A theater group made appearances in two cities. The hotel charge before tax in the second city was $ 1500 higher than in the first. The tax in the first city was
3%, and the tax in the second city was 9.5%. The total hotel tax paid for the two cities was $580. How much was the hotel charge in each city before tax?
First city:
si
Х
5
?
Second city: 0
I’m
Answers
9514 1404 393
Answer:
first city: $3500second city: $5000Step-by-step explanation:
Let x and y represent the hotel charges in the two cities. The relations we are given are ...
x - y = -1500 . . . . . . second city hotel charge was 1500 more
0.03x +0.095y = 580 . . . . . total tax paid was 580
Using the first equation, we can write an expression for x:
x = y -1500
Substituting that into the second equation gives ...
0.03(y -1500) +0.095y = 580
0.125y = 625 . . . . . . . add 45 to both sides and simplify
y = 5000 . . . . . . . . . . multiply by 8
x = 5000 -1500 = 3500
The hotel charges were ...
First city: $3500
Second city: $5000
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Answer: Answer below.
Step-by-step explanation:
Because one child was not hungry and the cake was cut into 6 equal pieces, this mean one person got 2/6 or 1/3 of the cake and the rest got 1/6 other than the child who was not hungry. I can't draw this but you want to draw a child with 2 equal slices of the cake, one with no slices, and 4 more with 1/6 slices of the cake. Sorry I couldn't draw it, but hope this helps!
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커
O isosceles
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acute
obtuse
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Answer:
obtuse isosceles
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Step-by-step explanation:
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Step-by-step explanation:
According to the February 2008 Federal Trade Commission report on consumer fraud and identity theft, 23% of all complaints in 2007 were for identity theft. In that year, Alaska had 321 complaints of identity theft out of 1,432 consumer complaints ("Consumer fraud and," 2008). Does this data provide enough evidence to show that Alaska had a lower proportion of identity theft than 23%? State the random variable, population parameter, and hypotheses.
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Testing the hypothesis, it is found that since the p-value of the test is of 0.3015, which is greater than the significance level of 0.05, the data does not provide enough evidence to show that Alaska had a lower proportion of identity theft than 23%.
At the null hypothesis, we test if the proportion is of 23%, that is:
[tex]H_0: p = 0.23[/tex]
At the alternative hypothesis, we test if this proportion is lower than 23%, that is:
[tex]H_1: p < 0.23[/tex]
The test statistic is given by:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
In which:
[tex]\overline{p}[/tex] is the sample proportion. p is the proportion tested at the null hypothesis. n is the sample size.In this problem, the parameters are: [tex]p = 0.23, n = 1432, \overline{p} = \frac{321}{1432} = 0.2242[/tex]
Then, the value of the test statistic is:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
[tex]z = \frac{0.2242 - 0.23}{\sqrt{\frac{0.23(0.77)}{1432}}}[/tex]
[tex]z = -0.52[/tex]
The p-value of the test is the probability of finding a sample proportion of 0.2242 or below, which is the p-value of z = -0.52.
Looking at the z-table, z = -0.52 has a p-value of 0.3015.Since the p-value of the test is of 0.3015, which is greater than the significance level of 0.05, the data does not provide enough evidence to show that Alaska had a lower proportion of identity theft than 23%.
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Gomenosai
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D
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(-1,1)
(0,-1)
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A.
{-1, 1}
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{-1, 0, 5, 10}
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