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rodikova [14]
3 years ago
8

Evaluate the function.

ula1" title="f(x) = - x^{2} - 7x - 5" alt="f(x) = - x^{2} - 7x - 5" align="absmiddle" class="latex-formula">
Find
f( - 5)
​
Mathematics
1 answer:
Nat2105 [25]3 years ago
6 0
<h3>Answer:  5</h3>

===================================================

Explanation:

Everywhere you see an x, replace it with -5. Then use the order of operations PEMDAS to simplify

f(x) = -x^2 - 7x - 5\\\\f(-5) = -(-5)^2 - 7(-5) - 5\\\\f(-5) = -(25) - 7(-5) - 5\\\\f(-5) = -25 + 35 - 5\\\\f(-5) = 10 - 5\\\\f(-5) = 5\\\\

Keep in mind that for the x^2 term, you are squaring all of x = -5, which includes the negative as well. So that means

x^2 = (-5)^2 = (-5)*(-5) = 25

but sticking a negative out front makes

-x^2 = -(-5)^2 = -25

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Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive
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Answer:

(a) 0.343

(b) 0.657

(c) 0.189

(d) 0.216

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Step-by-step explanation:

Let P(a vehicle passing the test) = p

                        p = \frac{70}{100} = 0.7  

Let P(a vehicle not passing the test) = q

                         q = 1 - p

                         q = 1 - 0.7 = 0.3

(a) P(all of the next three vehicles inspected pass) = P(ppp)

                           = 0.7 × 0.7 × 0.7

                           = 0.343

(b) P(at least one of the next three inspected fails) = P(qpp or qqp or pqp or pqq or ppq or qpq or qqq)

      = (0.3 × 0.7 × 0.7) + (0.3 × 0.3 × 0.7) + (0.7 × 0.3 × 0.7) + (0.7 × 0.3 × 0.3) + (0.7 × 0.7 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.3)

      = 0.147 + 0.063 + 0.147 + 0.063 + 0.147 + 0.063 + 0.027

      = 0.657

(c) P(exactly one of the next three inspected passes) = P(pqq or qpq or qqp)

                 =  (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7)

                 = 0.063 + 0.063 + 0.063

                 = 0.189

(d) P(at most one of the next three vehicles inspected passes) = P(pqq or qpq or qqp or qqq)

                 =  (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7) + (0.3 × 0.3 × 0.3)

                 = 0.063 + 0.063 + 0.063 + 0.027

                 = 0.216

(e) Given that at least one of the next 3 vehicles passes inspection, what is the probability that all 3 pass (a conditional probability)?

P(at least one of the next three vehicles inspected passes) = P(ppp or ppq or pqp or qpp or pqq or qpq or qqp)

=  (0.7 × 0.7 × 0.7) + (0.7 × 0.7 × 0.3) + (0.7 × 0.3 × 0.7) + (0.3 × 0.7 × 0.7) + (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7)

= 0.343 + 0.147 + 0.147 + 0.147 + 0.063 + 0.063 + 0.063

                  = 0.973  

With the condition that at least one of the next 3 vehicles passes inspection, the probability that all 3 pass is,

                         = \frac{P(all\ of\ the\ next\ three\ vehicles\ inspected\ pass)}{P(at\ least\ one\ of\ the\ next\ three\ vehicles\ inspected\ passes)}

                         = \frac{0.343}{0.973}

                         = 0.353

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