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klio [65]
3 years ago
6

Use the graph of f(x) = log x to obtain the graph of g(x) = log x + 5.

Mathematics
2 answers:
galina1969 [7]3 years ago
8 0

The graph of g(x) = f(x) + 5 will have the graph of f(x) moved 5 units upward. The best choice is the last one.

_____

Adding 5 to each y-value (the output of f(x)) moves it upward by 5 units.

creativ13 [48]3 years ago
6 0

Answer:

Option D

Step-by-step explanation:

Given : f(x) = log x

             g(x) = log x + 5.

Solution:

Rule : f(x)→f(x)+b

Graph f(x) is translated up by b units

Now using this rule

f(x) = log x→g(x) = log x + 5.

So, f(x) is translated up by 5 units to obtain g(x).

In option D the graph is translated up by 5 units.

Thus Option D is correct.

Refer the attached graph for the translation.

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How many of the numbers from the set $\{1,\ 2,\ 3,\ldots,\ 50\}$ have a perfect square factor other than one?
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Answer:

2^2 ( 1,2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)  =  4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44,48

3^2(1,2,3, 5)   =  9, 18, 27, 45

5^2(1, 2)  = 25, 50

7^2  = 49

19  numbers

Step-by-step explanation:

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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
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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

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