Answer:
<h3>
The percentage is <u>
60%</u>
of the total amount paid with the gift card. </h3>
Step-by-step explanation:
Given:
Total worth of clothing is $178.25 and the amount paid with cash is $71.30.
Let us find the part of amount paid with the gift card:
Amount of gift card(G) = Total worth(T) - Amount paid in cash(C).

.
So, the amount paid with the gift card is $106.95.
Now, according to question:


%.
Therefore, the percentage is <u>60%</u> of the total amount paid with the gift card.
Answer:
S = {0,2,3,4}
P(X=0) = 0.573 , P(X=2) = 0.401 , P(x=3) = 0.025, P(X=4) = 0.001
Mean = 0.879
Standard Deviation = 1.033
Step-by-step explanation:
Let the number of people having same birth month be = x
The number of ways of distributing the birthdays of the 4 men = (12*12*12*12)
The number of ways of distributing their birthdays = 12⁴
The sample space, S = { 0,2,3,4} (since 1 person cannot share birthday with himself)
P(X = 0) = 
P(X=0) = 0.573
P(X=2) = P(2 months are common) P(1 month is common, 1 month is not common)
P(X=2) = 
P(X=2) = 0.401
P(X=3) = 
P(x=3) = 0.025
P(X=4) = 
P(X=4) = 0.001
Mean, 

Standard deviation, ![SD = \sqrt{\sum x^{2} P(x) - \mu^{2}} \\SD =\sqrt{ [ (0^{2} * 0.573) + (2^{2} * 0.401) + (3^{2} * 0.025) + (4^{2} * 0.001)] - 0.879^{2}}](https://tex.z-dn.net/?f=SD%20%3D%20%5Csqrt%7B%5Csum%20x%5E%7B2%7D%20P%28x%29%20-%20%5Cmu%5E%7B2%7D%7D%20%20%5C%5CSD%20%3D%5Csqrt%7B%20%5B%20%280%5E%7B2%7D%20%2A%200.573%29%20%2B%20%282%5E%7B2%7D%20%20%2A%200.401%29%20%2B%20%283%5E%7B2%7D%20%2A%200.025%29%20%2B%20%284%5E%7B2%7D%20%2A%200.001%29%5D%20-%200.879%5E%7B2%7D%7D)
SD = 1.033
Answer:

Step-by-step explanation:


Answer:
Lynnette can wash 2111109 in 11 days
Step-by-step explanation:
Have a good day :)
let's bear in mind that sin(θ) in this case is positive, that happens only in the I and II Quadrants, where the cosine/adjacent are positive and negative respectively.
![\bf sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{hypotenuse}{6}}\qquad \impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{6^2-5^2}=a\implies \pm\sqrt{36-25}\implies \pm \sqrt{11}=a \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20sin%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Bopposite%7D%7B5%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B6%7D%7D%5Cqquad%20%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Badjacent%20side%7D%7D%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20%5Cpm%5Csqrt%7Bc%5E2-b%5E2%7D%3Da%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3Dadjacent%5C%5C%20b%3Dopposite%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20%5Cpm%5Csqrt%7B6%5E2-5%5E2%7D%3Da%5Cimplies%20%5Cpm%5Csqrt%7B36-25%7D%5Cimplies%20%5Cpm%20%5Csqrt%7B11%7D%3Da%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
