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Mila [183]
2 years ago
10

Which is the best description for this graph? Someone Please help me :)

Mathematics
2 answers:
Stella [2.4K]2 years ago
8 0

Answer: the graph is decreasing everywhere

wlad13 [49]2 years ago
4 0
The answer would be the second one, the graph is decreasing everywhere
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Divided 85 lbs 2 oz by 6? PLEASE HELP its 12:00 am and I can't think right now
Ostrovityanka [42]
1 pound = 16 ounces
So we have 85 lbs and 2 oz, that is if we convert them to ounces:
85(16) + 2 = 1362 oz
so we have 1362 oz total due to the conversion factor stated above, now we only need to divide:
1362/6 = 227
that is 227 oz
or if we want the result in pounds (lbs), we can by dividing again by 16
227/16 = 14.19 lbs
6 0
3 years ago
i need help with this and i just copy and paste it.Beth's mother made some cookies. She put 4 cookies on each of 8 plates. Which
mr_godi [17]

Answer:

4x8=32

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
-105=7(1+2n)<br> A:-8<br> B-4<br> C-10<br> D:-9<br> What’s the value of n in this equation?
Marat540 [252]
The answer is -8 or a
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2 years ago
Zora saved of her paycheck and to that she added $25.30. If the total that went into her savings was $101.68, what was the amoun
vovangra [49]
The answer would be 76.38
6 0
2 years ago
You play two games against the same opponent. The probability you win the first game is 0.4. If you win the first game, the prob
Ilia_Sergeevich [38]

Answer:

a) No

b) 42%

c) 8%

d) X               0                 1                2

   P(X)           42%            50%         8%

e) 0.62

Step-by-step explanation:

a) No, the two games are not independent because the the probability you win the second game is dependent on the probability that you win or lose the second game.

b) P(lose first game) = 1 - P(win first game) = 1 - 0.4 = 0.6

P(lose second game) = 1 - P(win second game) = 1 - 0.3 = 0.7

P(lose both games) = P(lose first game)  × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%

c)   P(win first game)  = 0.4

P(win second game) = 0.2

P(win both games) = P(win first game)  × P(win second game) = 0.4 × 0.2 = 0.08 = 8%

d) X               0                 1                2

   P(X)           42%            50%         8%

P(X = 0)  =  P(lose both games) = P(lose first game)  × P(lose second game) = 0.6 × 0.7 = 0.42 = 42%

P(X = 1) = [ P(lose first game)  × P(win second game)] + [ P(win first game)  × P(lose second game)] = ( 0.6 × 0.3) + (0.4 × 0.8) = 0.18 + 0.32 = 0.5 = 50%

e) The expected value  \mu=\Sigma}xP(x)= (0*0.42)+(1*0.5)+(2*0.08)=0.66

f) Variance \sigma^2=\Sigma(x-\mu^2)p(x)= (0-0.66)^2*0.42+ (1-0.66)^2*0.5+ (2-0.66)^2*0.08=0.3844

Standard deviation \sigma=\sqrt{variance} = \sqrt{0.3844}=0.62

8 0
3 years ago
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