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Virty [35]
3 years ago
12

The range of the function y=secx-2 is all reals except -1 1 -3 -2

Mathematics
1 answer:
Umnica [9.8K]3 years ago
6 0

Answer:

For a function y = f(x), the range is the set of all the possible values of y.

In the question you wrote:

y = secx - 2

This can be interpreted as:

y = sec(x - 2)

or

y = sec(x) - 2

So let's see each case (these are kinda the same)

If the function is:

y = sec(x - 2)

Firs remember that:

sec(x) = 1/cos(x)

then we can rewrite:

y = 1/cos(x - 2)

notice that the function cos(x) has the range -1 ≤ y ≤ 1

Then for the two extremes we have:

y = 1/1 = 1

y = 1/-1 = -1

Notice that for:

y = 1/cos(x - 2)

y can never be in the range  -1 < x < 1

As the denominator cant be larger, in absolute value, than 1.

Then we can conclude that the range is all reals except the interval:

-1 < y < 1

If instead the function was:

y = sec(x) - 2

y = 1/cos(x) - 2

Then with the same reasoning, the range will be the set of all real values except:

-1 - 2 < y < 1 - 2

-3 < y < -1

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Answer:

9x + 25 + 3x = 34 + 54

9x + 3x = 12x

34 + 54 = 88

12x + 25 = 88

      - 25    - 25

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<em><u>x = 5.25</u></em>

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3 years ago
A sample of 11 students using a Ti-89 calculator averaged 31.5 items identified, with a standard deviation of 8.35. A sample of
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Answer:

We reject H₀, with CI = 90 % we can conclude that the mean numbers of times identified with Ti-84 does not exceed that of the Ti-89 by more than 1,80

Step-by-step explanation:

Ti-89 Calculator

Sample mean     x = 31,5

Sample standard deviation   s₂  = 8,35

Sample size        n₂ = 11

Ti-84 Calculator

Sample mean     y = 46,2

Sample standard deviation   s₁  = 9,99

Sample size        n₁ = 12

t(s)  = ( y - x - d ) / √s₁²/n₁ + s₂²/n₂

t(s)  = 12,9 / √(99,8/12) + (69,72/11)

t(s)  = 12,9 / √8,32 + 6,34

t(s)  = 12,9 / 3,83

t(s) = 3,37

Test Hypothesis

Null Hypothesis               H₀      y - x > 1,80

Altenative Hypothesis       Hₐ      y - x ≤ 1,80

We have a t(s) = 3,37

We need to compare with t(c)   critical value for

t(c) α; n₁ +n₂-2            df = 12 +11 -2     df  = 21

If we choose  CI = 95 %    then  α = 5 %   α = 0,05

From t-student table

t(c) = 1,72

t(s) = 3,37

t(s) >t(c)

t(s) is in the rejection region therefore we accept Hₐ with CI = 95 %

8 0
3 years ago
6 x 20 = 6 x ___ tens= ___tens = ____
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6x20=6x2 tens=12tens=120
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3 years ago
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