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3 years ago

Domain and Range: determine the domain and Range for each graph

Mathematics

1 answer:

bija089 [108]3 years ago

7 0

Given:

The graph of a function.

To find:

The domain and range of the given graph.

Solution:

Domain is the set of x-values or input values.

From the given graph it is clear that, the graph is defined for the interval (-4,4).

But there is a closed circle at x = -4 and an open circle at x = 4. It means, -4 is included and 4 is not included in the solution set.

Domain: [-4,4)

Domain: $-4\leq x$

Range is the set of y-values or output values.

From the given graph it is clear that, the output values are lie in the interval (-3,4).

But there closed circle at x = -3 and x = 4. It means, -3 and 4 are is included in the set of output values.

Range: [-3,4]

Range: $-3\leq y\leq 4$

Therefore, domain is $-4\leq x$ and range is $-3\leq y\leq 4$ .

You might be interested in

The speed-time graph shows the first 30 minutes of another train journey. The distance travelled is 100km. The maximum speed of

aalyn [17]

The question is missing the graph. So, the graph is attached below.

Answer:

The maximum speed of the train is $V=240\ km/h$

Step-by-step explanation:

Given:

Distance travelled is 100 km.

Time interval is 30 minutes.

We know that, area under the curve of a speed-time graph gives the total distance travelled by the train.

The area under the curve is the area of the given trapezium shape shown in the graph below.

The height of the trapezium is equal to the maximum speed of the train represented by 'V' km/h. The parallel sides are of lengths equal to (25 - 5 =) 20 min and (30 - 0 =) 30 min.

Converting the units of parallel sides from minutes to hours we get:

$30\ min=\frac{30}{60}=\frac{1}{2}\ h\\20\ min=\frac{20}{60}=\frac{1}{3}\ h$

Now, area of a trapezium is given as:

$Area=\frac{1}{2}\times\textrm{Sum of parallel sides}\times \textrm{Height}\\\\Area=\frac{1}{2}(\frac{1}{2}+\frac{1}{3})(V)\\\\Area=\frac{1}{2}(\frac{3+2}{6})(V)\\\\Area=\frac{1}{2}(\frac{5}{6})(V)\\\\Area=(\frac{1\times 5}{2\times6})(V)\\\\Area=\frac{5}{12}(V)$

Now, area under the curve is equal to the distance travelled. So,

$\frac{5}{12}(V)=100\\V=\frac{100\times12}{5}\\V=240\ km/h$

Therefore, the maximum speed of the train is $V=240\ km/h$

4 0

3 years ago

Suppose that 20% of the subscribers of a cable television company watch the shopping channel at least once a week. The cable com

riadik2000 [5.3K]

Answer:

4.01% probability that the cable company will keep the shopping channel.

Step-by-step explanation:

For each subscriber, there are only two possible outcomes. Either they watch the shopping channel at least once a week, or they do not. This means that we can solve this problem using the binomial probability distribution.

However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal.

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 $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$n = 100, p = 0.20$

$E(X) = 100*0.20 = 20$

$\sqrt{V(X)} = \sqrt{100*0.20*0.80} = 4$

The cable company has decided to keep the shopping channel if the sample proportion is greater than 0.27. What is the approximate probability that the cable company will keep the shopping channel, even though the true proportion who watch it is only 0.20?

This probability is 1 subtracted by the pvalue of Z when $X = 0.27*100 = 27$ . So