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aivan3 [116]
3 years ago
10

A full year membership to a gym costs $325 upfront with no monthly charge. A monthly membership costs $100 upfront and $30 per m

onth. For what numbers of months is it less expensive to have a monthly membership?
Mathematics
1 answer:
kykrilka [37]3 years ago
6 0
Its cheaper to pay for a full year. a monthly charge will cost u 100$ upfront and 12 months at 30$ per month is 360 all together coming to 460
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Please help me out!!!!!!!!
qaws [65]

Answer:

x = - 6

Step-by-step explanation:

Given that y varies directly as x then the equation relating them is

y = kx ← k is the constant of variation

To find k use the condition y = - 9 when x = 3, thus

k = \frac{y}{x} = \frac{-9}{3} = - 3

y = - 3x ← equation of variation

When y = 18, then

18 = - 3x ( divide both sides by - 3 )

x = - 6

5 0
3 years ago
ANSWERS
Anna [14]
1) No
2) Yes
3) Yes
4) Yes

if you need me to explain i will.

have a great rest of your day! :)
4 0
3 years ago
Ill give 50 points ik i just asked but i rlly need help
belka [17]

Answer:

2 runners will be needed

Step-by-step explanation:

1/2 = 2/4

2/4 / 1/4 = 2

2 runners, for 1/2 mile

8 0
3 years ago
Read 2 more answers
Which answer best describes the solution to the equation
german
Let's check
3m - 9 = 9+3m
collect like terms
3m -3m = 9 + 9
0m = 18
which is not possible, so it will not have any solutions 
7 0
4 years ago
Read 2 more answers
The brand manager for a brand of toothpaste must plan a campaign designed to increase brand recognition. He wants to first deter
JulsSmile [24]

Answer:

1) n=114

2) n=59

3) On this case no, because if we survey just the adults of the nearest college that would be a convenience sample. And when we use "convenience sample" we have some problems associated to bias. This methodology it's not appropiate in order to have a good estimation of the parameter of interest. It's better use a random, cluster or stratified sampling.

Step-by-step explanation:

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".  

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 80% of confidence, our significance level would be given by \alpha=1-0.80=0.2 and \alpha/2 =0.1. And the critical value would be given by:

z_{\alpha/2}=-1.28, z_{1-\alpha/2}=1.28

Part 1

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

And on this case we have that ME =\pm 0.06 or 6% points, and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

Since we don't have a prior estimate of \het p we can use 0.5 as a good estimate, replacing into equation (b) the values from part a we got:

n=\frac{0.5(1-0.5)}{(\frac{0.06}{1.28})^2}=113.77  

And rounded up we have that n=114

Part 2

On this case we have a prior estimate for the population proportion and is \hat p =0.85 so replacing the values into equation (b) we got:

n=\frac{0.85(1-0.85)}{(\frac{0.06}{1.28})^2}=58.027

And rounded up we have that n=59

Part 3

On this case no, because if we survey just the adults of the nearest college that would be a convenience sample. And when we use "convenience sample" we have some problems associated to bias. This methodology it's not appropiate in order to have a good estimation of the parameter of interest. It's better use a random, cluster or stratified sampling.

5 0
3 years ago
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