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svp [43]
3 years ago
13

help me on the following, in the picture IK U WANT THE POINTS JUST HELLPPP , ONLY IF UKNOW THE ANSWER/ NO LINKS and swrong answe

rs

Mathematics
1 answer:
earnstyle [38]3 years ago
3 0

Answer:

any coordinate outside the parabola is a reasonable answer.

Step-by-step explanation:

test the solution (0, 0) and if if that number is greater than 0 then the rest of the points outside on the inside are.

x = 0

y = 0

(0)^2 - 2(0) - 3 = -3

-3 is less than 0 so (0, 0) is not a possibility so any coordinate outside the parabola is a reasonable answer.

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Samantha has 124 ft of yarn. She wants to cut the yarn into equal pieces to give 17 students. How much yarn will she give each s
xenn [34]
She can give 7.29 ft or 7.3 ft of yarn to her 17 students given that the total measurement of the yarn is 124 ft.

solution:

124ft/17students   - the operation used is division, dived 124ft w/c is the                                        measurement of the yarn by 17 students. 
7.294117647058824 - simplify the answer by rounding it off, at least one or                                       two decimal points; making it 7.29 ft or 7.3ft
6 0
3 years ago
A bag contains 150 marbles. Some are red and the rest are black. There are 14 red marbles for every black marble. How many red m
REY [17]

Answer:

114

Step-by-step explanation:

Ratio of red marbles to black marbles is 19:1. (Data from 3rd sentence)

19+1= 20

20 units= 120 (Data from 1st sentence)

1 unit= 120/20= 6

19 units= 120-6= 114 (Values solved in working above)

Ans: There are 114 red marbles.

i hope i cleared your doubt

8 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
2x-y+52=16<br> X-6у +2z=-9<br> 3x+4y- z=32
Archy [21]

The solution to the given system of equation is (3, 3, 3)

<h3>Solving system of equations</h3>

Given the following system of equation expressed as:

2x-y+5z=16

X-6у +2z=-9

3x+4y- z=32

Multiply equation 2 by 2 to have:

2x-y+5z=16

2X-12у +4z=-18

Subtract

10y+z = 34 .........4

Similarly

3X-18у +6z=-27

3x+4y- z=32

Subtract

-22y+7z = -59 ........... 5

Equate 4 and 5

10y+z = 34 .........4 * 7

-22y+7z = -59 ........... 5 * 1

_______________________

70y+7z = 238

-22y+7z = -59

Subtract

92y = 297

y = 3

Recall that

10y+z = 34

10(3) + z = 34

z = 3

Since x - 6y +2z = -9

x-6(3)+2(3) = -9
x - 18 + 6 = -9
x = 3

Hence the solution to the given system of equation is (3, 3, 3)

Learn more on system of equation here: brainly.com/question/25976025

#SPJ1

4 0
2 years ago
A plane is traveling due north. The wind is blowing 30° east of north at 60 miles/hour. At what speed must the plane fly to trav
oksano4ka [1.4K]
The answer is 10 Miles per hour

I would use this proportion
60/x 30/5
60 times 5=300
300 divided by 30=10
X=10
5 0
3 years ago
Read 2 more answers
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