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3 years ago

12 A) Find the nth term of sequency. 1. 7 17 31 49.....

Mathematics

1 answer:

Sladkaya [172]3 years ago

3 0

Answer:

Step-by-step explanation:

ndlcdjwl

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Consider randomly selecting a student at a large university. Let A be the event that the selected student has a Visa card, let B

wariber [46]

Answer:

a. 0.76

b. 0.23

c. 0.5

d. p(B/A) is the probability that given that a student has a visa card, they also have a master card

p(A/B) is the probability that given a student has a master card, they also have a visa card

e. 0.35

f. 0.31

Step-by-step explanation:

a. p(AUBUC)= P(A)+P(B)+P(C)-P(AnB)-P(AnC)-P(BnC)+P(AnBnC)

=0.6+0.4+0.2-0.3-0.11-0.1+0.07= 0.76

b. P(AnBnC')= P(AnB)-P(AnBnC)

=0.3-0.07= 0.23

c. P(B/A)= P(AnB)/P(A)

=0.3/O.6= 0.5

e. P((AnB)/C))= P((AnB)nC)/P(C)

=P(AnBnC)/P(C)

=0.07/0.2= 0.35

f. P((AUB)/C)= P((AUB)nC)/P(C)

=(P(AnC) U P(BnC))/P(C)

=(0.11+0.1)/0.2

=0.21/0.2 = 0.31

7 0

3 years ago

Select the correct answer.

Arlecino [84]

I’m pretty sure it’s linear!

7 0

3 years ago

What is the prime factorization of 293 in expanded form

lutik1710 [3]

293 is a prime number, so you can't divide it in factors

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3 years ago

Lily Quinn makes $12.50 and hour. She works four hours on Monday, six hours on Tuesday, nine hours on Wednesday, three hours on

Bas_tet [7]

The answer for your question is C. $362.50

8 0

3 years ago

The planets move around the sun in elliptical orbits with the sun at one focus. The point in the orbit at which the planet is cl

gladu [14]

Consider the attached ellipse. Let the sun be at the right focus. Then perihelion is at right vertex on the x-axis and aphelion is at the left vertex on the x-axis.

The distances:

from perihelion to the sun in terms of ellipse is a-c;
from aphelion to the sun in terms of ellipse is a+c.

Then

$\left\{\begin{array}{l}a-c=741,000,000\\a+c=817,000,000\end{array}\right.$

Add these two equations:

$2a=1,558,000,000 \\ \\a=779,000,000$

and subtract first equation from the second:

$2c=76,000,000 \\ \\c=38,000,000.$

Note that $b=\sqrt{a^2-c^2},$ thus

$b=\sqrt{779,000,000^2-38,000,000^2}=\sqrt{605,397,000,000,000,000}.$

The equation for the planet's orbit is

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\Rightarrow \dfrac{x^2}{606,841,000,000,000,000}+\dfrac{y^2}{605,397,000,000,000,000}=1.$

7 0

3 years ago