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2 years ago

F(x)= 2-1/xg(x)= x/2what is f(g(x))?

1 answer:

8 0

To find f(g(x)), plug g(x) into f(x) and simplify.

f(x/2) = 2 - 1/(x/2)

All you have to do is simplify the right side of the equation.

Take it from here.

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Word problem:

Jlenok [28]

Answer:

Below.

Step-by-step explanation:

30 kg 500g + 28kg 700g

= 58kg + 1200g

= 59kg 200g.

Difference

= 30 kg 500g - 28kg 700g

= 2kg + 500 - 700

= 2kg - 200g

= 1kg 800g.

3 0

3 years ago

Mr. Swanson has $350 to order music books for the new school year. Music books cost $10 each and come in packages of 5. Mr. Swan

Talja [164]

Answer:
$300

Explanation:
$10(cost of 1 book) x 5(how many comes in packages) = 50 ( total cost of 1 package)
Mr. Swanson ordered 6 so...
$50 x 6 = $300
Mr. Swanson would have $50 let's over

3 0

2 years ago

A parabola intersects the x-axis at x = 3 and x = 9.

Novay_Z [31]

Answer:

Step-by-step explanation:

This function corresponds to 'even' function, then

in order to calculate the 'x' of the vertex: (3+9)/2=6.

3 0

3 years ago

Read 2 more answers

Here 1 more than BIG PRIZE

Firlakuza [10]

Hey there! :D

Use the distributive property.

a(b+c)= ab+ac

6(9x+2)+2x

54x+12+2x

56x+ 12 <== equivalent expression

I hope this helps!
~kaikers

4 0

2 years ago

According to a study done by Wakefield Research, the proportion of Americans who can order a meal in a foreign language is 0.47.

UNO [17]

Answer:

Probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is 0.19766.

Step-by-step explanation:

We are given that according to a study done by Wake field Research, the proportion of Americans who can order a meal in a foreign language is 0.47.

Suppose a random sample of 200 Americans is asked to disclose whether they can order a meal in a foreign language.

Let $\hat p$ = sample proportion of Americans who can order a meal in a foreign language

The z-score probability distribution for sample proportion is given by;

Z = $\frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion

p = population proportion of Americans who can order a meal in a foreign language = 0.47

n = sample of Americans = 200

Probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is given by = P( $\hat p$ > 0.50)

P( $\hat p$ > 0.06) = P( $\frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ > $\frac{0.5-0.47}{\sqrt{\frac{0.5(1-0.5)}{200} } }$ ) = P(Z > 0.85) = 1 - P(Z $\leq$ 0.85)

= 1 - 0.80234 = 0.19766

Now, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 0.85 in the z table which has an area of 0.80234.

Therefore, probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is 0.19766.

4 0

3 years ago