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charle [14.2K]
3 years ago
13

The number of taxi accidents is approximated by y=−143.3x2+1823.3x+6820 and the number of injuries is approximated by y=−416.72x

^2+4416.7x+8500. Let x=0 represents​ 1990, x=1 represents​ 1991, and so on. According to the equation for the number of​ accidents, if the trend​ continued, how many taxi accidents occurred in 1998​?
Mathematics
1 answer:
antiseptic1488 [7]3 years ago
7 0

To get for the taxi accidents in 1998, we must note that x = 8. Therefore using the equation y=−143.3x2+1823.3x+6820 , we can calculate for the number of taxi accidents.

y = -143.3 * (8)^2 + 1823.3 * 8 + 6820

y = 12,235.2

There were about 12,235 taxi accidents in 1998.

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Elements are of the form

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Step-by-step explanation:

Given intervals are,

(i) [a,a] (ii) [a,a) (iii) (a,a] (iv) (a,a) (v) (a,b) where a>b (vi)  [a,b] where a>b.

To show all its elements,

(i) [a,a]

Imply the set including aa from left as well as right side.

Its elements are of the form.

\{[a,a] : a\in \mathbb R\}=\{[0,0],[1, 1],[-1,-1],[2,2],[-2,-2],[3,3],[-3,-3],........\}

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(ii) [a,a)

This means given interval containing a by left and exclude a by right.

Its elements are of the form.

[ 1, 1),[-1,-1),[2,2),[-2,-2),[3,3),[-3,-3),........

Since there is a singleton element a of real numbers withis the set, this set is empty.

Because there is no increment so if a\in \mathbb R then the set  [a,a) represents singleton sets, and singleton sets are empty so is [a.a).

(iii) (a,a]

It means the interval not taking a by left and include a by right.

Its elements are of the form.

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Since there is a singleton element a of real numbers, this set is empty.

Because there is no increment so if a\in \mathbb R then the set  (a,a] represents singleton sets, and singleton sets are empty so is (a,a].

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Since there is a singleton element a of real numbers, this set is empty.

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That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,

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(vi) [a,b] where a\leq b.

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That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,

[a,b]=\{[5,0],[5,1],[5,2].....\} e.t.c

So this set is connected and we know singletons are connected in \mathbb R. Hence given set is empty.

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