Let us assume the cost of 1 apple = x dollars
Let us also assume the cost of 1 pear = y dollars
Then we can form two equations from the details given in the question. Based on those details the required answer to the question can be easily deduced.
3x + 8y = 14.50
And
6x + 4y = 14
Dividing both sides of the equation by 2 we get
3x + 2y = 7
2y = 7 - 3x
y = (7 - 3x)/2
Putting the value of y from the second equation in the first equation we get
3x + 8y = 14.50
3x + 8[(7 - 3x)/2] = 14.50
3x + 4 (7 - 3x) = 14.50
3x + 28 - 12x = 14.50
- 9x = 14.50 - 28
- 9x = - 13.5
9x = 13.5
x = 13.5/9
= 1.5
Putting the value of x in the second equation we get
6x + 4y = 14
(6 * 1.5) + 4y = 14
9 + 4y = 14
4y = 14 - 9
4y = 5
y = 5/4
= 1.25
So we can find from the above deduction that the cost of 1 apple is 1.5 dollars and the cost of 1 pear is 1.25 dollars
Then
Cost of 2 apples = 2 * 1.5 dollars
= 3.0 dollars
So the cost of 2 apples is $3 and the cost of 1 pear is $1.25.
Answer:
SLOPE IS -3
Step-by-step explanation:
4-7 over 0-(-1)
Answer:
The standard error of the mean is 4.5.
Step-by-step explanation:
As we don't know the standard deviation of the population, we can estimate the standard error of the mean from the standard deviation of the sample as:
![\sigma_{\bar{x}}\approx\frac{s}{\sqrt{n}}](https://tex.z-dn.net/?f=%5Csigma_%7B%5Cbar%7Bx%7D%7D%5Capprox%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D)
The sample is [30mins, 40 mins, 60 mins, 80 mins, 20 mins, 85 mins]. The size of the sample is n=6.
The mean of the sample is:
![\bar{x}=\frac{1}n} \sum x_i =\frac{30+40+60+80+20+85}{6}=52.5](https://tex.z-dn.net/?f=%5Cbar%7Bx%7D%3D%5Cfrac%7B1%7Dn%7D%20%5Csum%20x_i%20%3D%5Cfrac%7B30%2B40%2B60%2B80%2B20%2B85%7D%7B6%7D%3D52.5)
The standard deviation of the sample is calculated as:
![s=\sqrt{\frac{1}{n-1}\sum (x_i-\bar x)^2} \\\\ s=\sqrt{\frac{1}{5}\cdot ((30-52.5)^2+(40-52.5)^2+(60-52.5)^2+(80-52.5)^2+(20-52.5)^2+(85-52.5)^2}\\\\s=\sqrt{\frac{1}{5} *3587.5}=\sqrt{717.5}=26.8](https://tex.z-dn.net/?f=s%3D%5Csqrt%7B%5Cfrac%7B1%7D%7Bn-1%7D%5Csum%20%28x_i-%5Cbar%20x%29%5E2%7D%20%5C%5C%5C%5C%20s%3D%5Csqrt%7B%5Cfrac%7B1%7D%7B5%7D%5Ccdot%20%28%2830-52.5%29%5E2%2B%2840-52.5%29%5E2%2B%2860-52.5%29%5E2%2B%2880-52.5%29%5E2%2B%2820-52.5%29%5E2%2B%2885-52.5%29%5E2%7D%5C%5C%5C%5Cs%3D%5Csqrt%7B%5Cfrac%7B1%7D%7B5%7D%20%2A3587.5%7D%3D%5Csqrt%7B717.5%7D%3D26.8)
Then, we can calculate the standard error of the mean as:
![\sigma_{\bar{x}}\approx\frac{s}{\sqrt{n}}=\frac{26.8}{6}= 4.5](https://tex.z-dn.net/?f=%5Csigma_%7B%5Cbar%7Bx%7D%7D%5Capprox%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%3D%5Cfrac%7B26.8%7D%7B6%7D%3D%204.5)
<h3><u>Answer</u><u>:</u><u>-</u></h3>
x=17
<h3><u>step-by</u><u>-</u><u>step</u><u> </u><u>Explanation</u><u>:</u><u>-</u></h3>
![{\boxed{\quad\:\mapsto\rm Firstly\:Let's\:understand\:the\:concept:-}}](https://tex.z-dn.net/?f=%7B%5Cboxed%7B%5Cquad%5C%3A%5Cmapsto%5Crm%20Firstly%5C%3ALet%27s%5C%3Aunderstand%5C%3Athe%5C%3Aconcept%3A-%7D%7D)
<h3 />
This is a isosceles triangle. As it is a triangle we can apply sum theory. we have to take the sum of given unknown polynomials as 180° .Then by solving it we can find the value of x.
<h3><u>Solution</u><u>:</u><u>-</u></h3>
Given angles
According to sum theory
![{\boxed{\sf The \:sum\:of\:angles=180°}}](https://tex.z-dn.net/?f=%7B%5Cboxed%7B%5Csf%20The%20%5C%3Asum%5C%3Aof%5C%3Aangles%3D180%C2%B0%7D%7D)
![\qquad \quad{:}\longmapsto\tt (6x+10)+(x+17)+(4x-34)=180](https://tex.z-dn.net/?f=%5Cqquad%20%5Cquad%7B%3A%7D%5Clongmapsto%5Ctt%20%286x%2B10%29%2B%28x%2B17%29%2B%284x-34%29%3D180%20)
![\qquad \quad{:}\longmapsto\tt 6x+10+x+17+4x-34=180](https://tex.z-dn.net/?f=%5Cqquad%20%5Cquad%7B%3A%7D%5Clongmapsto%5Ctt%206x%2B10%2Bx%2B17%2B4x-34%3D180%20)
- Together like polynomials and constants
![\qquad \quad{:}\longmapsto\tt 6x+x+4x+10+17-34=180](https://tex.z-dn.net/?f=%5Cqquad%20%5Cquad%7B%3A%7D%5Clongmapsto%5Ctt%206x%2Bx%2B4x%2B10%2B17-34%3D180%20)
![\qquad \quad{:}\longmapsto\tt 11x-7=180](https://tex.z-dn.net/?f=%5Cqquad%20%5Cquad%7B%3A%7D%5Clongmapsto%5Ctt%2011x-7%3D180%20)
![\qquad \quad{:}\longmapsto\tt 11x=180+7](https://tex.z-dn.net/?f=%5Cqquad%20%5Cquad%7B%3A%7D%5Clongmapsto%5Ctt%2011x%3D180%2B7%20)
![\qquad \quad{:}\longmapsto\tt 11x=187](https://tex.z-dn.net/?f=%5Cqquad%20%5Cquad%7B%3A%7D%5Clongmapsto%5Ctt%2011x%3D187%20)
![\qquad \quad{:}\longmapsto\tt x=\cancel{\dfrac{187}{11}}](https://tex.z-dn.net/?f=%5Cqquad%20%5Cquad%7B%3A%7D%5Clongmapsto%5Ctt%20x%3D%5Ccancel%7B%5Cdfrac%7B187%7D%7B11%7D%7D)
![\qquad \quad{:}\longmapsto\tt x=17](https://tex.z-dn.net/?f=%5Cqquad%20%5Cquad%7B%3A%7D%5Clongmapsto%5Ctt%20x%3D17)
![\therefore\sf The \:value\:of\:x\;is\:17.](https://tex.z-dn.net/?f=%5Ctherefore%5Csf%20The%20%5C%3Avalue%5C%3Aof%5C%3Ax%5C%3Bis%5C%3A17.)
Tom bought 768 pounds of candy because 333 + 435 = 768