The answer is 24, hopes this helps!
The formula to use is the nCr combination formula
n C r = (n!)/(r!*(n-r)!)
where the exclamation marks indicate factorial notation
Plug n = 10 and r = 2 into the formula
n C r = (n!)/(r!*(n-r)!)
10 C 2 = (10!)/(2!*(10-2)!)
10 C 2 = (10!)/(2!*8!)
10 C 2 = (10*9*8!)/(2!*8!)
10 C 2 = (10*9)/(2!)
10 C 2 = (10*9)/(2*1)
10 C 2 = 90/2
10 C 2 = 45
Answer: 45
Answer:
10 pounds of trial mix that costs $2.45 and 20 pounds of trial mix that costs $2.30
Step-by-step explanation:
![\\\text{Let x pounds of the trial mix that costs \$2.45 per pound, and y pounds of the trial mix}\\\text{that costs \$2.30 per pound are added to get a 30 pounds of trial mix }\\\text{that costs \$2.35 per pound. since the total mixture is 30 pounds, so we get}\\\\x+y=30\ \ \ \ \ \ \ \ \ \ \ \ \ ...... (i)\\\\\text{and the total cost equation is given by}\\\\\$2.45 x + \$2.30 y=\$2.35 (30)\\\\2.45 x + 2.30 y=70.5 \ \ \ \ \ \ \ \ \ \ \ \ \ ...... (ii)\\](https://tex.z-dn.net/?f=%5C%5C%5Ctext%7BLet%20x%20pounds%20of%20the%20trial%20mix%20that%20costs%20%5C%242.45%20per%20pound%2C%20and%20y%20pounds%20of%20the%20trial%20mix%7D%5C%5C%5Ctext%7Bthat%20costs%20%5C%242.30%20per%20pound%20are%20added%20to%20get%20a%2030%20pounds%20of%20trial%20mix%20%7D%5C%5C%5Ctext%7Bthat%20costs%20%5C%242.35%20per%20pound.%20since%20the%20total%20mixture%20is%2030%20pounds%2C%20so%20we%20get%7D%5C%5C%5C%5Cx%2By%3D30%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%20%5C%20%5C%20%5C%20%5C%20%5C%20%20%5C%20%20......%20%28i%29%5C%5C%5C%5C%5Ctext%7Band%20the%20total%20cost%20equation%20is%20given%20by%7D%5C%5C%5C%5C%5C%242.45%20x%20%2B%20%5C%242.30%20y%3D%5C%242.35%20%2830%29%5C%5C%5C%5C2.45%20x%20%2B%202.30%20y%3D70.5%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20......%20%28ii%29%5C%5C)
![\text{now to solve the equation (i) and (ii), we'll substitute }y=30-x \text{ from (i) into (ii)}\\\text{so we get}\\\\2.45x+2.30(30-x)=70.5\\\\2.45x+69-2.30x=70.5\\\\\Rightarrow 2.45x-2.30x=70.5-69\\\\\Rightarrow 0.15x=1.5\\\\\Rightarrow x=\frac{1.5}{0.15}\\\\\Rightarrow x=10\\\\\text{plug this value of x in (i), we get}\\\\y=30-x=30-10=20\\\\\text{so we must add 10 Pounds of trial mix that costs \$ 2.45 and 20 pounds of}\\\text{the trial mix that costs \$2.30}](https://tex.z-dn.net/?f=%5Ctext%7Bnow%20to%20solve%20the%20equation%20%28i%29%20and%20%28ii%29%2C%20we%27ll%20substitute%20%7Dy%3D30-x%20%5Ctext%7B%20from%20%28i%29%20into%20%28ii%29%7D%5C%5C%5Ctext%7Bso%20we%20get%7D%5C%5C%5C%5C2.45x%2B2.30%2830-x%29%3D70.5%5C%5C%5C%5C2.45x%2B69-2.30x%3D70.5%5C%5C%5C%5C%5CRightarrow%202.45x-2.30x%3D70.5-69%5C%5C%5C%5C%5CRightarrow%200.15x%3D1.5%5C%5C%5C%5C%5CRightarrow%20x%3D%5Cfrac%7B1.5%7D%7B0.15%7D%5C%5C%5C%5C%5CRightarrow%20x%3D10%5C%5C%5C%5C%5Ctext%7Bplug%20this%20value%20of%20x%20in%20%28i%29%2C%20we%20get%7D%5C%5C%5C%5Cy%3D30-x%3D30-10%3D20%5C%5C%5C%5C%5Ctext%7Bso%20we%20must%20add%2010%20Pounds%20of%20trial%20mix%20that%20costs%20%5C%24%202.45%20and%2020%20pounds%20of%7D%5C%5C%5Ctext%7Bthe%20trial%20mix%20that%20costs%20%5C%242.30%7D)
Answer: 1.300228
Step-by-step explanation:
The t-score we use for the confidence interval is two-tailed , i.e. the t-score should be used to find a (
) is given by :-
, where n is the sample size.
Given : Level of confidence: ![1-\alpha: 0.80](https://tex.z-dn.net/?f=1-%5Calpha%3A%200.80)
Significance level : ![\alpha: 1-0.80=0.20](https://tex.z-dn.net/?f=%5Calpha%3A%201-0.80%3D0.20)
Sample size : n= 47
Then, degree of freedom : ![n-1=46](https://tex.z-dn.net/?f=n-1%3D46)
Now by using standard normal t-distribution table,
![t_{n-1, \alpha/2}=t_{46, 0.10}=1.300228](https://tex.z-dn.net/?f=t_%7Bn-1%2C%20%5Calpha%2F2%7D%3Dt_%7B46%2C%200.10%7D%3D1.300228)
Hence, the t-score should be used to find a 80% confidence interval estimate for the population mean = 1.300228
Answer:
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