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2 years ago

X +2y =6 2x - 3y =26 solve each system. Tell how many solutions each system has

Mathematics

1 answer:

valina [46]2 years ago

4 0

(10, -2)
X+2y=6
2x-3y=26

X=6-2y
2x-3y=26
Solve the y
2(6-2y)-3y=26
Y= -2
X= 6-2•(-2)
Solve the x
X=10

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3 years ago

The cost of a pen at the office supply store is $2.65. the school store sells the pen for $3.45. How much does the school store

denis23 [38]

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30.1%

Step-by-step explanation:

The increase in price (markup) is ($3.45 - $2.65) = $0.80.

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2 years ago

Suppose that the mean time that visitors stay at a museum is 94.2 minutes with a standard deviation of 15.5 minutes. The standar

Maurinko [17]

Answer:

$94.2 -0.994*3.1 = 91.1186$

$94.2 +0.994*3.1 = 97.2814$

And the 68% confidence interval is given by (91.1186, 97.2814)

Step-by-step explanation:

For this case we know that mean time that visitors stay at a museum is given by:

$\bar X = 94.2$

The standard deviation is given by:

$s= 15.5$

And the standard error is given by:

$SE = \frac{s}{\sqrt{n}} =3.1$

And we want to interval captures 68% of the means for random samples of 25 scores and for this case the critical value can be founded like this using the normal standard distribution or excel:

$z_{\alpha/2}= \pm 0.994$

We can find the interval like this:

$\bar X \pm ME$

And replacing we got:

$94.2 -0.994*3.1 = 91.1186$

$94.2 +0.994*3.1 = 97.2814$

And the 68% confidence interval is given by (91.1186, 97.2814)

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3 years ago

Which of these constructions is impossible using only a compass and straightedge

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Answer:

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Step-by-step explanation:

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2 years ago

For many years businesses have struggled with the rising cost of health care. But recently, the increases have slowed due to les

kaheart [24]

Answer:

The confidence interval would be given by this formula

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

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

The margin of error for this case is given by:

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

And replacing we got:

$ME = 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.0259$

And replacing into the confidence interval formula we got:

$0.52 - 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.4941$

$0.52 + 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.5459$

And the 95% confidence interval would be given (0.4941;0.5459).

Step-by-step explanation:

Data given and notation

n=1000 represent the random sample taken

$\hat p=0.52$ estimated proportion of of U.S. employers were likely to require higher employee contributions for health care coverage

$\alpha=0.05$ represent the significance level (no given, but is assumed)

Solution to the problem

The confidence interval would be given by this formula

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

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

The margin of error for this case is given by:

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

And replacing we got:

$ME = 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.0259$

And replacing into the confidence interval formula we got:

$0.52 - 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.4941$

$0.52 + 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.5459$

And the 95% confidence interval would be given (0.4941;0.5459).

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2 years ago