The answer is 42! hope this helps!
Answer:
0.147
Step-by-step explanation:
NB: check exclusive solution on the file attached.
This is a limit theorem problem. To solve this problem, we have to make series of symbol assumptions.
We say that Xi is equal to the ith round off error. Therefore, Xi~ UNIF (−0.5,0.5). Also, the f(Xi)= 1( that is the density). When Xi is an element of -0.5,0.5; i=1,...,70.
Step one(1): calculate the E(Xi). And, in order to calculate E(Xi), we use;
E(Xi) = 0.5+(-0.5)/2
E(Xi)= 0.
Step two(2) : calculate the var(Xi). Which is; 0.5-(0.5^2)/12 = 1/12.
X(1) to X(70) are independent. Therefore, using the central limit theorem we know approximately that normal distribution is U= 50∗0=0. Also, σ^2= 70 × 1/12 = 5.83.
Step three(3): we find our P from the Z values(z values can be gotten from the z-standard deviation from cut off.
2p(Z>1.66)
= .147
The constants, -3 and -8, are like terms
The terms 3p and p are like terms
The terms in the expression are p^2, -3, 3p, -8, p, p^3
The expression contains 6 terms.
Like terms have the same variables raised to the same powers.
Answer:
y<-2
Step-by-step explanation:
8y-3y+6>2y+8y+16
-3y-2y>16-6
-5y>10
y<-2