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3 years ago

What are the solutions of the equation (2x + 3)2 + 8(2x + 3) + 11 = 0? Use u substitution and the quadratic formula to solve

Mathematics

2 answers:

Y_Kistochka [10]3 years ago

8 0

Amswer:x= -41/20
Solutions:
(2x + 3)2 + 8(2x + 3) + 11 = 0
4x+6+16x+24+11=0
20x+41=0
20x=-41
x= -41/20

olga_2 [115]3 years ago

3 0

Answer:

The given expression in single variable is:

$(2 x +3)^2 +8 \times (2 x +3) +11=0\\\\ 4x^2 +9 +2 \times 2 x \times 3+8 \times 2 x+8 \times 3+11=0\\\\4 x^2+12 x+16 x+9+24+11=0\\\\4x^2+28 x+44=0\\\\4 \times (x^2+7 x+11)=0\\\\x^2+7 x+11=0\\\\ x=\frac{-7\pm \sqrt{7^2-4 \times 11 \times 1}}{2 \times 1}\\\\x=\frac{-7\pm \sqrt{49-44}}{2}\\\\x=\frac{-7\pm \sqrt{5}}{2}$

⇒Used Distributive property of Multiplication with respect to Addition:

a × (b+c)=a ×b+a × c

⇒For, a Quadratic expression of the form

Ax² +Bx +C=0

Discriminant=D=B²-4 AC

$x=\frac{-B\pm \sqrt{D}}{2 A}$

There are two zeroes which are

$x_{1}=\frac{-7+ \sqrt{5}}{2}\\\\x_{2}=\frac{-7- \sqrt{5}}{2}$

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A manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective

andriy [413]

Answer:

(a) P(X $\leq$ 20) = 0.9319

(b) Expected number of defective light bulbs = 15

(c) Standard deviation of defective light bulbs = 3.67

Step-by-step explanation:

We are given that a manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective independently. A box of 150 bulbs is selected at random.

Firstly, the above situation can be represented through binomial distribution, i.e.;

$P(X=r) = \binom{n}{r} p^{r} (1-p)^{2} ;x=0,1,2,3,....$

where, n = number of samples taken = 150

r = number of success

p = probability of success which in our question is % of bulbs that

are defective, i.e. 10%

Now, we can't calculate the required probability using binomial distribution because here n is very large(n > 30), so we will convert this distribution into normal distribution using continuity correction.

So, Let X = No. of defective bulbs in a box

Mean of X, $\mu$ = $n \times p$ = $150 \times 0.10$ = 15

Standard deviation of X, $\sigma$ = $\sqrt{np(1-p)}$ = $\sqrt{150 \times 0.10 \times (1-0.10)}$ = 3.7

So, X ~ N( $\mu = 15, \sigma^{2} = 3.7^{2})$

Now, the z score probability distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

(a) Probability that this box will contain at most 20 defective light bulbs is given by = P(X $\leq$ 20) = P(X < 20.5) ---- using continuity correction

P(X < 20.5) = P( $\frac{X-\mu}{\sigma}$ < $\frac{20.5-15}{3.7}$ ) = P(Z < 1.49) = 0.9319

(b) Expected number of defective light bulbs found in such boxes, on average is given by = E(X) = $n \times p$ = $150 \times 0.10$ = 15.

Standard deviation of defective light bulbs is given by = S.D. = $\sqrt{np(1-p)}$ = $\sqrt{150 \times 0.10 \times (1-0.10)}$ = 3.67

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Answer:

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