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3 years ago

If f(x) - 4x2 - 3 and g(x) = x+1, find (f- g)(x).

Mathematics

1 answer:

igomit [66]3 years ago

4 0

Answer:

Given f(x) = 2x + 3 and g(x) = –x2 + 5, find ( f o f )(x). ... Given h(x) = sqrt(4x + 1), determine two functions f (x) and g(x) which, when composed, generate h(x).

Step-by-step explanation:

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Write an absolute value function that shifted 2<br> units right.<br><br> HELPPPP

iVinArrow [24]

Answer: (a)

Step-by-step explanation:

Suppose the original function is $y=\left | x \right |$

To shift it 2 units right, replace x by x-2 such that

$\Rightarrow x-2=0\\\Rightarrow x=2$

So, the function becomes $y=\left | x -2\right |$

4 0

3 years ago

The table below shows field goal statistics for three different kickers

FromTheMoon [43]

Answer:

<h2>From least to greatest: C, B and A</h2>

Step-by-step explanation:

To order each Kicker correctly, we have to find the success rate of each one, which is obtained by dividing and then multiplying by 100 to obtained the percentage:

$R=\frac{made}{attempted}$

<h3>Success rate of A:</h3>

$R_{A}=\frac{12}{14}.100=86\%$

<h3>Success rate of B:</h3>

$R_{B}=\frac{15}{18}.100=83\%$

<h3>Success rate of C:</h3>

$R_{C}=\frac{19}{24}.100=79\%$

So, comparing all three, from least to greatest we have:

C with 79%, then B with 83%, and A with 86%.

3 0

3 years ago

Read 2 more answers

Question Help Suppose that the lifetimes of light bulbs are approximately normally distributed, with a mean of 5656 hours and a

koban [17]

Answer:

a)3.438% of the light bulbs will last more than 6262 hours.

b)11.31% of the light bulbs will last 5252 hours or less.

c) 23.655% of the light bulbs are going to last between 5858 and 6262 hours.

d) 0.12% of the light bulbs will last 4646 hours or less.

Step-by-step explanation:

Normally distributed problems can be solved by the z-score formula:

On a normaly distributed set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a value X is given by:

$Z = \frac{X - \mu}{\sigma}$

After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.

In this problem, we have that:

The lifetimes of light bulbs are approximately normally distributed, with a mean of 5656 hours and a standard deviation of 333.3 hours.

So $\mu = 5656, \sigma = 333.3$

(a) What proportion of light bulbs will last more than 6262 hours?

The pvalue of the z-score of $X = 6262$ is the proportion of light bulbs that will last less than 6262. Subtracting 100% by this value, we find the proportion of light bulbs that will last more than 6262 hours.