Answer attached in photo
Hope this helped :)
Hope this helped :)
Groceries cost $75. The tax is 5%. Will you have enough money to pay for the groceries if you have $80. Please explain.
1 answer:
Answer:
Yes, we have enough money to pay for the groceries.
Step-by-step explanation:
We are given that Groceries cost $75. The tax rate is 5%.
We have to check that will the amount of $80 enough to pay for the groceries or not.
<em>For this we will firstly calculate the total cost of groceries after applying tax rate of 5% on groceries. </em>
Cost of groceries after applying tax rate is given by;
Since we know that by applying a tax on any item it's cost gets increased.
So, <u>Cost of groceries after tax</u> = Cost of groceries before tax + Amount of tax
Amount of tax = Tax rate Cost of groceries
= 5% $75
= = $3.75
So, Cost of groceries after tax = $75 + $3.75 = $78.75
Since we have an amount of $80 with us which means that we are able to buy the groceries after giving tax on it as $78.75 < $80.
Therefore, yes we have enough money to pay for the groceries.
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Answer:
The answer is 0.31
Step-by-step explanation:
<u>Old Method New Method .</u>
Mean 73.5625 Mean 75.70588
Standard Error 3.143736 Standard Error 2.923994
Median 72 Median 75
Mode 90 Mode 64
Standard deviation 12.57494 Standard deviation 12.05594
Sample Variance 158.1292 Sample Variance 145.3456
Kurtosis -1.14544 Kurtosis -0.76646
Skewness 0.171025 Skewness 0.091008
Range 39 Range 41
Minimum 55 Minimum 56
Maximum 94 Maximum 97
Sum 1177 Sum 1287
Count 16 Count 17
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: μNew< μOld
Alternative hypothesis: μNew > μOld
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the mean difference between sample means is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
DF = 31
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.
The observed difference in sample means produced a t statistic of - 0.499. We use the t Distribution Calculator to find P(t < - 0.499) = 0.311
Therefore, the P-value in this analysis is 0.311.
Interpret results. Since the P-value (0.311) is greater than the significance level (0.05), we cannot reject the null hypothesis.
From the above test we do have sufficient evidence in the favor of the claim that new method is efficient than the old method.
