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2 years ago

give an example on an addition problem in which you would and would not group the addends differently to add

1 answer:

3 0

Addends are any of the numbers added together in an equation.

The only time their grouping would matter would be if there were parentheses used to alter the normal Order of Operations.

For ex:
2 - (8 + 3) here, the 8 and 3 have to be grouped together before doing the subtraction.

Any addition problem without parentheses can be used for one where the grouping doesn't matter

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Two old rsm students were standing twenty feet apart when they spotted mrs. rifkin. if they immediately run in opposite directio

Black_prince [1.1K]

Speed of student A is 180 ft/min
Speed of student B is 120 ft/min
Relative speed=(120+180)=300 ft/min
Distance between them is 2420 ft
Time taken for them to meet will be:
time=distance/speed
=(2420-20)/300
=8 min

3 0

2 years ago

Read 2 more answers

A player of a video game is confronted with a series of four opponents and an 80% probability of defeating each opponent. Assume

mixer [17]

Answer:

(a) 0.4096

(b) 0.64

Step-by-step explanation:

The probability that the player wins is,

$P(W)=0.80$

Then the probability that the player losses is,

$P(L)=1-P(W)=1-0.80=0.20$

The player is playing the video game with 4 different opponents.

It is provided that when the player is defeated by an opponent the game ends.

All the possible ways the player can win is: {L, WL, WWL, WWWL and WWWW)

(a)

The results from all the 4 opponents are independent, i.e. the result of a game played with one opponent is unaffected by the result of the game played with another opponent.

The probability that the player defeats all four opponents in a game is,

P (Player defeats all 4 opponents) = $P(W)\times P(W)\times P(W)\times P(W)=[P(W)]^{4} =(0.80)^{4}=0.4096$

Thus, the probability that the player defeats all four opponents in a game is 0.4096.

(b)

The probability that the player defeats at least two opponents in a game is,

P (Player defeats at least 2) = 1 - P (Player losses the 1st game) - P (Player losses the 2nd game) = $1-P(L)-P(WL)$

$=1-(0.20)-(0.80\times0.20)\\=1-0.20-0.16\\=0.64$

Thus, the probability that the player defeats at least two opponents in a game is 0.64.

(c)

Let X = number of times the player defeats all 4 opponents.

The probability that the player defeats all four opponents in a game is,

P(WWWW) = 0.4096.

Then the random variable $X\sim Bin(n=3, p=0.4096)$

The probability distribution of binomial is:

$P(X=x)={n\choose x}p^{x} (1-p)^{n-x}$

The probability that the player defeats all the 4 opponents at least once is,

P (X ≥ 1) = 1 - P (X < 1)

= 1 - P (X = 0)

$=1-[{3\choose 0}(0.4096)^{0} (1-0.4096)^{3-0}]\\=1-[1\times1\times (0.5904)^{3}\\=1-0.2058\\=0.7942$

Thus, the probability that the player defeats all the 4 opponents at least once is 0.7942.

3 0

2 years ago

ALGEBRA 2/ TRIG QUESTION ATTACHED Please help ❤️❤️

ss7ja [257]

Let's calculate the value of angle A and B

sin(A) =-4/5 → sin⁻¹(- 4/5) = A → A = - 53.13

cos(B) = -5/13 → cos⁻¹ (- 5/13) = B → B = 112.62

tan (A+B) = sin(A+B)/cos(A+B) with A+B = -53.13 + 112.62 = 59.49

tan (A+B) = sin(59.49)/cos(59.49) = 0.86154/0.507688 = 1.6969.

(Answer H = 56/33 = 1.6969)

7 0

2 years ago

Max's cell phone plan charges a monthly

ludmilkaskok [199]

Answer:

Step-by-step explanation:

7.41 / 0.13 = 57

57 - 15 = 42

42 / 2 = 21

21, 36

6 0

2 years ago

Consider a set of 7500 scores on a national test whose score is known to be distributed normally with a mean of 510 and a standa

german

$\mathbb P(X>600)=\mathbb P\left(\dfrac{X-510}{85}>\dfrac{600-510}{85}\right)=\mathbb P(Z>1.059)\approx0.145$

So approximately 14.5% of the scores are higher than 600. This means in a sample of 7500, one could expect to see $0.145\times7500\approx10.86$ scores above 600.

5 0

2 years ago