When two characters are next to each other (like p6) this means multiplication. Therefore it can be written as p x 6. If you substitute p for 3 you have 3 x 6 which equals 18.
4/5 of 60 p = 48 p
probability of a machine component failing = 2/7 P(at least 4 failed) = P( 4 failed)+P(5 failed) +P(6 failed)+P(7 failed) Using binomial probability: P(4 failed ) =7C4 (2/7)^4 ((5/7)^3 = 0.084987 P(5 failed) = 7C5 (2/7)^5 (5/7)^2 = 0.020395 P(6 failed ) = 7C6 (2/7)^6 (5/7) = 0.002719 P(7 failed) = (2/7)^7 = 0.000155 Adding, P(at least 4 failures) = 0.108257
p = 4
The equation is as follows: p - 5/8 (1/4 + p). This equation expands to (1/4)p + p^2 - 5/32 - (5/8)p, which simplifies to p^2 - (3/8)p - 5/32. Now either solve for p using the quadratic formula or solve by factoring.
-1507
When two characters are next to each other (like p6) this means multiplication. Therefore it can be written as p x 6. If you substitute p for 3 you have 3 x 6 which equals 18.
4/5 of 60 p = 48 p
-4(3 - p) > 5(p + 1) => -12 + 4p > 5p + 5-17 > p, that is p < -17.
This is a simple algebraic equation. To solve for p, you need to isolate the variable (p) on one side of the equation. Here are the steps to solve the equation: p + 5 = 9 Subtract 5 from both sides p = 4 So the value of p that makes the equation true is 4.
probability of a machine component failing = 2/7 P(at least 4 failed) = P( 4 failed)+P(5 failed) +P(6 failed)+P(7 failed) Using binomial probability: P(4 failed ) =7C4 (2/7)^4 ((5/7)^3 = 0.084987 P(5 failed) = 7C5 (2/7)^5 (5/7)^2 = 0.020395 P(6 failed ) = 7C6 (2/7)^6 (5/7) = 0.002719 P(7 failed) = (2/7)^7 = 0.000155 Adding, P(at least 4 failures) = 0.108257
p = 4
42p-1 = 5p+4 42p = 5p+5 37p = 5 p = 5/37 p = 0.135135...
The equation is as follows: p - 5/8 (1/4 + p). This equation expands to (1/4)p + p^2 - 5/32 - (5/8)p, which simplifies to p^2 - (3/8)p - 5/32. Now either solve for p using the quadratic formula or solve by factoring.
4 with a remainder of 4
A prime triplet contains a pair of twin primes (p and p + 2, or p + 4 and p+ 6), a pair of cousin primes (pand p + 4, or p + 2 and p + 6), and a pair of sexy primes (p and p + 6).
PQR P=2 Q=4 R=5 2 x 4 x 5 = 40