The equation 45p take away 6p equals too 5p plus 4. This is taught in math.
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