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3 years ago

Solve the equation. 12x – 15 = 3(2x +3)

Mathematics

1 answer:

MA_775_DIABLO [31]3 years ago

8 0

Answer:

x = 4

Step-by-step explanation:

Solve for x:

12 x - 15 = 3 (2 x + 3)

Hint: | Write the linear polynomial on the left hand side in standard form.

Expand out terms of the right hand side:

12 x - 15 = 6 x + 9

Hint: | Move terms with x to the left hand side.

Subtract 6 x from both sides:

(12 x - 6 x) - 15 = (6 x - 6 x) + 9

Hint: | Combine like terms in 12 x - 6 x.

12 x - 6 x = 6 x:

6 x - 15 = (6 x - 6 x) + 9

Hint: | Look for the difference of two identical terms.

6 x - 6 x = 0:

6 x - 15 = 9

Hint: | Isolate terms with x to the left hand side.

Add 15 to both sides:

6 x + (15 - 15) = 15 + 9

Hint: | Look for the difference of two identical terms.

15 - 15 = 0:

6 x = 9 + 15

Hint: | Evaluate 9 + 15.

9 + 15 = 24:

6 x = 24

Hint: | Divide both sides by a constant to simplify the equation.

Divide both sides of 6 x = 24 by 6:

(6 x)/6 = 24/6

Hint: | Any nonzero number divided by itself is one.

6/6 = 1:

x = 24/6

Hint: | Reduce 24/6 to lowest terms. Start by finding the GCD of 24 and 6.

The gcd of 24 and 6 is 6, so 24/6 = (6×4)/(6×1) = 6/6×4 = 4:

Answer: x = 4

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Step-by-step explanation:

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3 years ago

Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1

alexandr1967 [171]

Answer:

$a) \simeq 0.3012$ $b) \simeq 0.0494$ c) $\simeq 0.2438$

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, $\frac{1.2}{4}$

= 0.3 collisions per month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

P(X =x) = $\frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}
$

for x ∈ N ∪ {0}

= 0 otherwise --------------------------------------(1)

here, $\lambda = 0.3$ collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

P(X = 0) = $\frac{e^{-0.3}\times {\0.3}^{0}}{0!}
$

$\simeq 0.7408182207$ ------------------------------------(2)

Now, since we are calculating this for 4 months,

so, P(No collision in 4 month period)

= $0.7408182207^{4}$

$\simeq 0.3012$ -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

P(X =1) = $\frac{e^{-0.3}\times {\0.3}^{1}}{1!}
$

$\simeq 0.2222454662$ ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

= $0.2222454662^{2}$

$\simeq 0.0494$ -----------------------------------------(5)

1 collision in 6 months period means

$\frac{1}{6}$ collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6] (which is to be estimated)

= $\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}$

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

$\simeq 0.6543894283$ -------------------------------------------(6)

So,

P(1 collision in 6 month period)

= $0.6543894283^{6}$

$\simeq 0.0785267444$ ------------------------------------------------(7)

So,

P(No collision in 6 months period)

= $(P(X =0)^{6}$

$\simeq 0.1652988882$ ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

$\simeq 0.2438$ ---------------------------------------------(9)

7 0

3 years ago

Find the coordinates of the midpoint of the segment whose endpoints are R(9, 3) and S(-1, -9).

amm1812

Answer:

$(4, -3 )$

General Formulas and Concepts:

Pre-Algebra I

Order of Operations: BPEMDAS

Algebra I

Midpoint Formula: $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} )$

Step-by-step explanation:

Step 1: Define

R (9, 3)

S (-1, -9)

Step 2: Find midpoint

Substitute: $(\frac{9-1}{2}, \frac{3-9}{2} )$
Subtract: $(\frac{8}{2}, \frac{-6}{2} )$
Divide: $(4, -3 )$

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Millie had $450 in a savings account. She spent $171 on new luggage. What percent of Millie's money is left in her savings accou

Rudiy27

Amount left = $450 - $171 = $279

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Answer: (C) 62%

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