You would do 2 · 26 · 26 = 1,352 different arrangements
let's convert firstly the mixed fractions to improper fractions and then add up.
![\bf \stackrel{mixed}{2\frac{3}{8}}\implies \cfrac{2\cdot 8+3}{8}\implies \stackrel{improper}{\cfrac{19}{8}}~\hfill \stackrel{mixed}{1\frac{1}{4}}\implies \cfrac{1\cdot 4+1}{4}\implies \stackrel{improper}{\cfrac{5}{4}} \\\\\\ \stackrel{mixed}{2\frac{7}{8}}\implies \cfrac{2\cdot 8+7}{8}\implies \stackrel{improper}{\cfrac{23}{8}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Bmixed%7D%7B2%5Cfrac%7B3%7D%7B8%7D%7D%5Cimplies%20%5Ccfrac%7B2%5Ccdot%208%2B3%7D%7B8%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B19%7D%7B8%7D%7D~%5Chfill%20%5Cstackrel%7Bmixed%7D%7B1%5Cfrac%7B1%7D%7B4%7D%7D%5Cimplies%20%5Ccfrac%7B1%5Ccdot%204%2B1%7D%7B4%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B5%7D%7B4%7D%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7Bmixed%7D%7B2%5Cfrac%7B7%7D%7B8%7D%7D%5Cimplies%20%5Ccfrac%7B2%5Ccdot%208%2B7%7D%7B8%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B23%7D%7B8%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
Answer:
63° (same as Z)
Step-by-step explanation:
The question indicates that ΔXYZ is similar to ΔEDF.
By definition, when we indicate that triangle XYZ is similar to the triangle EDF, the order of the letters is important and match the congruent angles.
So, ∠X = ∠E, ∠Y = ∠D and ∠Z = ∠F
It's hard to see because the angles are quite similar in opening (56°, 63° and 61°), and the drawing doesn't seems to be to scale.
But according to the naming convention: ∠Z = ∠F
Since ∠Z = 63°, ∠F = 63°
Answer:
Step-by-step explanation:
Subtract p on both sides.
Divide both sides by pr.
Answer:
95% confidence level should be used for a confidence interval.
The given confidence interval contains the value of 85 sec, so there is not sufficient evidence to support the claim that the mean is greater than 85 sec.
Step-by-step explanation:
0.05 significance level
1 - 0.05 = 0.95
0.95*100% = 95%
This means that a 95% confidence level should be used for a confidence interval.
Confidence interval is found to be −1.8 sec<μ<213.5 sec, what should we conclude about the claim?
Contains the value of 85 sec, thus there is not sufficient evidence to support the claim that the mean is greater than 85 sec.
