Answer:
1) the constant of variation is 1/6
2) b=80 when A=10
3)the value of k is 40
4) x=0.032 when y = 10
Step-by-step explanation:
1)m varies directly as y
![\Rightarrow m \propto y \\\Rightarrow m =ky](https://tex.z-dn.net/?f=%5CRightarrow%20m%20%5Cpropto%20y%20%5C%5C%5CRightarrow%20m%20%3Dky)
k is the constant of variation
We are given that m is 6 when y is 36
![\Rightarrow 6=k(36)\\\Rightarrow \frac{6}{36}=k\\\Rightarrow \frac{1}{6}=k](https://tex.z-dn.net/?f=%5CRightarrow%206%3Dk%2836%29%5C%5C%5CRightarrow%20%5Cfrac%7B6%7D%7B36%7D%3Dk%5C%5C%5CRightarrow%20%5Cfrac%7B1%7D%7B6%7D%3Dk)
Hence the constant of variation is 1/6
2)A varies directly as b.
![\Rightarrow A \propto b\\\Rightarrow A =kb](https://tex.z-dn.net/?f=%5CRightarrow%20A%20%5Cpropto%20b%5C%5C%5CRightarrow%20A%20%3Dkb)
k is the constant of variation
We are given that A = 3 when b = 24
![\Rightarrow 3=k(24)\\\Rightarrow \frac{3}{24}=k\\\Rightarrow \frac{1}{8}=k\\So,A=\frac{1}{8}b](https://tex.z-dn.net/?f=%5CRightarrow%203%3Dk%2824%29%5C%5C%5CRightarrow%20%5Cfrac%7B3%7D%7B24%7D%3Dk%5C%5C%5CRightarrow%20%5Cfrac%7B1%7D%7B8%7D%3Dk%5C%5CSo%2CA%3D%5Cfrac%7B1%7D%7B8%7Db)
Substitute A=10
![10=\frac{1}{8}b](https://tex.z-dn.net/?f=10%3D%5Cfrac%7B1%7D%7B8%7Db)
80=b
So, b=80 when A=10
3)y varies inversely with x
![\Rightarrow y \propto \frac{1}{x}\\\Rightarrow y = \frac{k}{x}](https://tex.z-dn.net/?f=%5CRightarrow%20y%20%5Cpropto%20%5Cfrac%7B1%7D%7Bx%7D%5C%5C%5CRightarrow%20y%20%3D%20%5Cfrac%7Bk%7D%7Bx%7D)
k is the constant of variation
We are given that y = 5 when x = 8
![\Rightarrow 5 = \frac{k}{8}\\\Rightarrow 40=k](https://tex.z-dn.net/?f=%5CRightarrow%205%20%3D%20%5Cfrac%7Bk%7D%7B8%7D%5C%5C%5CRightarrow%2040%3Dk)
So, the value of k is 40
4)y varies inversely with x
![\Rightarrow y \propto \frac{1}{x}\\\Rightarrow y = \frac{k}{x}](https://tex.z-dn.net/?f=%5CRightarrow%20y%20%5Cpropto%20%5Cfrac%7B1%7D%7Bx%7D%5C%5C%5CRightarrow%20y%20%3D%20%5Cfrac%7Bk%7D%7Bx%7D)
k is the constant of variation
We are given that k=0.32
![\Rightarrow y = \frac{0.32}{x}](https://tex.z-dn.net/?f=%5CRightarrow%20y%20%3D%20%5Cfrac%7B0.32%7D%7Bx%7D)
Substitute y = 10
![\Rightarrow 10 = \frac{0.32}{x}\\\Rightarrow x = 0.032](https://tex.z-dn.net/?f=%5CRightarrow%2010%20%3D%20%5Cfrac%7B0.32%7D%7Bx%7D%5C%5C%5CRightarrow%20x%20%3D%200.032)
So, x=0.032 when y = 10
Answer:
v=-4
Step-by-step explanation:
isolate the variable by dividing each side by factors that don't contain the variable.
The amount of money that Suzie will make over the four weeks is; $1856
<h3>How to find amount of profit made?</h3>
Suzie will make a profit of:
The first week = $300
The second week = (300 * 30%) + 300 = $390
The third week = 390 * (1430%) = $507
The fourth week: 507 × (1430%) = $659.1
Thus, Sum of profit = $300 + $390 + $507 + $659.1
Sum of Profit = $1856.1 ≈ $1856
Read more about Profit at; brainly.com/question/23916300
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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.
All to red 13:8
Green to all 5:13