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kozerog [31]
3 years ago
7

What is 3 5/10 - 4/8 answer quickly please and a explanation

Mathematics
2 answers:
Harrizon [31]3 years ago
7 0
I hope this helps you

Bad White [126]3 years ago
5 0
The answer is 3. First simply 5/10 to 1/2. Then convert 3 1/2 to an improper fraction, by multiplying 2*3 and then adding 1, then put answer 7 over 2. Next simply 4/8 into 1/2. Subtract 7 and 1 and get 6. put 6 over 2 and dic=vide to get your answer which is 3.
You might be interested in
Use the Distributive Property to rewrite the expression. 9(y + 4)
Ulleksa [173]

Answer:

Answer would be 9y+36

Step-by-step explanation:

Because if you distribute the 9 inside the parenthesis, you'd get

9*y=9y and 9*4=36

so 9y+36

Hope my answer was helpful to you!

7 0
4 years ago
For each of these word problems, determine if the answer can be found using the division problem 125 ÷ 12.
olasank [31]

Answer:

125/12

Step-by-step explanation:

1st problem-no

2nd problem-yes

3rd problem-yes

4th problem-no

3 0
3 years ago
Help please!!
IRINA_888 [86]
I think it’s QT EJ and HA
7 0
3 years ago
Read 2 more answers
HELPPPP
MissTica

Answer:

$41.0625

Step-by-step explanation:

Divide $054.75 by 4

Multiply the answer to that (13.6875) by 3 or subtract 054.5 by 13.6873

The answer to that step is your answer.

8 0
3 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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