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3 years ago

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The high school, middle school, and elementary school are all on the same road. The distance from the high school to the middle

school can be represented by the expression 8.7x2+11.9x+3.5 . The distance from the middle school to the elementary school can be represented by the expression 3.23x2+2.7x+0.75 . What is the total distance from the high school to the elementary school?

2 answers:

lara31 [8.8K]3 years ago

8 0

I know how to work it out. The only problem is that your expressions don't make sense because you put the multiplication sign next to the addition sign and that can't work.

Margarita [4]3 years ago

6 0

Answer:

what do you mean in this question plus what are you asking I don't get it at all.

Step-by-step explanation:

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Answer:

First, plot the y-intercept. The y-intercept is 1, so plot the point (0,1).

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Draw a line through the two points.

Step-by-step explanation:

y = mx+b

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belka [17]

Answer:

Slope: $-\frac{3}{4}$

Y-intercept: 399

Slope-intercept form: $y=-\frac{3}{4}x+399$

Step-by-step explanation:

1. The equation in Standard form is given in the problem:

$3y+4y=1596$

2. Then, to write the equation of the line in slope-intercept form $y=mx+b$ , where <em>m</em> is the slope and <em>b</em> is the y-intercept, you must solve for <em>y</em> as you can see below:

$4y=-3x+1596\\y=\frac{-3x+1596}{4}\\y=-\frac{3}{4}x+399$

3. Therefore, the slope and the y-intercept are:

$m=-\frac{3}{4}\\b=399$

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3 years ago

There are three power plants [X, Y, Z] that at any given time each one either generates electricity or idles. Event A is that pl

insens350 [35]

We're told that

$P(A\cap B)=0.15$

$P(A\cup B)^C=0.06\implies P(A\cup B)=0.94$

$P(B\mid A)=P(B^C\mid A)=0.5$

where the last fact is due to the law of total probability:

$P(A)=P(A\cap B)+P(A\cap B^C)$

$\implies P(A)=P(B\mid A)P(A)+P(B^C\mid A)P(A)$

$\implies 1=P(B\mid A)+P(B^C\mid A)$

so that $B\mid A$ and $B^C\mid A$ are complementary.

By definition of conditional probability, we have

$P(B\mid A)=P(B^C\mid A)$

$\implies\dfrac{P(A\cap B)}{P(A)}=\dfrac{P(A\cap B^C)}{P(A)}$

$\implies P(A\cap B)=P(A\cap B^C)$

We make use of the addition rule and complementary probabilities to rewrite this as

$P(A\cap B)=P(A\cap B^C)$

$\implies P(A)+P(B)-P(A\cup B)=P(A)+P(B^C)-P(A\cup B^C)$

$\implies P(B)-[1-P(A\cup B)^C]=[1-P(B)]-P(A\cup B^C)$

$\implies2P(B)=2-[P(A\cup B)^C+P(A\cup B^C)]$

$\implies2P(B)=[1-P(A\cup B)^C]+[1-P(A\cup B^C)]$

$\implies2P(B)=P(A\cup B)+P(A\cup B^C)^C$

$\implies2P(B)=P(A\cup B)+P(A^C\cap B)\quad(*)$

By the law of total probability,

$P(B)=P(A\cap B)+P(A^C\cap B)$

$\implies P(A^C\cap B)=P(B)-P(A\cap B)$

and substituting this into $(*)$ gives

$2P(B)=P(A\cup B)+[P(B)-P(A\cap B)]$

$\implies P(B)=P(A\cup B)-P(A\cap B)$

$\implies P(B)=0.94-0.15=\boxed{0.79}$

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3 years ago