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Orlov [11]
4 years ago
13

How to solve ×+2y=11. 2×+3y=50

Mathematics
1 answer:
9966 [12]4 years ago
5 0

Answer:

x=−2y+11. x=

−3/2y+25

Step-by-step explanation:

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3 years ago
The U.S. Energy Information Administration (US EIA) reported that the average price for a gallon of regular gasoline is $2.94. T
Anit [1.1K]

Answer:

a) 25

b) 67

c) 97

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.025 = 0.975, so z = 1.96

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample. In this problem, \sigma = 0.25

(a) The desired margin of error is $0.10.

This is n when M = 0.1. So

M = z*\frac{\sigma}{\sqrt{n}}

0.1 = 1.96*\frac{0.25}{\sqrt{n}}

0.1\sqrt{n} = 1.96*0.25

\sqrt{n} = \frac{19.6*0.25}{0.1}

(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.1})^{2}

n = 24.01

Rounding up to the nearest whole number, 25.

(b) The desired margin of error is $0.06.

This is n when M = 0.06. So

M = z*\frac{\sigma}{\sqrt{n}}

0.06 = 1.96*\frac{0.25}{\sqrt{n}}

0.06\sqrt{n} = 1.96*0.25

\sqrt{n} = \frac{19.6*0.25}{0.06}

(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.06})^{2}

n = 66.7

Rounding up, 67

(c) The desired margin of error is $0.05.

This is n when M = 0.05. So

M = z*\frac{\sigma}{\sqrt{n}}

0.05 = 1.96*\frac{0.25}{\sqrt{n}}

0.05\sqrt{n} = 1.96*0.25

\sqrt{n} = \frac{19.6*0.25}{0.05}

(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.05})^{2}

n = 96.04

Rounding up, 97

8 0
3 years ago
Would someone be able to answer this before 3 PM central?
Margaret [11]

Answer:

yes

Step-by-step explanation:

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The equation of line m is 3x−5y=−4.
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Paul [167]

Answer:

Step-by-step explanation:

You haven't went into the assignment enough to say. Its asking for the date you spoke to the teacher and for you to submit that with the assignment.

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