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3 years ago

Which of the following is an open statement?

Mathematics

1 answer:

nignag [31]3 years ago

5 0

Answer:

an angel with measure 40 degree is acute because 42 is less than 90

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Convert the subtraction problem into addition problem -3.7-(-10.1)

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3 years ago

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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

kipiarov [429]

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.

$P(15$

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

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3 years ago

3. Is AABC a right triangle? Explain

lukranit [14]

Answer:

no its bbac

Step-by-step explanation:

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3 years ago

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Henry is 4 years younger than Linda. Linda is 20 years old. After correctly carrying out the Solve step of the five-step problem

ehidna [41]

C. 16.

20 - 4 = 16

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4 years ago

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(a) A lamp has two bulbs of a type with an average lifetime of 1600 hours. Assuming that we can model the probability of failure

lara [203]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The probability is $P_T= 0.4560$

The probability is $P_F= 0.0013$

Step-by-step explanation:

From the question we are told that

The mean for the exponential density function of bulbs failure is $\mu = 1600 \ hours$

Generally the cumulative distribution for exponential distribution is mathematically represented as

$1 - e^{- \lambda x}$

The objective is to obtain the p=probability of the bulbs failure within 1800 hours

So for the first bulb the probability will be

$P_1(x < 1800)$

And for the second bulb the probability will be

$P_2 (x< 1800)$

So from our probability that we are to determine the area to the left of 1800 on the distribution curve

Now the rate parameter $\lambda$ is mathematically represented as

$\lambda$ $= \frac{1}{\mu}$

$\lambda = \frac{1}{1600}$

The probability of the first bulb failing with 1800 hours is mathematically evaluated as

$P_1(x < 1800) = 1 - e^{\frac{1}{1600} * 1800 }$

$= 0.6753$

Now the probability of both bulbs failing would be

$P_T=P_1(x < 1800) * P_2(x < 1800)$

$= 0.6375 * 06375$

$P_T= 0.4560$

Let assume that one bulb failed at time $T_a$ and the second bulb failed at time $T_b$ then

$T_a + T_b = 1800\ hours$

The mathematical expression to obtain the probability that the first bulb failed within between zero and $T_a$ and the second bulb failed between $T_a \ and \ 1800$ is represented as

$P_F=\int_{0}^{1800}\int_{0}^{1800-x} \f{\lambda }^{2}e^{-\lambda x}* e^{-\lambda y}dx dy$

$=\int_{0}^{1800} {\lambda }e^{-\lambda x}\int_{0}^{1800-x} {\lambda } e^{-\lambda y}dx dy$

$=\int_{0}^{1800} {\frac{1}{1600} }e^{-\lambda x}\int_{0}^{1800-x} \frac{1}{1600 } e^{-\lambda y}dx dy$

$=\int_{0}^{1800} {\frac{1}{1600} }e^{-\lambda x}[e^{- \lambda y}]\left {1800-x} \atop {0}} \right. dx$

$=\int_{0}^{1800} {\frac{1}{1600} }e^{-\frac{x}{1600} }[e^{- \frac{1800 -x}{1600} }-1] dx$

$=[ {\frac{1}{1600} }e^{-\frac{1800}{1600} }-\frac{1}{1600}[e^{- \frac{x}{1600} }] \left {1800} \atop {0}} \right.$

$=[ {\frac{1}{1600} }e^{-\frac{1800}{1600} }-\frac{1}{1600}[e^{- \frac{1800}{1600} }] -[[ {\frac{1}{1600} }e^{-\frac{1800}{1600} }-\frac{1}{1600}[e^{-0}]$

$=[\frac{1}{1600} e^{-\frac{1800}{1600} } - \frac{1}{1600} e^{-0} ]$

$=0.001925 -0.000625$

$P_F= 0.0013$

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3 years ago

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