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3 years ago

If 3x-5=-11, find x+5

Mathematics

1 answer:

Misha Larkins [42]3 years ago

6 0

3x - 5 = -11

isolate x:

3x = -11 + 5

3x = - 6

divide by coefficient:

x = -6/3

x = -2

plug in x:

-2 + 5 = 3

Your answer is 3

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Opposite reciprocal of -9/1 and 11/-7

PtichkaEL [24]

Answer:

1/9 and 7/11

Step-by-step explanation:

hope this is helpful :)

7 0

2 years ago

the partial factorization of 2 x - 3 x - 10 is modeled with algebra tiles which unit tiles are needed to complete the factorizat

nikklg [1K]

The partial factorization of 2 x - 3 x - 10 is modeled with algebra tiles is that you will need 2 x tiles and 4 negative 1 tiles and 3 positive 1 tiles.

3 0

3 years ago

Its middle school math i need help

skad [1K]

Answer:

x = 3

Step-by-step explanation:

4x+2 = x+11

Subtract x from each side

4x+2-x = x+11-x

3x+2 =11

Subtract 2 from each side

3x+2-2 =11-2

3x = 9

Divide by 3

3x/3 =9/3

x = 3

5 0

3 years ago

Read 2 more answers

A ball is dropped from the top of the stairs and reaches a velocity of 25 m/s, in 4 seconds. What is the acceleration of the bal

kodGreya [7K]

The acceleration of the ball would be 6.25

3 0

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