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bagirrra123 [75]

2 years ago

6

Of the coffee makers sold in an appliance store, 4.0% have either a faulty switch or a defective cord, 2.5% have a faulty switch

, and 0.1% have both defects. What is the probability that a coffee maker will have a defective cord? Express the answer as a percentage.

1 answer:

Lelechka [254]2 years ago

3 0

Add 4.0% and 0.1 percent and what do you get?

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Vijay has an annual salary of 47,000, and his company pays him twice a month. What is the gross income in each paycheck that Vij

leonid [27]

Alright, so he has an <em>annual </em>salary of 47,000 dollars. Which means that he is paid 47,000 dollars in 12 months. You'll first have to calculate the pay Vijay receives <em>each month</em>, which is $\frac{47,000}{12 months}$ , or about $3,916.67 (I'll round up to 3917 for simplicity).

Now, he gets paid twice a month. So each paycheck is half of $3917. 3917 x (1/2) = $1958.50.

So each paycheck should be $1958.50 (this is a rounded figure).

6 0

2 years ago

if you are adding two fractions that are both greater than 1/2 what must be true about the sum? Give three examples to support y

Sunny_sXe [5.5K]

Answer:

If you are adding to fractions that are both greater than 1/2, the sum will be greater than 1.

Step-by-step explanation: 1/2 + 1/2 =1

Since both fractions are greater, obviously the sum will too.

Examples: 2/3+3/4= 1 5/12 4/6+6/9= 1 1/3

Hope this helps

6 0

2 years ago

F(x)=3x 2 +9f, left parenthesis, x, right parenthesis, equals, 3, x, squared, plus, 9 and g(x)=\dfrac{1}{3}x^2-9g(x)= 3 1 x 2

34kurt

Answer:

$f(g(x)) = \frac{1}{3}x^4 - 18x^2 + 252$

$g(f(x)) = 3x^4 + 18x^2 + 18$

<em>f(x) and g(x) and not inverse functions</em>

Step-by-step explanation:

Given

$f(x) = 3x^2 + 9$

$g(x) = \dfrac{1}{3}x^2 - 9$

Required

Determine f(g(x))

Determine g(f(x))

Determine if both functions are inverse:

Calculating f(g(x))

$f(x) = 3x^2 + 9$

$f(g(x)) = 3(\frac{1}{3}x^2 - 9)^2 + 9$

$f(g(x)) = 3(\frac{1}{3}x^2 - 9)(\frac{1}{3}x^2 - 9) + 9$

Expand Brackets

$f(g(x)) = (x^2 - 27)(\frac{1}{3}x^2 - 9) + 9$

$f(g(x)) = x^2(\frac{1}{3}x^2 - 9) - 27(\frac{1}{3}x^2 - 9) + 9$

$f(g(x)) = \frac{1}{3}x^4 - 9x^2 - 9x^2 + 243 + 9$

$f(g(x)) = \frac{1}{3}x^4 - 18x^2 + 252$

Calculating g(f(x))

$g(x) = \dfrac{1}{3}x^2 - 9$

$g(f(x)) = \frac{1}{3}(3x^2 + 9)^2 - 9$

$g(f(x)) = \frac{1}{3}(3x^2 + 9)(3x^2 + 9) - 9$

$g(f(x)) = (x^2 + 3)(3x^2 + 9) - 9$

Expand Brackets

$g(f(x)) = x^2(3x^2 + 9) + 3(3x^2 + 9) - 9$

$g(f(x)) = 3x^4 + 9x^2 + 9x^2 + 27 - 9$

$g(f(x)) = 3x^4 + 18x^2 + 18$

Checking for inverse functions

$f(x) = 3x^2 + 9$

Represent f(x) with y

$y = 3x^2 + 9$

Swap positions of x and y

$x = 3y^2 + 9$

Subtract 9 from both sides

$x - 9 = 3y^2 + 9 - 9$

$x - 9 = 3y^2$

$3y^2 = x - 9$

Divide through by 3

$\frac{3y^2}{3} = \frac{x}{3} - \frac{9}{3}$

$y^2 = \frac{x}{3} - 3$

Take square root of both sides

$\sqrt{y^2} = \sqrt{\frac{x}{3} - 3}$

$y = \sqrt{\frac{x}{3} - 3}$

Represent y with g(x)

$g(x) = \sqrt{\frac{x}{3} - 3}$

Note that the resulting value of g(x) is not the same as $g(x) = \dfrac{1}{3}x^2 - 9$

<em>Hence, f(x) and g(x) and not inverse functions</em>

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Hello!

Your answer is 7.

Toodles~

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8 books with 2 sheets each and 1 extra sheet as leftover

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