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ira [324]
3 years ago
12

Math problem. The price of a community pool membership has a one time sign up fee and monthly fee. The price can be modeled by t

he function y=20x+50, where x is the numer of months. Question: what is the slope, and what does it represent.
Mathematics
2 answers:
Ugo [173]3 years ago
8 0
The Slope is 20, it represents that the pool membership will cost $20 every month
inessss [21]3 years ago
6 0

Answer: In the given function the slope (m)=20

It represents the change in price of a community pool membership with respect to the change in number of months.


Step-by-step explanation:

Given: The price of a community pool membership has a one time sign up fee and monthly fee.

The price can be modeled by the functiony=20x+50 , where x is the number of months.

We know that the equation of line is  y=mx+c , where m is the slope of the line which shows the rate of change of y with respect to the change of x.

In the given function the slope (m)=20

It represents the change in price of a community pool membership with respect to the change in number of months.


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A triangle has angles that measure 46 and 84<br><br> what is the third angle?
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The degenerative disease osteoarthritis most frequently affects weight-bearing joints such as the knee. Two random samples from
Vinvika [58]

Answer:

t=\frac{(810-770)-0}{\sqrt{\frac{67^2}{13}+\frac{56^2}{19}}}}=1.771  

df=n_1 +n_2 -2=13+19-2=30  

Since is a right tailed test the p value would be:  

p_v =P(t_{30}>1.771)=0.0434  

Comparing the p value with the significance level \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for group 1 is significantly higher than the mean for the group 2

Step-by-step explanation:

Data given

\bar X_{1}=810 represent the mean for sample 1  

\bar X_{2}=770 represent the mean for sample 2  

s_{1}=67 represent the sample standard deviation for 1  

s_{2}=56 represent the sample standard deviation for 2  

n_{1}=13 sample size for the group 2  

n_{2}=19 sample size for the group 2  

\alpha=0.05 Significance level provided

t would represent the statistic (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the mean for category 1 is higher than the mean for category 2, the system of hypothesis would be:  

Null hypothesis:\mu_{1}-\mu_{2}\leq 0  

Alternative hypothesis:\mu_{1} - \mu_{2}> 0  

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:  

t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}} (1)  

And the degrees of freedom are given by df=n_1 +n_2 -2=13+19-2=30  

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

With the info given we can replace in formula (1) like this:  

t=\frac{(810-770)-0}{\sqrt{\frac{67^2}{13}+\frac{56^2}{19}}}}=1.771  

P value  

Since is a right tailed test the p value would be:  

p_v =P(t_{30}>1.771)=0.0434  

Comparing the p value with the significance level \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for group 1 is significantly higher than the mean for the group 2

4 0
3 years ago
a. 0.98 – 0.053 b. 0.67 – 0.4 c. 0.3 – 0.002 d. 3.2 – .789 e. 6.53 – 4.298 f. 6 – 4.32 g. 7 – 3.574 h. 4.83 – 1.8 i. 3.7 – 1.8 j
Pavlova-9 [17]

a. 0.927

We have:

0.98 – 0.053

We can re-write it as:

0. 9 8 0 -

0. 0 5 3

Moving digits to the right:

0. 9 7 10 -

0. 0 5 3

Digit-per-digit subtraction:

0. 9 2 7


b. 0.27

We have:

0.67 – 0.4

We can re-write it as:

0. 6 7 -

0. 4 0

Digit-per-digit subtraction:

0. 2 7


c. 0.298

We have:

0.3 – 0.002

We can re-write it as:

0. 3 0 0 -

0. 0 0 2

Moving digits to the right:

0. 2 10 0 -

0. 0 0 2

Again:

0. 2 9 10 -

0. 0 0 2

Digit-per-digit subtraction:

0. 2 9 8


d. 2.411

We have:

3.2 – .789

We can re-write it as:

3. 2 0 0 -

0. 7 8 9

We need to rewrite the first term by moving digits to the right several times:

3. 2 0 0 = 2. 12 0 0 = 2. 11 10 0 = 2. 11 9 10

So now we have:

2. 11 9 10 -

0. 7 8 9

Digit-per-digit subtraction:

2. 4 1 1


e. 2.232

We have:

6.53 – 4.298

We can re-write it as:

6. 5 3 0 -

4. 2 9 8

We need to rewrite the first term by moving digits to the right several times:

6. 5 3 0 = 6. 5 2 10 = 6. 4 12 10

So now we have:

6. 4 12 10 -

4. 2 9 8

Digit-per-digit subtraction:

2. 2 3 2


f. 1.68

We have:

6 – 4.32

We can re-write it as:

6. 0 0 -

4. 3 2

We need to rewrite the first term by moving digits to the right several times:

6. 0 0 = 5. 10 0 = 5. 9 10

So now we have:

5. 9 10 -

4. 3 2

Digit-per-digit subtraction:

1. 6 8


g. 4.426

We have:

7 – 3.574

We can re-write it as:

7. 0 0 0 -

3. 5 7 4

We need to rewrite the first term by moving digits to the right several times:

7. 0 0 0 = 6. 10 0 0 = 6. 9 10 0 = 6. 9 9 10

So now we have:

6. 9 9 10 -

3. 5 7 4 =

Digit-per-digit subtraction:

3. 4 2 6


h. 3.03

We have:

4.83 – 1.8

We can re-write it as:

4. 8 3 -

1. 8 0

We can immediately do the digit-per-digit subtraction:

3. 0 3


i. 2.9

We have:

3.7 – 1.8

We can re-write it as:

3. 7 -

1. 8

We need to rewrite the first term by moving digits to the right:

3. 7 = 2. 17

So now we have:

2. 17 -

1. 8 =

Digit-per-digit subtraction:

2. 9


j. 4.538

We have:

16.17 – 11.632

We can re-write it as:

1 6 . 1 7 0 -

1 1 . 6 3 2

We need to rewrite the first term by moving digits to the right:

1 6. 1 7 0 = 1 6. 1 6 10 = 1 5. 11 6 10  

So now we have:

1 5. 11 6 10 -

1 1. 6 3 2 =

Digit-per-digit subtraction:

0 4. 5 3 8

3 0
3 years ago
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