All categories

3 years ago

ONLY ANSWER IF UR CORRECT PLZZ!

Mathematics

1 answer:

anyanavicka [17]3 years ago

8 0

Answer:

∠4 & ∠8

∠3 & ∠7

Step-by-step explanation:

When the transversal (the line that crosses the other 2) intersects through 2 lines, it forms 2 pairs of alternate interior angles.

The alternate interior angles are always located in the inner corners of the intersections and are on opposite ends of the transversal.

You might be interested in

GIVING BRAINLIEST PLS HELP :(

Kryger [21]

Answer:

d) not proportional

Step-by-step explanation:

the rate of change between the first two pairs of x-and-y values is 12/3, between the second pair it is 21/3, and 15/3 between the last pair

4, 7, and 5 are all different values so there is not one constant of proportionality

4 0

3 years ago

How many positive factors of 200 are divisible by 4?

Stella [2.4K]

Answer:

C. 6, D. 20, B. 18, B. 9

Step-by-step explanation:

#1. Factors of 200 are 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100 & 200, of which 6 are divisible by 4 (4, 8, 20, 40, 100 & 200)

#2. Prime factors of 35 are 7 & 5, twice the sum of these factors, k = 24. k-4 = 20

#3. Non-multiples of 4 nor 5 between 16 and 45 are 17, 18, 19, 21, 22, 23, 26, 27, 29, 31, 33, 34, 37, 38, 39, 41, 42 & 43...thus 18 non-multiples

#4. Even factors of 120 greater than 1 and less than 40 are 2, 4, 6, 8, 10, 12, 20, 24 & 30...thus there are 9 even factors

3 0

3 years ago

What is the slope of the line that passes through the point A (-4,3) and B (-2,-5)

Ghella [55]

Answer:

-1

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

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.

$P(15$

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

3 0

3 years ago

Suppose that someone is offering to sell you raffle tickets. There are blue, green, yellow, and red tickets available.

Ghella [55]

Answer:

Step-by-step explanation:

↑

8 0

3 years ago