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3 years ago

QUICK! EASY! 50 POINTS!

Mathematics

2 answers:

bekas [8.4K]3 years ago

5 0

Answer:

Create a circle with the origin as its center and a radius of the origin and point A, then locate a point on the circle that is 90° clockwise from point A.

Step-by-step explanation:

90° clockwise rotation of point A (1, 4) around the origin

Answer options

Create a line perpendicular to the y-axis from point A, and locate a point on the perpendicular line that is equidistant to the distance between the y-axis and A.

Incorrect. It describes the reflection over y axis

Create a circle with the origin as its center and a radius of the origin and point A, then locate a point on the circle that is 90° clockwise from point A.

Correct

Create a line perpendicular to the x‒axis from point A, and locate a point on the perpendicular line that is equidistant to the distance between the y-axis and A.
Incorrect. It is the transformation by 1 unit down.

Create a circle with point A as its center and a radius of the origin and point A, then locate a point on the circle that is 90° counterclockwise from point A.
Incorrect. It is the transformation of origin but not point A

Paha777 [63]3 years ago

4 0

Answer:

create a circle with the origin as its centre and a radius of the origin and point a then locate a point on a circle that is 90 degrees clockwise from point a.

Please mark me as brainliest and follow me.

You might be interested in

Tell whether this function is quadratic. {(10, 50), (11, 71), (12, 94), (13, 119), (14, 146)} .

irga5000 [103]

When we have consecutive values, f(10), f(11), f(12), f(13), f(14), we can make a difference table to determine the degree of f as a polynomial. A quadratic will have a constant second difference:

x 10 11 12 13 14

f(x) 50 71 94 119 146

1st diff 21 23 25 27

2nd diff 2 2 2

We got a constant second difference, so f is a polynomial of degree two.

Answer: This function is quadratic

6 0

4 years ago

I dont know how to do this one

-BARSIC- [3]

Answer:

$2\sqrt{34}$

Step-by-step explanation:

We can add a line to create a triangle and a rectangle. The triangle will have side lengths 10 and 6. Furthermore, it will be a right triangle, meaning we can apply the Pythagorean Theorem to find x, the hypotenuse.

$6^2 + 10^2 = x^2\\x = \sqrt{6^2 + 10^2}\\ = \sqrt{136} \\= 2\sqrt34}$

hope this helped! :)

7 0

2 years ago

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

The number of new songs that a rock group needs to stay at the top each year is inversely proportional to the year and that the

sweet-ann [11.9K]

Answer: a) $3\times50=5 \times x$

b) 30

Step-by-step explanation:

Given: The number of new songs that a rock group needs to stay at the top each year is inversely proportional to the year and that the group has been in the industry.

Let x be the number of songs will be needed after 5 years.

Now the inverse variation equation for the given question will be:-

$3\times50=5 \times x\\\\\Rightarrow x=\dfrac{3\times50}{5}\\\\\Rightarrow\ x=30$

Hence , the number songs will be needed after 5 years = 30

3 0

4 years ago

What is the length of the line segment with endpoints (5,−7) and (5, 11) ?

aliya0001 [1]

11 because the greatest you value was 11 so 11 should be the right awnswer

6 0

3 years ago

Read 2 more answers