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babymother [125]
3 years ago
10

Alley bought 1 pound of potatoes costing $4 from a local grocery store last week. Her final bill included an additional sales ta

x 8%. The next week alley found that the price for a pound of potatoes had gone up to $6. What is the percentage increase in the cost of a pound of potatoes, including the sales tax?
Mathematics
2 answers:
Sergeeva-Olga [200]3 years ago
7 0

Answer: 54%

Step-by-step explanation:

Given: The cost of 1 pound of  potato in  a local grocery store last week= $4

The cost of 1 pound of  potato in  a local grocery store last week after 8 % tax= $4+8% of $4

=4+0.08\times4=4+0.32=\$4.32

The cost of 1 pound of  potato in  a local grocery store next week= $6

The cost of 1 pound of  potato in  a local grocery store next week after 8 % tax= $6+8% of $6

=6+0.08\times6=6+0.48=\$6.48

The percentage increase in the cost of a pound of potatoes, including the sales tax will be given by :-

P=\frac{\text{ increase in cost}}{\text{cost last week}}\times100\\\\\Rioghtarrow\ P=\frac{6.48-4.32}{4}\times100\\\\\Rightarrow\ P=\frac{2.16}{4}\times100\\\\\Rightarrow\ P=54\%

Paul [167]3 years ago
5 0
$6.00 X 8%  $6.48

this is the answer. hope this helps
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A fish swims over the waterfall. The distance y (in feet) that the fish falls is given by the function y=16t2 , where t is the t
inessss [21]

Answer:

See Explanation

Step-by-step explanation:

Given

y = 16t^2

The equation is incomplete. However, a possible question could be to calculate the distance of the fish at a particular time

e.g. Time, t = 4,

The distance is:

y = 16t^2

y = 16 * 4^2

y = 16 * 16

y = 256ft

Another possible question is to determine the time taken to cover a certain distance

e.g.

Distance, y = 400

This gives:

y = 16t^2

400 = 16t^2

Divide by 16

25 = t^2

Take square roots

5 = t

t = 5\ seconds

8 0
3 years ago
Please help and show all work so I can fully understand it, thanks.
seraphim [82]

Answer:

\frac{1}{m-4}

Step-by-step explanation:

\frac{\frac{4m-5}{m^4 -7m^3 +12m^2}}{\frac{4m-5}{m^3 -3m^2}}

Factor the equation:

\frac{\frac{4m-5}{m^2(m^2 -7m+12)}}{\frac{4m-5}{m^2(m-3)}}

\frac{\frac{4m-5}{m^2(m-3)(m-4)}}{\frac{4m-5}{m^2(m-3)}}

Rewrite to suit the format of multiplying two fractions. Remember, dividing two fractions is the same as multiplying the first fraction by the reciprocal of the second. A reciprocal of a fraction is when one switches the place of the numerator and the denominator, that is, the value on top (numerator), and the value on the bottom (denominator).

\frac{4m-5}{m^2(m-3)(m-4)}*\frac{m^2(m-3)}{4m-5}

Simplify, take out common terms that are found on both the numerator and denominator

\frac{4m-5}{m^2(m-3)(m-4)}*\frac{m^2(m-3)}{4m-5}

\frac{1}{m-4}

4 0
3 years ago
Is education related to programming preference when watching TV? From a poll of 80 television viewers, the following data have b
Luda [366]

Answer:

a) H0:  There is no association between level of education and TV station preference (Independence)

H1: There is association between level of education and TV station preference (No independence)

b) \chi^2 = \frac{(15-10)^2}{10}+\frac{(15-20)^2}{20}+\frac{(10-10)^2}{10}+\frac{(5-10)^2}{10}+\frac{(25-10)^2}{10}+\frac{(10-20)^2}{20} =33.75

c) \chi^2_{crit}=5.991

d) Since the p value is lower than the significance level we enough evidence to reject the null hypothesis at 5% of significance, and we can conclude that we have dependence between the two variables analyzed.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

                                  High school   Some College   Bachelor or higher  Total

Public Broadcasting       15                       15                          10                     40

Commercial stations      5                         25                         10                     40  

Total                                20                      40                          20                    80

We need to conduct a chi square test in order to check the following hypothesis:

Part a

H0:  There is no association between level of education and TV station preference (Independence)

H1: There is association between level of education and TV station preference (No independence)

The level os significance assumed for this case is \alpha=0.05

The statistic to check the hypothesis is given by:

\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}

Part b

The table given represent the observed values, we just need to calculate the expected values with the following formula E_i = \frac{total col * total row}{grand total}

And the calculations are given by:

E_{1} =\frac{20*40}{80}=10

E_{2} =\frac{40*40}{80}=20

E_{3} =\frac{20*40}{80}=10

E_{4} =\frac{20*40}{80}=10

E_{5} =\frac{40*40}{80}=20

E_{6} =\frac{20*40}{80}=10

And the expected values are given by:

                                  High school   Some College   Bachelor or higher  Total

Public Broadcasting       10                       20                         10                     40

Commercial stations      10                        10                         20                     40  

Total                                20                      30                          30                    80

Part b

And now we can calculate the statistic:

\chi^2 = \frac{(15-10)^2}{10}+\frac{(15-20)^2}{20}+\frac{(10-10)^2}{10}+\frac{(5-10)^2}{10}+\frac{(25-10)^2}{10}+\frac{(10-20)^2}{20} =33.75

Now we can calculate the degrees of freedom for the statistic given by:

df=(rows-1)(cols-1)=(2-1)(3-1)=2

Part c

In order to find the critical value we need to look on the right tail of the chi square distribution with 2 degrees of freedom a value that accumulates 0.05 of the area. And this value is \chi^2_{crit}=5.991

Part d

And we can calculate the p value given by:

p_v = P(\chi^2_{3} >33.75)=2.23x10^{-7}

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(33.75,2,TRUE)"

Since the p value is lower than the significance level we enough evidence to reject the null hypothesis at 5% of significance, and we can conclude that we have dependence between the two variables analyzed.

7 0
3 years ago
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tiny-mole [99]

Answer:

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Step-by-step explanation:

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Answer:

8 square units is the answer

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