Okay, since they choose the same number, but divide by a different number. we know that if you divide by a number, you just times the number upside down (sorry i forgot which word is that) but, so 5/8 devide by 1/16 = 5/8 times 16. and Robyn's is 5/8 times 32. so ofc Robyn will be bigger since he is times it by a bigger number than Susan. ^ ^
-15 would be your answer a negative plus a negative the signs together would turn negative 15 to positive
Answer: the greatest number of trees that ElizaBeth can have in a row is 30
Step-by-step explanation:
given data:
citrus trees = 150
palm trees = 180
As we have to find the greatest number of trees that Elizabeth can have in a row, we will find the GCF or greatest common factor.
150 = 2 x 3 x 5 x 5
180 = 2 x 2 x 3 x 3 x 5
Greatest common factor is = 2 x 3 x 5 = 30
Therefore, the largest number of trees that ElizaBeth can have in a row is 30.
there will be 26 blue marbles
Step-by-step explanation:
41-15=26
Answer: (a) (602.95,705.37)
(b) 33
Step-by-step explanation:
(a) Given : Sample size : ![n=40](https://tex.z-dn.net/?f=n%3D40)
Sample mean : ![\overline{x}=654.16](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%3D654.16)
Standard deviation : ![\sigma= 165.23](https://tex.z-dn.net/?f=%5Csigma%3D%20165.23)
Significance level :![\alpha=1-0.95=0.05](https://tex.z-dn.net/?f=%5Calpha%3D1-0.95%3D0.05)
Critical value : ![z_{\alpha/2}=1.96](https://tex.z-dn.net/?f=z_%7B%5Calpha%2F2%7D%3D1.96)
The confidence interval for population mean is given by :-
![\mu\ \pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=%5Cmu%5C%20%5Cpm%20z_%7B%5Calpha%2F2%7D%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
![=654.16\pm(1.96)\dfrac{165.23}{\sqrt{40}}\\\\\approx654.16\pm51.21\\\\=(654.16-51.21,\ 654.16+51.21)=(602.95,705.37)](https://tex.z-dn.net/?f=%3D654.16%5Cpm%281.96%29%5Cdfrac%7B165.23%7D%7B%5Csqrt%7B40%7D%7D%5C%5C%5C%5C%5Capprox654.16%5Cpm51.21%5C%5C%5C%5C%3D%28654.16-51.21%2C%5C%20654.16%2B51.21%29%3D%28602.95%2C705.37%29)
Hence, the 95% (two-sided) confidence interval for true average
level in the population of all homes from which the sample was selected.
(b) Given : Standard deviation : ![s= 167\text{ ppm}](https://tex.z-dn.net/?f=s%3D%20167%5Ctext%7B%20ppm%7D)
Margin of error : ![E=\pm57\text{ ppm}](https://tex.z-dn.net/?f=E%3D%5Cpm57%5Ctext%7B%20ppm%7D)
Significance level :![\alpha=1-0.95=0.05](https://tex.z-dn.net/?f=%5Calpha%3D1-0.95%3D0.05)
Critical value : ![z_{\alpha/2}=1.96](https://tex.z-dn.net/?f=z_%7B%5Calpha%2F2%7D%3D1.96)
The formula to calculate the sample size is given by :-
![n=(\dfrac{z_{\alpha/2}s}{E})^2\\\\\Rightarrow\ n=(\dfrac{(1.96)(167)}{57})^2=32.9758025239\approx33](https://tex.z-dn.net/?f=n%3D%28%5Cdfrac%7Bz_%7B%5Calpha%2F2%7Ds%7D%7BE%7D%29%5E2%5C%5C%5C%5C%5CRightarrow%5C%20n%3D%28%5Cdfrac%7B%281.96%29%28167%29%7D%7B57%7D%29%5E2%3D32.9758025239%5Capprox33)
Hence, the minimum required sample size would be 33.