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riadik2000 [5.3K]
1 year ago
5

Find an explicit formula for the arithmetic sequence -5,13,31,49,...−5,13,31,49,...minus, 5, comma, 13, comma, 31, comma, 49, co

mma, point, point, point.
b(n)=b(n)=b, left parenthesis, n, right parenthesis, equals
Mathematics
1 answer:
dolphi86 [110]1 year ago
4 0

Answer:

aₙ = a₁ + (n-1) * 18        a₁ = -5

Step-by-step explanation:

aₙ = a₁ + (n-1) * d            a₁ = -5     d = 13 - (-5) = 18

a₂ = -5 + (2 - 1) * 18 = -5 + 18 = 13

a₃ = -5 + (3 - 1) * 18 = -5 + 36 = 31

a₄ = -5 + (4 - 1) * 18 = -5 + 54 = 49

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According to a report for veterinarians in the United States, 36.5 percent of households in the United States own dogs and 30.4
aleksandrvk [35]

Answer :E) Not enough information is given to determine the probability.

Step-by-step explanation:

Le A denotes the event that households in the United States own dogs .

and B denotes the event that households in the United States own cats.

As per given , we have

P(A)=36.5%= 0.365

P(B)=30.4% = 0.304

To find the probability that the selected household will own a dog or a cat, we apply the following formula :

P(A or B)=P(A)+P(B)+P(A and B)

But P(A and B) is not given to us.

i..e the probability that a house hold own both a cat and adg is not given to us.

Therefore, The correct option is (E) Not enough information is given to determine the probability.

3 0
1 year ago
The blue M&M was introduced in 1995. Before then, the color mix in a bag of plain M&Ms was (30% Brown, 20% Yellow, 20% R
IRISSAK [1]

Answer:

The probability that the yellow M&M came from the 1994 bag is 0.07407 or 7.407%

Step-by-step explanation:

Given

Before 1995

(Br) Brown = 30%

(Y) Yellow = 20%  =0.2

(R) Red = 20%

(G) Green =10%  =0.1

(O) Orange = 10%

(T) Tan = 10%

 

After 1995

(Br) Brown = 13%

(Y) Yellow = 14%  =0.14

(R) Red = 13%

(G) Green = 20% = 0.2

(O) Orange = 16%

(Bl) Blue = 24%

Since there are two bags, let A be the bag from 1994, and B be the bag from 1996

Then let AY imply we drew a yellow M&M from the 1994 bag

AG implies we drew a green M&M from the 1994 bag

BY implies imply we drew a yellow M&M from the 1996 bag

BG implies we drew a green M&M from the 1996 bag

P(AY) =0.2

P (BY) = 0.14

P(AG) =0.1

P(BG) =0.2

Since the draws from the 1994 and 1996 bag are independent,

therefore

P(AY n BG) = 0.2 * 0.2 = 0.04  -------(1)\\P(AG n BY) =0.1 * 0.14 =0.014   --------(2)\\

The draws can happen in either of the 2 ways in (1) and (2) above

therefore total probability E is given as

E =P( AY n BG) u P(AG n BY)\\=0.04 + 0.014 =0.O54

For the yellow one to be from 1994, it implies that the event to be chosen is

P(AYnBG) = 0.2*0.2

Since the total probability is given as E=0.054

then P((AYnBG) /E) =\frac{0.04}{0.054} = 0.07407

Concluding statement: This is the condition for the Yellow one to come from 1994 and green from 1996 provided that they obey the condition from E

4 0
2 years ago
Write two expressions where the solution is 41
Bogdan [553]
Two consecutive numbers can be defined as x and y, but a better choice is x and x+1. . The product of the two numbers is: x(x+1). We are told this 41 more than the sum. The sum of the two numbers is: x + (x+1). . So the equation we need to solve is: . x(x+1) = x+(x+1)+41 . x^2 +x = 2x + 42 . subtract 2x from both sides . x^2 +x -2x = 2x-2x + 42 x^2 -x = 42 . subtract 42 from both sides . x^2 -x -42 = 42-42 = 0
6 0
2 years ago
laura picks 7 apples. some apples are red and some are green. fewer than 4 of the apples are red. how many of lauras apples coul
Schach [20]
4 or more apples could be green because if there are 3 or less red then there must be 4-7 green
8 0
1 year ago
Read 2 more answers
Which system of linear inequalities has the point (2, 1) in its solution set?
AVprozaik [17]
You haven't provided the choices, therefore, I cannot provide an exact answer. However, I will help you with the concept.

For an order pair to be a solution to a system of equations, it has to satisfy <u>BOTH</u> equations. If it satisfies only one equation of the system or satisfy neither of the equations, the, it is not a solutions

<u><em>Examples:</em></u>
<u>System 1:</u>
x = y + 1
2x + 3y = 7
Let's check (2,1)
2 = 1 + 1 ........> equation 1 is satisfied
2(2) + 3(1) = 7 ......> equation 2 is satisfied
<u>(2,1) is a solution to this system</u>

<u>System 2:</u>
y = x + 3
y = x - 1
Let's check (2,1):
1 ≠ 2 + 3 ........> equation 1 isn't satisfied
1 = 2 - 1 ..........> equation 2 is satisfied
<u>(2,1) isn't a solution to this system</u>

<u>System 3:</u>
2y = 9 - 3x
3x + 2y = 9
Let's ceck (2,1):
2(1) ≠ 9 - 3(2) ..........> equation 1 isn't satisfied
3(2) + 2(1) ≠ 9 .........> equation 2 isn't satisfied
<u>(2,1) isn't a solution to this system
</u>
<u><em>Based on the above,</em></u> all you have to do is substitute with (2,1) in the system you have and pick the one where both equations are satisfied

Hope this helps :)
4 0
1 year ago
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