Answer:
2x + 4y = 32
4x + 3y = 44
Step-by-step explanation:
From the information supplied in the question, we can see that 2 adult tickets and 4 child tickets cost $32. This means we multiply the cost of an adult ticket by 2 and add it to the product of 4 child tickets and its price I.e y.
We can also see that to get a total cost of $44, 4 adult tickets and 3 child tickets were bought. Hence, we simply multiply the cost of an adult ticket by 4 and add it to the product of 3 child ticket and its price
My best guess is a line graph because it shows the consistent numbers and then you can lay them all out after you have completed the graph
<span>Part A
</span>f(x)=30∗0.2xand<span>f(x)=45+3x</span><span>
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Part B</span>
f(x)=30+0.2(5)f(x)=45+3(5)<span>
Neighborhood A and neighborhood B both have 60 houses after 5 years
</span>
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Part C
1 year
A=36
B=48
2 years
A=42
B=51
3 years
A=48
B=54
4 years
A = 54
B = 57
5 years
A = 60
B= 60
2 gallons is equal to 256 oz and a cup is equal to 8 oz so your total is 270 fl oz.
Answer:
And on this case if we see the significance level given 
we see that 
so we fail to reject the null hypothesis that the observed outcomes agree with the expected frequencies at 10% of significance.
Step-by-step explanation:
A chi-square goodness of fit test determines if a sample data obtained fit to a specified population.
represent the p value for the test
O= obserbed values
E= expected values
The system of hypothesis for this case are:
Null hypothesis: ![O_i = E_i[/tex[Alternative hypothesis: [tex]O_i \neq E_i](https://tex.z-dn.net/?f=O_i%20%3D%20E_i%5B%2Ftex%5B%3C%2Fp%3E%3Cp%3EAlternative%20hypothesis%3A%20%5Btex%5DO_i%20%5Cneq%20E_i%20)
The statistic to check the hypothesis is given by:
On this case after calculate the statistic they got: 
And in order to calculate the p value we need to find first the degrees of freedom given by:
, where k represent the number of levels (on this cas we have 10 categories)
And in order to calculate the p value we need to calculate the following probability:
And on this case if we see the significance level given we see that so we fail to reject the null hypothesis that the observed outcomes agree with the expected frequencies at 10% of significance.
