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2 years ago

For which value of a are all the solutions of 2(x-5)>3x+a less than or equal 5?

1 answer:

6 0

$2(x-5)>3x+a \\
2x-10 > 3x+a \\
2x-3x>a+10 \\
-x>a+10 \ \ \ |\times (-1) \\
x$

The solutions must be less than or equal to 5.
$-a-10 \leq 5 \\
-a \leq 5+10 \\
-a \leq 15 \ \ \ |\times (-1) \\
a \geq -15$

All the solutions are less than or equal to 5 for a≥-15.

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Y = -5x + 15 2x + y = 12

Nana76 [90]

Answer:

x = 1, y = 10

Step-by-step explanation:

y = -5x + 15 --- Equation 1

2x + y = 12 --- Equation 2

Substitute y = -5x + 15 into Equation 2:

2x + y = 12

2x - 5x + 15 = 12

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15 - 3x = 12

Isolate -3x.

-3x = 12 - 15

Evaluate like terms.

-3x = -3

Find x.

x = -3 ÷ -3

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2x + y = 12

2(1) + y = 12

2 + y = 12

Isolate y.

y = 12 - 2

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3 years ago

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There are 2,000 eligible voters in a precinct. A total of 500 voters are randomly selected and asked whether they plan to vote f

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Answer:

$0.7 - 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.647$

$0.7 + 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.753$

And the 99% confidence interval would be given (0.647;0.753).

So the correct answer would be:

a. 0.647 and 0.753

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

The estimated population proportion for this case is:

$\hat p = \frac{350}{500}=0.7$

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 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.