If f(x)= 4x+3 and g(x)= 3x evalute f(x). g(x) for x=-1 over 2

1 answer:

6 0

$\bf \begin{cases}
f(x)=4x+3\\
g(x)=3x\\
------------\\
f(x)\cdot g(x)=(4x+3)(3x)\\
\qquad \qquad \qquad 12x^2+9x
\end{cases}
\\\\\\
\boxed{x=-\cfrac{1}{2}}\qquad \qquad 12\left( -\frac{1}{2} \right)^2+9\left( -\frac{1}{2} \right)\implies 12\left( -\frac{1}{2} \right)\left( -\frac{1}{2} \right)+9\left( -\frac{1}{2} \right)
\\\\\\
12\left( \cfrac{1}{4} \right)-\cfrac{9}{2}\implies 3-\cfrac{9}{2}\implies \cfrac{6-9}{2}\implies \cfrac{-3}{2}\implies -1\frac{1}{2}$

You might be interested in

The set of numbers 1, 7, 11, and 36 contains values of m. What value of m makes the equation below true? 4m + 8 = 36

german

Answer:

i think its 7

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

An article written for a magazine claims that 78% of the magazine's subscribers report eating healthily the previous day. Suppos

Schach [20]

Answer:

89.44% probability that less than 80% of the sample would report eating healthily the previous day

Step-by-step explanation:

We use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .