For this case we have the following statement:
<em>"You have 2 meters of fabric and you cut 85% to make curtains, 50% of the rest to make the strips that hold it. How many centimeters of fabric were left?"</em>
So, we have:
2 meters ---------> 100%
x ----------------------> 85%
Where "x" represents the amount of fabric meters equivalent to 85%.
![x = \frac {85 * 2} {100}\\x = 1.7](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%20%7B85%20%2A%202%7D%20%7B100%7D%5C%5Cx%20%3D%201.7)
Thus, 1.7 meters of fabric were used to make curtains.
There are 0.3 meters of fabric left, of which half (50%) is used to make the strips that hold the curtain. That is to say:
![\frac {0.3} {2} = 0.15m](https://tex.z-dn.net/?f=%5Cfrac%20%7B0.3%7D%20%7B2%7D%20%3D%200.15m)
Thus, 0.15 meters of fabric remain.
![0.15 m * \frac {100 cm} {1 m} = 15\ centimeters.](https://tex.z-dn.net/?f=0.15%20m%20%2A%20%5Cfrac%20%7B100%20cm%7D%20%7B1%20m%7D%20%3D%2015%5C%20centimeters.)
Thus, 15 cm of fabric remain
Answer;
15 cm
Answer:
4 bags of takis
Step-by-step explanation:
take the $22 and subtract the $3.50 for the oreos, this leaves you with $18.50.
now, take the $18.50 and divide that by $4 (the amount takis cost) and you will find out how many bas he will he able to buy.
18.50/4 = 4.6 he will be able to buy only 4 bags and will have $2.50 left over
Answer:
$ 7.50
Step-by-step explanation:
The cost of the picnic cooler would be 100%.
With the 6% sales tax, Anton would pay 106% of the cost if the picnic cooler.
(since 100% +6%= 106%)
106%(picnic cooler)= $7.95
Cost of picnic cooler= 100%(picnic cooler).
Cost of picnic cooler
![= \frac{7.95}{106} \times 100 \\ = 7.50](https://tex.z-dn.net/?f=%20%3D%20%20%5Cfrac%7B7.95%7D%7B106%7D%20%20%5Ctimes%20100%20%5C%5C%20%20%3D%207.50)
Thus, he paid $7.50 for the cooler alone.
Answer:
27?
Step-by-step explanation:
45-18
= 27
Is it the correct answer? Or I understand your question in wrong way?
Answer:
Step-by-step explanation:
Given that:
A simple random sample n = 28
sample standard deviation S = 12.65
standard deviation
= 11.53
Level of significance ∝ = 0.05
The objective is to test the claim that the number of pieces in a set has a standard deviation different from 11.53.
The null hypothesis and the alternative hypothesis can be computed as follows:
Null hypothesis:
![H_0: \sigma^2 = \sigma_0^2](https://tex.z-dn.net/?f=H_0%3A%20%5Csigma%5E2%20%3D%20%5Csigma_0%5E2)
Alternative hypothesis:
![H_1: \sigma^2 \neq \sigma_0^2](https://tex.z-dn.net/?f=H_1%3A%20%5Csigma%5E2%20%5Cneq%20%5Csigma_0%5E2)
The test statistics can be determined by using the following formula in order to test if the claim is statistically significant or not.
![X_0^2 = \dfrac{(n-1)S^2}{\sigma_0^2}](https://tex.z-dn.net/?f=X_0%5E2%20%3D%20%5Cdfrac%7B%28n-1%29S%5E2%7D%7B%5Csigma_0%5E2%7D)
![X_0^2 = \dfrac{(28-1)(12.65)^2}{(11.53)^2}](https://tex.z-dn.net/?f=X_0%5E2%20%3D%20%5Cdfrac%7B%2828-1%29%2812.65%29%5E2%7D%7B%2811.53%29%5E2%7D)
![X_0^2 = \dfrac{(27)(160.0225)}{132.9409}](https://tex.z-dn.net/?f=X_0%5E2%20%3D%20%5Cdfrac%7B%2827%29%28160.0225%29%7D%7B132.9409%7D)
![X_0^2 = \dfrac{4320.6075}{132.9409}](https://tex.z-dn.net/?f=X_0%5E2%20%3D%20%5Cdfrac%7B4320.6075%7D%7B132.9409%7D)
![X_0^2 = 32.5002125](https://tex.z-dn.net/?f=X_0%5E2%20%3D%2032.5002125)
![X^2_{1- \alpha/2 , df} = X^2_{1- 0.05/2 , n-1}](https://tex.z-dn.net/?f=X%5E2_%7B1-%20%5Calpha%2F2%20%2C%20df%7D%20%3D%20X%5E2_%7B1-%200.05%2F2%20%2C%20n-1%7D)
![X^2_{1- \alpha/2 , df} = X^2_{1- 0.025 , 28-1}](https://tex.z-dn.net/?f=X%5E2_%7B1-%20%5Calpha%2F2%20%2C%20df%7D%20%3D%20X%5E2_%7B1-%200.025%20%2C%2028-1%7D)
From the chi-square probabilities table at 0.975 and degree of freedom 27;
= 14.573
![X^2_{\alpha/2 , df} = X^2_{ 0.05/2 , n-1}](https://tex.z-dn.net/?f=X%5E2_%7B%5Calpha%2F2%20%2C%20df%7D%20%3D%20X%5E2_%7B%200.05%2F2%20%2C%20n-1%7D)
![X^2_{\alpha/2 , df} = X^2_{0.025 , 28-1}](https://tex.z-dn.net/?f=X%5E2_%7B%5Calpha%2F2%20%2C%20df%7D%20%3D%20X%5E2_%7B0.025%20%2C%2028-1%7D)
From the chi-square probabilities table at 0.975 and degree of freedom 27;
43.195
Decision Rule: To reject the null hypothesis if
; otherwise , do not reject the null hypothesis:
The rejection region is ![X^2_0 \ > 43.195 \ \ \ or \ \ \ X^2_0 \ < \ 14.573](https://tex.z-dn.net/?f=X%5E2_0%20%20%5C%20%3E%20%2043.195%20%5C%20%5C%20%20%5C%20or%20%5C%20%5C%20%5C%20%20%20X%5E2_0%20%5C%20%20%3C%20%5C%20%2014.573)
Conclusion:
We fail to reject the null hypothesis since test statistic value 32.5002125 lies between 14.573 and 43.195.