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Lunna [17]
3 years ago
16

Which function is negative for the interval [–1, 1]? On a coordinate plane, a curved line with a minimum value of (negative 1, n

egative 2), crosses the x-axis at (negative 2.5, 0) and (0.5, 0), and the y-axis at (0, negative 1). On a coordinate plane, a curved line with a minimum value of (0, negative 2), crosses the x-axis at (negative 1.5, 0) and (1.5, 0). On a coordinate plane, a curved line with a maximum value of (0, 2) and a minimum value of (1.5, negative 2.75), crosses the x-axis at (negative 0.6, 0) and (0.6, 0). On a coordinate plane, a curved line with a maximum value of (negative 1.25, 4) and a minimum value of (1.5, negative 3.5), crosses the x-axis at (negative 1.75, 0), (0.75, 0), and (1.75, 0), and crosses the y-axis at (0, 2).
Mathematics
2 answers:
timofeeve [1]3 years ago
9 0

Answer:it’s b on edg

Step-by-step explanation:

Doss [256]3 years ago
7 0

Answer:

A. On a coordinate plane, a curved line with a minimum value of (negative 1, negative 2), crosses the x-axis at (negative 2.5, 0) and (0.5, 0), and the y-axis at (0, negative 1).

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

A. On a coordinate plane, a curved line with a minimum value of (negative 1, negative 2), crosses the x-axis at (negative 2.5, 0) and (0.5, 0), and the y-axis at (0, negative 1).

B. On a coordinate plane, a curved line with a minimum value of (0, negative 2), crosses the x-axis at (negative 1.5, 0) and (1.5, 0).

C.  On a coordinate plane, a curved line with a maximum value of (0, 2) and a minimum value of (1.5, negative 2.75), crosses the x-axis at (negative 0.6, 0) and (0.6, 0).

D. On a coordinate plane, a curved line with a maximum value of (negative 1.25, 4) and a minimum value of (1.5, negative 3.5), crosses the x-axis at (negative 1.75, 0), (0.75, 0), and (1.75, 0), and crosses the y-axis at (0, 2).

Apologies if I am wrong, I am not great with graphs but I hoped I helped.

<3 Enjoy,

    Dea

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According to the American Lung Association, 7% of the population has lung disease. Of those having lung disease, 90% are smokers
postnew [5]

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Step-by-step explanation:

Let L = event that % of population having lung disease, P(L) = 0.07

So,% of population not having lung disease, P(L') = 1 - P(L) = 1 - 0.07 = 0.93

S = event that person is smoker

% of population that are smokers given they are having lung disease, P(S/L) = 0.90

% of population that are smokers given they are not having lung disease, P(S/L') = 0.25

We know that, conditional probability formula is given by;

                        P(S/L) = \frac{P(S\bigcap L)}{P(L)}  

                        P(S\bigcap L) = P(S/L) * P(L)

                                      = 0.90 * 0.07 = 0.063

So,  P(S\bigcap L) = 0.063 .

Now, probability that a smoker has lung disease is given by = P(L/S)

      P(L/S) = \frac{P(S\bigcap L)}{P(S)}

P(S) = P(S/L) * P(L) + P(S/L') * P(L')

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Therefore, P(L/S) = \frac{0.063}{0.2955} = 0.2132

Hence, probability that a smoker has lung disease is 0.2132 .

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Answer:

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Step-by-step explanation:

In the given structure ΔKLM

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From right angle triangle KLM

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KM = 14×sin30° = 14×1/2 = 7

So the answer is KM = 7 Units

5 0
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