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3 years ago

Solve by the linear combination method (with or without multiplication) 2x+3y= -17, 5x+2y = -4

2 answers:

6 0

2x+3y= -17
5x+2y= -4
You must cancel either the x or y out to solve so i'm going to cancel the x.
-5(2x+3y= -17)
2(5x+2y= -4)
This simplifies to:
-15x -15y=85
15x+4y= -8
The -15x and 15x cancel out leaving you:
-15y=85
4y= -8
You add them top numbers by the bottom and get:
-11y=77
Then divide by -11 to get y by its self.
y= -7
Now that you know what y is just plug it in to either original problem. I will use the second one.
5x + 2(-7) = -4
5x -14 = -4
5x = 10
x = 2
So the answer is (2,-7) or x=2 and y=-7

lions [1.4K]3 years ago

6 0

Answer: (2,-7) or x=2 and y=-7

Step-by-step explanation: Took the test and can confirm this is the answer!

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As part of a drug trial, a survey was done to gather information about the participants' experiences with side effects. Here are

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The Venn diagram which represents the distribution of the participant in the drug trial is attached below. The Number of participants in the drug trial that has anxiety is 370

We can find the number of participants who has dizziness(D). Fatigue(F) and anxiety(A) can be calculated thus :

n(D) = n(D only) + (DnF only) + (DnA only) + (DnAnF)

271 = 36 + 86 + 23 + x

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n(DnAnF) = 126

Number who has atleast one of the three side effects can be expressed thus :

n(A only) + n(F only) + n(D only) + (DnF only) + (DnA only) + (DnAnF) + n(FnA only) = 585

36 + 23 + 62 + 86 + 126 + 93 + n(FnA) only = 585

n(FnA only) = 585 - 426 = 159

Number of participants who has anxiety can be calculated thus :

n(DnAnF) + (DnA only) + n(FnA only) + n(A only)

126 + 23 + 159 + 62 = 370 participants

Therefore, 370 of the total participants has Anxiety.

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2 years ago

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2 years ago

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Answer:

(A) The minimum sample size required achieve the margin of error of 0.04 is 601.

(B) The minimum sample size required achieve a margin of error of 0.02 is 2401.

Step-by-step explanation:

Let us assume that the percentage of support for the candidate, among voters in her district, is 50%.

(A)

The margin of error, MOE = 0.04.

The formula for margin of error is:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}$

The critical value of z for 95% confidence interval is: $z_{\alpha/2}=1.96$

Compute the minimum sample size required as follows:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.04=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.04}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=600.25\approx 601$

Thus, the minimum sample size required achieve the margin of error of 0.04 is 601.

(B)

The margin of error, MOE = 0.02.

The formula for margin of error is:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}$

The critical value of z for 95% confidence interval is: $z_{\alpha/2}=1.96$

Compute the minimum sample size required as follows:

$MOE=z_{\alpha /2}\sqrt{\frac{p(1-p)}{n}}\\0.02=1.96\times \sqrt{\frac{0.50(1-0.50)}{n}}\\(\frac{0.02}{1.96})^{2} =\frac{0.50(1-0.50)}{n}\\n=2401.00\approx 2401$

Thus, the minimum sample size required achieve a margin of error of 0.02 is 2401.

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3 years ago