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3 years ago

Use the data table provided to calculated the values requested below. Provide all answers to three decimal places.

Mathematics

1 answer:

babymother [125]3 years ago

5 0

Answer:

1) 0.700

2) 0.730

3) 0.030

4) 0.959

Step-by-step explanation:

1) proportion of support for the ban with at least one child =

$\frac{no of support atleast 1 child}{Total no of atleast 1 child\\}$

= $\frac{1739}{2485}$

= 0.700

2) proportion of support for the ban with no child =

= $\frac{no of support with no child}{Total of no child}$

= $\frac{3089}{4231}$

= 0.730

3) Difference in proportion of supporters for the ban between those with atleast one child and those with no child

= 0.700 - 0.730

= -0.03

4) Relative risk = $\frac{proportion with atleast on child}{proportion with no child}$

= $\frac{0.700}{0.730}$ = 0.959

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If c= the number of cars in a parking lot, which algebraic expression

DIA [1.3K]

Answer:

Option C) c+9 is correct

The sum of the number of cars and 9=c+9 is correct algebraic expression representation for given phrase

Step-by-step explanation:

Given that c=the number of cars in a parking lot.

The phrase given is "the sum of the number of cars and 9"

To find the algebraic expression which represents the given phrase :

The algebraic expression to the given phrase is c+9

That is option C) c+9 is correct

The sum of the number of cars and 9=c+9 is correct algebraic expression representation for given phrase

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3 years ago

Suppose x is a real number and epsilon > 0. Prove that (x - epsilon, x epsilon) is a neighborhood of each of its members; in

kaheart [24]

Answer:

See proof below

Step-by-step explanation:

We will use properties of inequalities during the proof.

Let $y\in (x-\epsilon,x+\epsilon)$ . then we have that $x-\epsilon$ . Hence, it makes sense to define the positive number delta as $\delta=\min\{x+\epsilon-y,y-(x-\epsilon)\}$ (the inequality guarantees that these numbers are positive).

Intuitively, delta is the shortest distance from y to the endpoints of the interval. Now, we claim that $(y-\delta,y+\delta)\subseteq (x-\epsilon,x+\epsilon)$ , and if we prove this, we are done. To prove it, let $z\in (y-\delta,y+\delta)$ , then $y-\delta$ . First, $\delta \leq y-(x-\epsilon)$ then $-\delta \geq -y+x-\epsilon$ hence $z>y-\delta \geq x-\epsilon$

On the other hand, $\delta \leq x+\epsilon-y$ then $z$ hence $z$ . Combining the inequalities, we have that $x-\epsilon$ , therefore $(y-\delta,y+\delta)\subseteq (x-\epsilon,x+\epsilon)$ as required.