let's firstly convert the mixed fraction to improper fraction and then take it from there, keeping in mind that the whole is "x".
![\stackrel{mixed}{5\frac{5}{6}}\implies \cfrac{5\cdot 6+5}{6}\implies \stackrel{improper}{\cfrac{35}{6}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{7}{3}x~~ = ~~5\frac{5}{6}\implies \cfrac{7}{3}x~~ = ~~\cfrac{35}{6}\implies 42x=105\implies x=\cfrac{105}{42} \\\\\\ x=\cfrac{21\cdot 5}{21\cdot 2}\implies x=\cfrac{21}{21}\cdot \cfrac{5}{2}\implies x=1\cdot \cfrac{5}{2}\implies x=2\frac{1}{2}](https://tex.z-dn.net/?f=%5Cstackrel%7Bmixed%7D%7B5%5Cfrac%7B5%7D%7B6%7D%7D%5Cimplies%20%5Ccfrac%7B5%5Ccdot%206%2B5%7D%7B6%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B35%7D%7B6%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7B7%7D%7B3%7Dx~~%20%3D%20~~5%5Cfrac%7B5%7D%7B6%7D%5Cimplies%20%5Ccfrac%7B7%7D%7B3%7Dx~~%20%3D%20~~%5Ccfrac%7B35%7D%7B6%7D%5Cimplies%2042x%3D105%5Cimplies%20x%3D%5Ccfrac%7B105%7D%7B42%7D%20%5C%5C%5C%5C%5C%5C%20x%3D%5Ccfrac%7B21%5Ccdot%205%7D%7B21%5Ccdot%202%7D%5Cimplies%20x%3D%5Ccfrac%7B21%7D%7B21%7D%5Ccdot%20%5Ccfrac%7B5%7D%7B2%7D%5Cimplies%20x%3D1%5Ccdot%20%5Ccfrac%7B5%7D%7B2%7D%5Cimplies%20x%3D2%5Cfrac%7B1%7D%7B2%7D)
Answer: a. 1.981 < μ < 2.18
b. Yes.
Step-by-step explanation:
A. For this sample, we will use t-distribution because we're estimating the standard deviation, i.e., we are calculating the standard deviation, and the sample is small, n = 12.
First, we calculate mean of the sample:


2.08
Now, we estimate standard deviation:


s = 0.1564
For t-score, we need to determine degree of freedom and
:
df = 12 - 1
df = 11
= 1 - 0.95
α = 0.05
0.025
Then, t-score is
= 2.201
The interval will be
± 
2.08 ± 
2.08 ± 0.099
The 95% two-sided CI on the mean is 1.981 < μ < 2.18.
B. We are 95% confident that the true population mean for this clinic is between 1.981 and 2.18. Since the mean number performed by all clinics has been 1.95, and this mean is less than the interval, there is evidence that this particular clinic performs more scans than the overall system average.
Answer:
13
Step-by-step explanation:
To solve this, lets first find the number of footballs he still needs:
75-24 = 51
Now to find the number of packages, divide by 4, since there are 4 balls per package:
51/4 = 12 Remainder 3
Since there is a remainder, and the Coach MUST have at least 75, then we need to add a package to include the remainder:
12 + 1 = 13 Packages
Answer:
2.
A. (P+h)(x)
2x/x+4 (x-1) + x/x-1 (x+4)
2x^2-1/x^2-4
+
X^2+4/x^2-4
= 3x^2+3/x^2-4
B. (F-g)(x)
X^2-7x+6-x - 6
= x^2 -8x
C. (Fg)(x)
(X^2-7x+6)(x-6)
= x^3-13x^2+48x-36
D. (H/p)(x)
X/x-1 / 2x/x+4
X/x-1 / x+4/2x
= X^2+4x/2x^2-2x
3.
A. (F+g)(3)
X^2+1 + x-4
3^2+1 + 3-4
10 -1
= 9
B. (f-g)(0)
X^2+1 - x-4
0+1 -0-4
1-4
= -3
C. (Fg)(-k)
(X^2+1) (x-4)
(-k^2+1) (-k-4)
K^3+4k^2-k-4
D. (F/g)(k-2)
X^2+1 /x-4
K-2^2+1 / k-2 -2
= K^2-4k+5 / k-4
Step-by-step explanation:
Step-by-step explanation:
25x ratio
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