I believe it is the first one sorry if I am wrong
Answer:2
Step-by-step explanation:
Answer:
Yes. Fertilization increases grape yields by more than 5 pounds.
Step-by-step explanation:
Let f-fertilized and o-old(unfertilized)
#First, we use our data to calculate the standard error:
![SE(\bar y_f-\bar y_o)=\sqrt{\frac{s_f^2}{n_f}+\frac{s_o^2}{n_o}}\\\\=\sqrt{\frac{3.7^2}{44}+\frac{3.4^2}{47}}\\\\=0.7464](https://tex.z-dn.net/?f=SE%28%5Cbar%20y_f-%5Cbar%20y_o%29%3D%5Csqrt%7B%5Cfrac%7Bs_f%5E2%7D%7Bn_f%7D%2B%5Cfrac%7Bs_o%5E2%7D%7Bn_o%7D%7D%5C%5C%5C%5C%3D%5Csqrt%7B%5Cfrac%7B3.7%5E2%7D%7B44%7D%2B%5Cfrac%7B3.4%5E2%7D%7B47%7D%7D%5C%5C%5C%5C%3D0.7464)
#State both null and alternative hypothesis:
![H_o:\mu_f-\mu_o=0\\\\H_A=\mu_f-\mu_o>0](https://tex.z-dn.net/?f=H_o%3A%5Cmu_f-%5Cmu_o%3D0%5C%5C%5C%5CH_A%3D%5Cmu_f-%5Cmu_o%3E0)
#We determine our degrees of freedom as 87.02(using R), we now compute the t-value as:
![t=\frac{\bar y_f-\bar y_o}{\sqrt{\frac{s_f^2}{n_f}+\frac{s_o^2}{n_o}}}\\\\\\t=\frac{53.4-52.1}{\sqrt{\frac{s_f^2}{n_f}+\frac{s_o^2}{n_o}}}\\\\\\t=\frac{1.3}{0.7464}\\\\t=1.7417\\\\P=P(t_{87.0>1.7417}=1-0.9575=0.0.0425](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B%5Cbar%20y_f-%5Cbar%20y_o%7D%7B%5Csqrt%7B%5Cfrac%7Bs_f%5E2%7D%7Bn_f%7D%2B%5Cfrac%7Bs_o%5E2%7D%7Bn_o%7D%7D%7D%5C%5C%5C%5C%5C%5Ct%3D%5Cfrac%7B53.4-52.1%7D%7B%5Csqrt%7B%5Cfrac%7Bs_f%5E2%7D%7Bn_f%7D%2B%5Cfrac%7Bs_o%5E2%7D%7Bn_o%7D%7D%7D%5C%5C%5C%5C%5C%5Ct%3D%5Cfrac%7B1.3%7D%7B0.7464%7D%5C%5C%5C%5Ct%3D1.7417%5C%5C%5C%5CP%3DP%28t_%7B87.0%3E1.7417%7D%3D1-0.9575%3D0.0.0425)
Since, the p-value is low, we Reject the null hypothesis. The is enough evidence suggesting that grape yields increase by more than 5 pounds than mean yields of unfertilized grapes.
10^3 = 1000 so you multiply the first number by 1000
Answer is 1546.3
Answer:
The volume of the empty space is ![476.67\ cm^{3}](https://tex.z-dn.net/?f=476.67%5C%20cm%5E%7B3%7D)
Step-by-step explanation:
we know that
The volume of the empty space inside the box is equal to the volume of the box minus the volume of the soccer ball
step 1
Find the volume of the box
The length side of the cube is equal to the diameter of the soccer ball
Let
b -----> the length side of the box
we have
![b=2r=2(5)=10\ cm](https://tex.z-dn.net/?f=b%3D2r%3D2%285%29%3D10%5C%20cm)
The volume of the box is equal to the volume of a cube
![V=b^{3}](https://tex.z-dn.net/?f=V%3Db%5E%7B3%7D)
substitute
![V=10^{3}](https://tex.z-dn.net/?f=V%3D10%5E%7B3%7D)
![V=1,000\ cm^{3}](https://tex.z-dn.net/?f=V%3D1%2C000%5C%20cm%5E%7B3%7D)
step 2
Find the volume of the soccer ball
The volume of the sphere (soccer ball) is equal to
![V=\frac{4}{3}\pi r^{3}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B4%7D%7B3%7D%5Cpi%20r%5E%7B3%7D)
we have
![r=5\ cm](https://tex.z-dn.net/?f=r%3D5%5C%20cm)
![\pi =3.14](https://tex.z-dn.net/?f=%5Cpi%20%3D3.14)
substitute
![V=\frac{4}{3}(3.14)(5)^{3}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B4%7D%7B3%7D%283.14%29%285%29%5E%7B3%7D)
![V=523.33\ cm^{3}](https://tex.z-dn.net/?f=V%3D523.33%5C%20cm%5E%7B3%7D)
step 3
Find the volume of the empty space
![V=1,000-523.33=476.67\ cm^{3}](https://tex.z-dn.net/?f=V%3D1%2C000-523.33%3D476.67%5C%20cm%5E%7B3%7D)