manufacturer reported a sample mean = 22.0 g and a sample standard deviation = 2.5 g based on a sample of 20 of their products

tester [92]

Answer:

$n=(\frac{1.640(2.5)}{2})^2 =4.2 \approx 5$

So the answer for this case would be n=5 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=22$ represent the sample mean

$\mu$ population mean (variable of interest)

s=2.5 represent the sample standard deviation

n represent the sample size

ME = 2 represent the margin of error accepted

Solution to the problem

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

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (2)

And on this case we have that ME =2 and we are interested in order to find the value of n, if we solve n from equation (2) we got:

$n=(\frac{z_{\alpha/2} s}{ME})^2$ (3)

We can use as estimator for the population deviation the sample deviation $\hat \sigma = s$

The critical value for 90% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.05;0;1)", and we got $z_{\alpha/2}=1.640$ , replacing into formula (3) we got: