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The graph of the step function g(x) = –⌊x⌋ + 3 is shown. On a coordinate plane, a step graph has horizontal segments that are ea

AleksAgata [21]

Answer:

{x| –2 ≤ x < 5}

Step-by-step explanation:

There is a box function plotted on the graph.

The function is g(x) = –⌊x⌋ + 3.

Now, we know that a box function represents a step graph having horizontal segments that are each 1 unit long. The left end of each segment is a closed circle. The right end of each segment is an open circle.

It is given that the left-most segment of the given graph goes from (-2,5) to (-1,5) and the rightmost segment goes from (4,-1) to (5,-1).

So, for the left most segment the domain is -2 ≤ x < -1

And for the right most segment the domain is 4 ≤ x < 5

Therefore, the total domain of g(x) will be {x| –2 ≤ x < 5} (Answer)

8 0

3 years ago

Read 2 more answers

How do I add this fraction 7 9/12 + 4 5/9

vovangra [49]

Make the same denominator (12) and then make them both non mixed numbers 93/12 and 54.6666.../12

so 147.666.../12

12.305/12

7 0

2 years ago

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Multiply: -3.2 (-13) x 6 OA) 249.6 OB) -249.6 OC) -234 OD) 234

Citrus2011 [14]

Possibly A)249.6

Explanation

4 0

3 years ago

suppose that you plan on taking a summer job selling magazines subscriptions . the magazine company will help you pay $0.75 for

alina1380 [7]

Answer:

Money earned by selling 175 magazine subscriptions will be 3176.25

Solution:

Given that you are planning to take a summer job selling magazine subscription.

For first subscription, magazine company will help you to pay $0.75

For each additional subscription sold, magazine company will pay you $0.20 more than what was paid for previous subscription.

So for second subscription, payments will be = 0.75 + 0.20 = 0.75 + 0.20 x 1

And for third subscription, payments will be = 0.75 + 0.20 = 0.75 + 0.20 x 2

Generalizing for nth subscription we get

Payments for nth subscription = 0.75 + 0.20(n-1)

so here we can see that arithmetic sequence is formed, having first term

$a_1 = 0.75$ and common difference = d = 0.20

We need to determine sum of above arithmetic sequence for n = 175

Formula for sum of arithmetic sequence is as follows 

$\mathrm{S}_{\mathrm{n}}=\frac{n}{2}[2 a+(n-1) d]$

In our case n = 175, a = 0.75, d = 0.20

$\begin{array}{l}{=>\mathrm{S}_{175}=\frac{175}{2}[(2 \times 0.75)+(175-1) 0.20]} \\\\ {=>\mathrm{S}_{175}=\frac{175}{2}[1.5+(174 \times 0.20)]} \\\\ {\Rightarrow \mathrm{S}_{175}=\frac{175}{2} \times 36.3=3176.25}\end{array}$

Hence money earn by selling 175 magazine subscriptions will be 3176.25

6 0

3 years ago

The weights of soy patties sold by a diner are normally distributed. A random sample of 25 patties yields a mean weight of 4.2 o

just olya [345]

Answer:

$t=\frac{4.2-4}{\frac{0.5}{\sqrt{25}}}=2$

The degrees of freedom are given by:

$df =n-1=25-1=24$

And the p value would be given by:

$p_v = P(t_{24}>2) =0.0285$

And since the p value is lower than the significance level we have enough evidence to conclude that the true mean for this case is significantly hiher than 4. And the claim for this case is not appropiate

Step-by-step explanation:

Data provided

$\bar X=4.2$ represent the sample mean for the weigths

$s=0.5$ represent the sample standard deviation

$n=25$ sample size

$\mu_o =4$ represent the value that we want to analyze

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic

$p_v$ represent the p value for the test

System of hypothesis

We want to conduct a hypothesis in order to check if the true mean weigth is less than 4 or not, the system of hypothesis would be:

Null hypothesis: $\mu \leq 4$

Alternative hypothesis: $\mu > 4$

The statistic for this case is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info given we got:

$t=\frac{4.2-4}{\frac{0.5}{\sqrt{25}}}=2$

The degrees of freedom are given by:

$df =n-1=25-1=24$

And the p value would be given by:

$p_v = P(t_{24}>2) =0.0285$

And since the p value is lower than the significance level we have enough evidence to conclude that the true mean for this case is significantly hiher than 4. And the claim for this case is not appropiate

5 0

3 years ago