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2 years ago

5(x-3) = 2(x+3) solve for x

Mathematics

2 answers:

sattari [20]2 years ago

4 0

Answer:

Step-by-step explanation:

5(x - 3) = 2(x + 3) Distribute

5x - 15 = 2x + 6

+ 15 + 15 Switch sides to the liked terms

5x = 2x + 6 + 15

-2x -2x Switch sides to the liked terms

5x - 2x = 6 + 15

3x = 21 Divide

3 3

x = 7

skelet666 [1.2K]2 years ago

3 0

Answer:

x = 7

Step-by-step explanation:

Equation: 5(x-3) = 2(x+3)

Step 1: Simplify the equation.

5 (x − 3) = 2(x + 3)

(5)(x) + (5)(−3) = (2)(x) + (2)(3) Distribute

5x + −15 = 2x + 6

5x − 15 = 2x + 6

Step 2: Subtract 2x.

5x − 15 −2x = 2x + 6 −2x

3x − 15 = 6

Step 3: Add 15.

3x − 15 + 15 = 6 + 15

3x = 21

Step 4: Divide by 3.

$\frac{3x}{3} = \frac{21}{3}$

Step 5: Divide 21 by 3.

The answer is x = 7.

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2 years ago

When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%. Let

sergiy2304 [10]

Answer:

(a) P(X=3) = 0.093

(b) P(X≤3) = 0.966

(c) P(X≥4) = 0.034

(d) P(1≤X≤3) = 0.688

(e) The probability that none of the 25 boards is defective is 0.277.

(f) The expected value and standard deviation of X is 1.25 and 1.089 respectively.

Step-by-step explanation:

We are given that when circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%.

Let X = the number of defective boards in a random sample of size, n = 25

So, X ∼ Bin(25,0.05)

The probability distribution for the binomial distribution is given by;

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

where, n = number of trials (samples) taken = 25

r = number of success

p = probability of success which in our question is percentage

of defectivs, i.e. 5%

(a) P(X = 3) = $\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}$

= $2300 \times 0.05^{3}\times 0.95^{22}$

= 0.093

(b) P(X $\leq$ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= $\binom{25}{0} \times 0.05^{0}\times (1-0.05)^{25-0}+\binom{25}{1} \times 0.05^{1}\times (1-0.05)^{25-1}+\binom{25}{2} \times 0.05^{2}\times (1-0.05)^{25-2}+\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}$

= $1 \times 1 \times 0.95^{25}+25 \times 0.05^{1}\times 0.95^{24}+300 \times 0.05^{2}\times 0.95^{23}+2300 \times 0.05^{3}\times 0.95^{22}$

= 0.966

(c) P(X $\geq$ 4) = 1 - P(X < 4) = 1 - P(X $\leq$ 3)

= 1 - 0.966

= 0.034

(d) P(1 ≤ X ≤ 3) = P(X = 1) + P(X = 2) + P(X = 3)

= $\binom{25}{1} \times 0.05^{1}\times (1-0.05)^{25-1}+\binom{25}{2} \times 0.05^{2}\times (1-0.05)^{25-2}+\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}$