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adell [148]
3 years ago
14

Factorise k^2 + 5k + 6​

Mathematics
1 answer:
fenix001 [56]3 years ago
5 0

Answer: (k+2) (k+3)

Step-by-step explanation:

Factor 5^2 +5k+6 using the AC method.

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PLEASE HELP ASAP <br> I need help finding the rate of change
Talja [164]
G(6)-g(2)/6-2
g(6) = 50
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50-14/6-2
36/4
=9
6 0
3 years ago
According to the American Lung Association, 7% of the population has lung disease. Of those having lung disease, 90% are smokers
postnew [5]

Answer:

Probability that a smoker has lung disease = 0.2132

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')

       = 0.90 * 0.07 + 0.25 * 0.93 = 0.2955

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

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

7 0
3 years ago
If x = -1, x cubed + x
forsale [732]
(-1)(-1)(-1)+(-1)=-1-1=-2

-2

Hope this helps!
6 0
3 years ago
Read 2 more answers
Assume that f is continuous on [-4,4] and differentiable on (-4,4). The table gives some values of f'(x) x: -4, -3, -2, -1, 0, 1
kondaur [170]
f will be increasing on the intervals where f'(x)>0 and decreasing wherever f'(x). Local extrema occur when f'(x)=0 and the sign of f'(x) changes to either side of that point.

f'(x) is positive when x is between -4 and some number between -2 and -1, and also 2 (exclusive) and 4, so you can estimate that f(x) is increasing on the intervals [-4, -2] and (2, 4].

f'(x) is negative when x is between some number between -2 and -1, up to some number less than 2. So f(x) is decreasing on the interval [-1, 1].

You then have two possible cases for extrema occurring. The sign of f'(x) changes for some x between -2 and -1, and again to either side of x=2.
4 0
3 years ago
Solv the following inequality.<br> p – 4 &lt; -11
bogdanovich [222]

Answer:

p < -7

Step-by-step explanation:

.................

6 0
3 years ago
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