Ask question

Ask question

All categories

3 years ago

12

Solve the equation for y. Then find the value of Y for each value of X.

2 answers:

Nataliya [291]3 years ago

6 0

Step-by-step explanation:

<h3>y + 7x = 9 </h3>

<h3>y = 9 - 7x , (x = - 1) </h3><h3>y = 9 - 7 (-1)</h3><h3>y= 9 - (-7)</h3><h3>y = - 16</h3>

<h3>y = 9 - 7x , (x = 0)</h3><h3>y = 9 - 7(0) </h3><h3>y = 9 - 0</h3><h3>y = 0</h3>

<h3>y= 9 - 7x , (x = 2) </h3><h3>y = 9 - 7 (2) </h3><h3>y = 9 - 14</h3><h3>y = - 5</h3>

ʕ •ᴥ•ʔ

<h3 />

Hunter-Best [27]3 years ago

4 0

Answer:

y=9-7x

If x= -1, y=16

If x= 0, y=9

If x=2, y= -5

Step-by-step explanation:

Because you are solving the equation for y, you want to have y by itself on one side. The only way to do that in this problem is by subtracting the 7x on both sides.

y+7x=9

-7x -7x ---> y=9-7x

The next part of the question says to find the value of Y for each value of X. Using the given X values, we plug them into the equation we just got in order to find Y.

y=9-7(-1) ----> y=16

y=9-7(0) ----> y=9

y=9-7(2) ----> y=-5

You might be interested in

I need help please.

makkiz [27]

Answer:

see explanation

Step-by-step explanation:

Using the rules of radicals/ exponents

$a^{m}$ × $a^{n}$ = $a^{(m+n)}$

$a^{\frac{m}{n} }$ ⇔ $\sqrt[n]{a^{m} }$

Simplifying each term

7 $\sqrt{x^{3} }$ = 7 $x^{\frac{3}{2} }$

x $\sqrt{9x}$

= x × $\sqrt{9}$ × $\sqrt{x}$

= x × 3 × $x^{\frac{1}{2} }$

= 3 × $x^{(1+\frac{1}{2}) }$

= 3 $x^{\frac{3}{2} }$

Subtracting the 2 simplified like terms, that is

7 $x^{\frac{3}{2} }$ - 3 $x^{\frac{3}{2} }$

= 4 $x^{\frac{3}{2} }$ ← return to radical form

= 4 $\sqrt{x^{3} }$

8 0

2 years ago

Suppose X has an exponential distribution with mean equal to 23. Determine the following:

e-lub [12.9K]

Answer:

a) P(X > 10) = 0.6473

b) P(X > 20) = 0.4190

c) P(X < 30) = 0.7288

d) x = 68.87

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

Mean equal to 23.

This means that $m = 23, \mu = \frac{1}{23} = 0.0435$

(a) P(X >10)

$P(X > 10) = e^{-0.0435*10} = 0.6473$

So

P(X > 10) = 0.6473

(b) P(X >20)

$P(X > 20) = e^{-0.0435*20} = 0.4190$

So

P(X > 20) = 0.4190

(c) P(X <30)

$P(X \leq 30) = 1 - e^{-0.0435*30} = 0.7288$

So

P(X < 30) = 0.7288

(d) Find the value of x such that P(X > x) = 0.05

So

$P(X > x) = e^{-\mu x}$

$0.05 = e^{-0.0435x}$

$\ln{e^{-0.0435x}} = \ln{0.05}$

$-0.0435x = \ln{0.05}$

$x = -\frac{\ln{0.05}}{0.0435}$

$x = 68.87$

5 0

3 years ago

Solve for x. 1 2 x + 3 2 (x + 1) − 1 4 = 5 A) 5 2 B) 15 4 C) 15 8 D) 17 8

Hitman42 [59]

Answer: x=-13/44 or -.2954

8 0

3 years ago

Read 2 more answers

Express the product as a whole number.<br><br> 312×28=

Monica [59]

The answer would be 8736

Hope this helps :)

3 0

3 years ago

Which of the following graphs shows a linear function?

Bad White [126]

Answer:

Graph W only. But be careful it is not directly proporotional.

3 0

3 years ago

Read 2 more answers