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2 years ago

Graph the function f(x) = -3/5x+8.

Mathematics

1 answer:

emmasim [6.3K]2 years ago

6 0

Answer:

The y-intercept is at (0,8), so you can graph that point. Then the slope is -3/5. So the change in y coordinate divide the change in x coordinate is always -3/5. Thus 8-y/0-x=-3/5. You can choose any value for y and x as long as 8-y=-3. And 0-x=5. The point on the graph will just be (x,y) for the values you find.

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Terry has $56 dollars but earlier spent $18 at Target. How much money did Terry have before he went to Target?

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Answer:

$74.00

Step-by-step explanation:

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3 years ago

Read 2 more answers

PLEASEE HELP!!

Alex777 [14]

Answer:

Step-by-step explanation:

<h3>to understand this</h3><h3>you need to know about:</h3>

quadratic equation
quadratic equation word problems
solving quadratic

<h3>given:</h3>

h(t) = -4t² + 12t + 100

<h3>to solve:</h3>

<h3>tips and formulas:</h3>

<u>the</u><u> </u><u>Ball</u><u> </u><u>will</u><u> </u><u>hit </u><u>the</u><u> ground</u><u> </u><u>when</u><u> </u><u>the</u><u> height</u><u> is</u><u> </u><u>0</u>
<u>solving</u><u> </u><u>quadratics </u><u>using</u><u> </u><u>quadratic</u><u> formula</u>
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<h3>let's solve: </h3>

$step - 1 : \: define$

$- 4 {t}^{2} + 12t + 100 = 0$

$step - 2 : \\ divide \: both \: sides \: by \: - 4$

${t}^{2} - 3t - 25 = 0$

$step - 3 : \\ solve \: the \: quadratic$

$formula : \\ x = \frac{ - b± \sqrt{ {b}^{2} - 4ac } }{2a}$

$define \: a ,b \: and \: c \\ which \: are \: 1, - 3 \: and \: - 25 \: respectively$

$t = \frac{ - ( - 3)± \sqrt{ {( - 3)}^{2} - 4.1. - 25 } }{2.1}$

$t = \frac{ 3± \sqrt{ 9 + 100} }{2}$

$t = \frac{3 + \sqrt{ 109 } }{2}$

$t = \frac{3 - \sqrt{ 109 } }{2}$

$t = 6.7 \\ t = - 3.7$

$as \: we \: know \: time \: cannot \: be \: negative$

$\huge \therefore \: t = 6.7$

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3 years ago

Suppose the population of a certain city is 5893 thousand. It is expected to decrease to 5218 thousand in 50 years. Find the per

leonid [27]

Answer:

% Decrease = 11.45%

Step-by-step explanation:

Original Population = 5893

Expected Population = 5218

% Decrease = Original Population - Expected Population / Original population

% Decrease = (5893 - 5218 ) / 5893

% Decrease = 675/5893

% Decrease = 0.11454268

% Decrease = 11.454268%

% Decrease = 11.45% approximately.

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3 years ago

Assuming the incident of fires for individual reactors can be described by a Poisson distribution, what is the probability that

Maksim231197 [3]

Complete Question

The Brown's Ferry incident of 1975 focused national attention on the ever-present danger of fires breaking out in nuclear power plants. The Nuclear Regulatory Commission has estimated that with present technology there will be on average, one fire for every 10 years for a reactor. Suppose that a certain state has two reactors on line in 2020 and they behave independently of one another. Assuming the incident of fires for individual reactors can be described by a Poisson distribution, what is the probability that by 2030 at least two fires will have occurred at these reactors?

Answer:

The value is $P(x_1 + x_2 \ge 2 )= 0.5940$

Step-by-step explanation:

From the question we are told that

The rate at which fire breaks out every 10 years is $\lambda = 1$

Generally the probability distribution function for Poisson distribution is mathematically represented as

$P(x) = \frac{\lambda^x}{ k! } * e^{-\lambda}$

Here x represent the number of state which is 2 i.e $x_1 \ \ and \ \ x_2$

Generally the probability that by 2030 at least two fires will have occurred at these reactors is mathematically represented as

$P(x_1 + x_2 \ge 2 ) = 1 - P(x_1 + x_2 \le 1 )$

=> $P(x_1 + x_2 \ge 2 ) = 1 - [P(x_1 + x_2 = 0 ) + P( x_1 + x_2 = 1 )]$

=> $P(x_1 + x_2 \ge 2 ) = 1 - [ P(x_1 = 0 , x_2 = 0 ) + P( x_1 = 0 , x_2 = 1 ) + P(x_1 , x_2 = 0)]$

=> $P(x_1 + x_2 \ge 2 ) = 1 - P(x_1 = 0)P(x_2 = 0 ) + P( x_1 = 0 ) P( x_2 = 1 )+ P(x_1 = 1 )P(x_2 = 0)$

=> $P(x_1 + x_2 \ge 2 ) = 1 - \{ [ \frac{1^0}{ 0! } * e^{-1}] * [[ \frac{1^0}{ 0! } * e^{-1}]] )+ ( [ \frac{1^1}{1! } * e^{-1}] * [[ \frac{1^1}{ 1! } * e^{-1}]] ) + ( [ \frac{1^1}{ 1! } * e^{-1}] * [[ \frac{1^0}{ 0! } * e^{-1}]]) \}$

=> $P(x_1 + x_2 \ge 2 )= 1- [[0.3678 * 0.3679] + [0.3678 * 0.3679] + [0.3678 * 0.3679] ]$

$P(x_1 + x_2 \ge 2 )= 0.5940$

3 0

2 years ago

Solve system by subsubstitution?<br> 14x-2y=46<br> -7x+y=-23

e-lub [12.9K]

$14x-2y=46 \\
-7x+y=-23 \\ \\
\hbox{solve the 2nd equation for y:} \\
-7x+y=-23 \\
y=7x-23 \\ \\
\hbox{substitute 7x-23 for y in the 1st equation:} \\
14x-2(7x-23)=46 \\
14x-14x+46=46 \\
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\boxed{\hbox{infinitely many solutions}}$

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3 years ago