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Radda [10]
2 years ago
14

Complete the sentence. the roots of the equation x^2 - 8x - 48 = 0 are -4 and *blank*

Mathematics
2 answers:
prisoha [69]2 years ago
8 0

Answer:

12 is the answer.

Step-by-step explanation:

The best way to get the roots of the equation is to factorize the equation but when the factors are not possible then we should apply the quadratic equations formula to get the value of x.

Now we will try to factorize

x² - 8x -48 = 0

Now we multiply coefficient of highest degree term and constant term

48×1 = 48 which may be factored in 12×4. 8×6, 24×2.

Now we will choose 12×4 out of three because the 12-4 = 8 which is the coefficient of x.

Now its simple.

x²-12x + 4x -48 = 0

x(x -12) + 4(x-12) = 0

(x +4)(x-12) = 0

Therefore roots are x + 4 = 0

x = -4

and x - 12 = 0

x = 12

So the answer is 12.

Likurg_2 [28]2 years ago
3 0

Answer:

12 is the answer.

Step-by-step explanation:

We have given an equation.

x²-8x-48 = 0

We have given one root of equation. We have to find the other root of equation.

We use method of factoring to solve this.

Splitting the middle term, we have

x²-12x+4x-48 = 0

Making groups, we have

x(x-12)+4(x-12) = 0

Taking (x-12) common, we have

(x-12)(x+4) = 0

Applying Zero-Product Property, we have

x-12 = 0 or x+4 = 0

x = 12 or x = -4.

Hence, the roots of the given equation are -4 and 12.

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Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a
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Answer:

a) P(k≤11) = 0.021

b) P(k>23) = 0.213

c) P(11≤k≤23) = 0.777

P(11<k<23) = 0.699

d) P(15<k<25)=0.687

Step-by-step explanation:

a) What is the probability that the number of drivers will be at most 11?

We have to calculate P(k≤11)

P(k\leq11)=\sum_0^{11} P(k

P(k=0) = 20^0e^{-20}/0!=1 \cdot 0.00000000206/1=0\\\\P(k=1) = 20^1e^{-20}/1!=20 \cdot 0.00000000206/1=0\\\\P(k=2) = 20^2e^{-20}/2!=400 \cdot 0.00000000206/2=0\\\\P(k=3) = 20^3e^{-20}/3!=8000 \cdot 0.00000000206/6=0\\\\P(k=4) = 20^4e^{-20}/4!=160000 \cdot 0.00000000206/24=0\\\\P(k=5) = 20^5e^{-20}/5!=3200000 \cdot 0.00000000206/120=0\\\\P(k=6) = 20^6e^{-20}/6!=64000000 \cdot 0.00000000206/720=0\\\\P(k=7) = 20^7e^{-20}/7!=1280000000 \cdot 0.00000000206/5040=0.001\\\\

P(k=8) = 20^8e^{-20}/8!=25600000000 \cdot 0.00000000206/40320=0.001\\\\P(k=9) = 20^9e^{-20}/9!=512000000000 \cdot 0.00000000206/362880=0.003\\\\P(k=10) = 20^{10}e^{-20}/10!=10240000000000 \cdot 0.00000000206/3628800=0.006\\\\P(k=11) = 20^{11}e^{-20}/11!=204800000000000 \cdot 0.00000000206/39916800=0.011\\\\

P(k\leq11)=\sum_0^{11} P(k

b) What is the probability that the number of drivers will exceed 23?

We can write this as:

P(k>23)=1-\sum_0^{23} P(k=x_i)=1-(P(k\leq11)+\sum_{12}^{23} P(k=x_i))

P(k=12) = 20^{12}e^{-20}/12!=8442485.238/479001600=0.018\\\\P(k=13) = 20^{13}e^{-20}/13!=168849704.75/6227020800=0.027\\\\P(k=14) = 20^{14}e^{-20}/14!=3376994095.003/87178291200=0.039\\\\P(k=15) = 20^{15}e^{-20}/15!=67539881900.067/1307674368000=0.052\\\\P(k=16) = 20^{16}e^{-20}/16!=1350797638001.33/20922789888000=0.065\\\\P(k=17) = 20^{17}e^{-20}/17!=27015952760026.7/355687428096000=0.076\\\\P(k=18) = 20^{18}e^{-20}/18!=540319055200533/6402373705728000=0.084\\\\

P(k=19) = 20^{19}e^{-20}/19!=10806381104010700/121645100408832000=0.089\\\\P(k=20) = 20^{20}e^{-20}/20!=216127622080213000/2432902008176640000=0.089\\\\P(k=21) = 20^{21}e^{-20}/21!=4322552441604270000/51090942171709400000=0.085\\\\P(k=22) = 20^{22}e^{-20}/22!=86451048832085300000/1.12400072777761E+21=0.077\\\\P(k=23) = 20^{23}e^{-20}/23!=1.72902097664171E+21/2.5852016738885E+22=0.067\\\\

P(k>23)=1-\sum_0^{23} P(k=x_i)=1-(P(k\leq11)+\sum_{12}^{23} P(k=x_i))\\\\P(k>23)=1-(0.021+0.766)=1-0.787=0.213

c) What is the probability that the number of drivers will be between 11 and 23, inclusive? What is the probability that the number of drivers will be strictly between 11 and 23?

Between 11 and 23 inclusive:

P(11\leq k\leq23)=P(x\leq23)-P(k\leq11)+P(k=11)\\\\P(11\leq k\leq23)=0.787-0.021+ 0.011=0.777

Between 11 and 23 exclusive:

P(11< k

d) What is the probability that the number of drivers will be within 2 standard deviations of the mean value?

The standard deviation is

\mu=\lambda =20\\\\\sigma=\sqrt{\lambda}=\sqrt{20}= 4.47

Then, we have to calculate the probability of between 15 and 25 drivers approximately.

P(15

P(k=16) = 20^{16}e^{-20}/16!=0.065\\\\P(k=17) = 20^{17}e^{-20}/17!=0.076\\\\P(k=18) = 20^{18}e^{-20}/18!=0.084\\\\P(k=19) = 20^{19}e^{-20}/19!=0.089\\\\P(k=20) = 20^{20}e^{-20}/20!=0.089\\\\P(k=21) = 20^{21}e^{-20}/21!=0.085\\\\P(k=22) = 20^{22}e^{-20}/22!=0.077\\\\P(k=23) = 20^{23}e^{-20}/23!=0.067\\\\P(k=24) = 20^{24}e^{-20}/24!=0.056\\\\

3 0
2 years ago
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