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Elden [556K]
3 years ago
11

How can you use two-step equations and inequalities to represent and solve real-world problems?

Mathematics
1 answer:
Mashcka [7]3 years ago
7 0

You can use two-step equations to represent and solve real-world problems by translating the words into an algebraic equation, solving the equation, and then interpreting the solution to the equation.

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Line Q??? You don't really have an image of it
5 0
3 years ago
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A student is getting ready to take an important oral examination and is concerned about the possibility of having an "on" day or
Tamiku [17]

Answer:

The students should request an examination with 5 examiners.

Step-by-step explanation:

Let <em>X</em> denote the event that the student has an “on” day, and let <em>Y</em> denote the

denote the event that he passes the examination. Then,

P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})

The events (Y|X) follows a Binomial distribution with probability of success 0.80 and the events (Y|X^{c}) follows a Binomial distribution with probability of success 0.40.

It is provided that the student believes that he is twice as likely to have an off day as he is to have an on day. Then,

P(X)=2\cdot P(X^{c})

Then,

P(X)+P(X^{c})=1

⇒

2P(X^{c})+P(X^{c})=1\\\\3P(X^{c})=1\\\\P(X^{c})=\frac{1}{3}

Then,

P(X)=1-P(X^{c})\\=1-\frac{1}{3}\\=\frac{2}{3}

Compute the probability that the students passes if request an examination with 3 examiners as follows:

P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})

        =[\sum\limits^{3}_{x=2}{{3\choose x}(0.80)^{x}(1-0.80)^{3-x}}]\times\frac{2}{3}+[\sum\limits^{3}_{x=2}{{3\choose x}(0.40)^{3}(1-0.40)^{3-x}}]\times\frac{1}{3}

       =0.715

The probability that the students passes if request an examination with 3 examiners is 0.715.

Compute the probability that the students passes if request an examination with 5 examiners as follows:

P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})

        =[\sum\limits^{5}_{x=3}{{5\choose x}(0.80)^{x}(1-0.80)^{5-x}}]\times\frac{2}{3}+[\sum\limits^{5}_{x=3}{{5\choose x}(0.40)^{x}(1-0.40)^{5-x}}]\times\frac{1}{3}

       =0.734

The probability that the students passes if request an examination with 5 examiners is 0.734.

As the probability of passing is more in case of 5 examiners, the students should request an examination with 5 examiners.

8 0
3 years ago
-7p (p+8)=21 is equal to
andrew11 [14]

We have the equation:

-7p(p+8)=21

So then we need to distribute -7p to the parentheses:

-7p^{2}+(-56p)=21

Then we need to set the equation equal to 0, so we must subtract 21 on both sides:

-7p^{2}-56p-21=0

Then we need to factor this. So we end up with:

7(-x^{2}-8p-3)=0

This is as much as this can be factored, so <u>we cannot go any further</u>.

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Write an equation in slope-intercept form for the line that passes through (-3, 5), and is perpendicular to the graph of 5x - 6y
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Y =-6/5x+7/5 Please vote me as brainliest if you think i helped and i attached a photo with work as well :) Have a great day

4 0
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giving brainliest to whoever explains best. don't answer if you don't speak english well. don't answer if you're 9 years old. do
Marta_Voda [28]

Answer:

Gillmans

Step-by-step explanation:

I feel like gillmans class did better cuz their class didnt have anyone get a score less than 60 but mr athens class had people who got below 50. Although athens class has people over 95 the most were 60-90 while gillmans was 70-90

8 0
3 years ago
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