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romanna [79]
3 years ago
9

3. x + 4y = 30 x - 2y = 0

Mathematics
1 answer:
kolbaska11 [484]3 years ago
5 0

Answer:

Step-by-step explanation:

Given that:

x + 4y = 30

x - 2y = 0

Subtracting both equations:

6y = 30

y = 30/6

y = 5

Putting in above equation

x + 4(5) = 30

x + 20 = 30

x = 10

i hope it will help you

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Find the area of this figure.
zheka24 [161]

Answer:

98

Step-by-step explanation:

11(4)+9(6)

44+54

98

8 0
3 years ago
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Write a linear equation given the point and slope or two points.
attashe74 [19]

Answer:

1) y = -2x - 1

2) y = -3/4 + 3

3) y = 4x + 9

4) y = - 5/3x - 2

Step-by-step explanation:

b = y - m*x

1) (-7,13) and slope: -2

b = 13 - (-2)*(-7)

b = 13 - 14

b = -1

2) (4,6) lope = -3/4

b = 6 - (-3/4)*(-4)

b = 6 - 3

b = 3

y = -3/4x + 3

3) (-5,-11) and (3,-7)

Slope: (1 - - 11)/(-2 - - 5) = 12/3

= 4

b = -11 - (4) (-5) = 9

b = 9

4) slope = (-7 - 8)/(3- - 6)

= -15/9 = - 5/3

b = 8 - (-5/3)*(-6)

b = 8 - 10

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5 0
3 years ago
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.

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Find the Area of the shaded region of the circle. Leave answers in terms of Pi. (Image attached). Thank you!!!
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Answer:

I've attached the Answer

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Drag the amounts to order them from greatest to least.
Debora [2.8K]
For easier comparison, convert all amounts to grams:

30 oz * 28.3 grams = 849 grams

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Comparing the three quantities, the highest is 30 oz (849 grams); next is 800 grams, then 1.6 lb (724.48 grams)
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