Well if you subtract 437.5 from 500 you get 62.5 that is the difference between them but in fraction form it would be 5/8.
        
             
        
        
        
Answer:

Step-by-step explanation:
We need to write  as a single logarithm.
 as a single logarithm.
We know that 

Therefore we have:

→ 
→ 
The solution is: 
 
        
             
        
        
        
Menjawab:
 [(√1-p²) -3√p] / 2
Penjelasan langkah demi langkah:
Dari identitas trigonometri, pemuaiannya benar:
Cos (A + B) = cosAcosB-sinAsinB
Menerapkan ini dalam memperluas cos (x + 60).
cos (x + 60) = cosxcos60 - sinxsin60
Jika sinx = p = berlawanan / sisi miring
opp = p, hyp = 1
adj² = 1²-p²
adj = √1-p²
Cos (x) = adj / hyp = √1-p² / 1
Cos (x) = √1-p²
Cos60 = 1/2 dan sin60 = √3 / 2
Mengganti nilai-nilai ini ke dalam rumus
cos (x + 60) = cosxcos60 - sinxsin60
cos (x + 60) = √1-p² (1/2) - p (√3 / 2)
cos (x + 60) = (√1-p²) / 2 - √3p / 2
Temukan KPK tersebut
cos (x + 60) = [(√1-p²) -3√p] / 2
Oleh karena itu cos (x + 60) = [(√1-p²) -3√p] / 2
 
        
             
        
        
        
Answer:
a
    
b
    
Step-by-step explanation:
From the question we are told that 
 The proportion that has outstanding balance is p = 0.20
 The sample size is n = 15
Given that the properties of the binomial distribution apply, for a randomly selected number(X) of credit card
 
 Generally the probability of finding 4 customers in a sample of 15 who have "maxed out" their credit cards is mathematically represented as 
 
 => 
Here C stand for combination 
 =>  
 
Generally the probability that 4 or fewer customers in the sample will have balances at the limit of the credit card is mathematically represented as 
 ![P(X \le 4) =  [ ^{15}C_0 * (0.20)^0 * (1 - 0.20)^{15-0}]+[ ^{15}C_1 * (0.20)^1 * (1 - 0.20)^{15-1}]+\cdots+[ ^{15}C_4 * (0.20)^4 * (1 - 0.20)^{15-4}]](https://tex.z-dn.net/?f=P%28X%20%5Cle%204%29%20%3D%20%20%5B%20%5E%7B15%7DC_0%20%2A%20%280.20%29%5E0%20%2A%20%281%20-%200.20%29%5E%7B15-0%7D%5D%2B%5B%20%5E%7B15%7DC_1%20%2A%20%280.20%29%5E1%20%2A%20%281%20-%200.20%29%5E%7B15-1%7D%5D%2B%5Ccdots%2B%5B%20%5E%7B15%7DC_4%20%2A%20%280.20%29%5E4%20%2A%20%281%20-%200.20%29%5E%7B15-4%7D%5D)
=>   