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3 years ago

Solve each system by substitution. y=-4x-9 y=3x+19

Mathematics

1 answer:

Alekssandra [29.7K]3 years ago

5 0

Answer:

x=-4, y=7

Step-by-step explanation:

According to the first equation, y = -4x - 9, so we can substitute y in the second equation for -4x - 9.

y = 3x + 19

-4x - 9 = 3x + 19

Add 9 to both sides

-4x = 3x + 28

Subtract 3x

-7x = 28

Divide by -7

x = -4

Plugging this into the equation, we have:

y = -4x - 9

y = -4(-4) - 9

y = 16 - 9

y = 7

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0.1 hope it helps you

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4 years ago

Read 2 more answers

Andre has a summer job selling magazine subscriptions. He earns $25 each week plus $3 for every subscription he sells. Andre hop

Strike441 [17]

Answer: 25 + 3n

Step-by-step explanation:

Hi, the answer is lacking the last part:

<em>Write an expression for the amount of money he makes this week. </em>

So, to answer this we have to write an expression:

The fixed amount that he earns per week (25) plus the product of the amount he earns per subscription (3) and the number of subscriptions sold (n) , must be equal to his weekly earnings.

Mathematically speaking:

25 + 3n

Feel free to ask for more if needed or if you did not understand something.

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3 years ago

Help I’m confused on this one

Zarrin [17]

If one foot on the model represents 70 feet on the real object, the scale is 1:70. That scale applies to all dimensions and units. In other words, you can divide any object dimension by 70 to find the corresponding model dimension in the same units.

model tail height = 392 inches / 70 = 5.6 inches

model wing span = 2537 inches / 70 ≈ 36.2 inches

model length = 2785 inches / 70 ≈ 39.8 inches

4 0

4 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

89 - 42 x 4 + 12<br> Somebody help me quick pls I’m timed

Gre4nikov [31]

89 - 168 + 12
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3 years ago