Find all integers $n$ such that the quadratic $7x^2 + nx - 11$ can be expressed as the product of two linear factors with integer coefficients.

The all integer values of n are -76, -4, 4, 76

Step-by-step explanation:

∵ The factors of 7 are 1 and 7

∵ The factors of 11 are 1 and 11

- The last term of the quadratic is negative, that means the two

    linear factors have different middle sign

∴ The factors are (7x ± 1)(x ± 11) OR (7x ± 11)(x ± 1)

Let us take one by one

The middle term of the quadratic is the sum of the product of nears and ext-reams

In (7x + 1)(x - 11)

∵ The product of ext-reams is 7x × -11 = -77x

∵ The product of nears is 1 × x = x

∵ Their sum is -77x + x = -76x

∴ n = -76

In (7x - 1)(x + 11)

∵ The product of ext-reams is 7x × 11 = 77x

∵ The product of nears is -1 × x = -x

∵ Their sum is 77x + -x = 76x

∴ n = 76

In (7x + 11)(x - 1)

∵ The product of ext-reams is 7x × -1 = -7x

∵ The product of nears is 11 × x = 11x

∵ Their sum is -7x + 11x = 4x

∴ n = 4

In (7x - 11)(x + 1)

∵ The product of ext-reams is 7x × 1 = 7x

∵ The product of nears is -11 × x = -11x

∵ Their sum is 7x + -11x = -4x

∴ n = -4

The all integer values of n are -76, -4, 4, 76

85.560%

step-by-step explanation:

conditional probability is defined as:

p(a|b) = p(a∩b) / p(b)

in other words, the probability that a occurs, given that b has occurred = the probability that both a and b occur / the probability that b occurs.

in this example:

p(turns 64 | turns 44) = p(turns 64 & turns 44) / p(turns 44).

if someone turns 64, they must have already turned 44, so:

p(turns 64 | turns 44) = p(turns 64) / p(turns 44)

on page 42, we see that of the original 100,000, 94944 turn 44, and 81234 turn 64.   therefore:

p(turns 64 | turns 44) = 81234 / 94944

p(turns 64 | turns 44) = 0.85560