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lesya692 [45]
3 years ago
10

Find The Sum. 6 2/3 + 1 5/6=

Mathematics
2 answers:
patriot [66]3 years ago
5 0
6 2/3 + 1 5/6 = 8 1/2

i am a mathematics teacher. if anything to ask please pm me
Svetlanka [38]3 years ago
4 0
You first turn both fractions into improper fractions. then you get 20/3 and 11/6. afterthat multiply 20/3 by 2 because you need the same denominator as 11/6. so that will be 40/6. then add 40/6 and 11/6 = 51/6. Lastly turn 51/6 into a mixed number by dividing 51÷6= 8.5 meaning the answer would be 8 and 1/2
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8(3a+6)=2a-4
NISA [10]
A=-26/11 —> you will solve for a by simplifying both sides of the equation, then isolating the variable.
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3 years ago
A frog jumps across a pond. Its height, as a function of time, is modeled by y
sdas [7]

Answer:The function is nonlinear

Step-by-step explanation:

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
Koji has 5,502 in savings this is 30 less than 6 times the amount in his checking, how much money?
omeli [17]

Answer:

$922

Step-by-step explanation:

5 0
3 years ago
A survey of teens suggested that 33% can name at least one professional baseball player, and 90% of those teens can also name at
loris [4]

Answer:

d.\ 0.33 + 0.64 – (0.90)(0.33) = 0.673

Step-by-step explanation:

Given

P(Baseball) = 33\%

P(Baseball\ n\ Football) = 90\%

P(Football) = 64\%

Required

P(Baseball\ or\ Football)

This is calculated as:

P(Baseball\ or\ Football) =P(Baseball) + P(Football) -P(Baseball\ n\ Football)

P(Baseball\ or\ Football) = 0.33 + 0.64 - (0.90)(0.33)

P(Baseball\ or\ Football) = 0.33 + 0.64 - 0.297

P(Baseball\ or\ Football) = 0.673

3 0
3 years ago
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