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Triss [41]
3 years ago
13

Which graph represents the function on the interval [-3,3]? f(x)=[x]-3

Mathematics
1 answer:
Svetlanka [38]3 years ago
5 0
<h2>Answer:</h2>

Shown below

<h2>Step-by-step explanation:</h2>

This function represents greatest integer function, which is the most famous of the step functions, which is denoted by [x]. So, this function is defined as:

f(x)=[x] \ the \ greatest \ integer \ less \ than \ or \ equal \ to \ x.. The graph of the greatest integer function has the following characteristics:

a) The domain of the function is the set of all real numbers.

b) The range of the function is the set of all integers.

c) The graph has a y-intercept at (0,0) and x-intercept in the interval [0,1)

d) The graph is constant between each pair of consecutive integers.

e) The graph jumps vertically one unit at each integer value.

f(x)=[x]-3 tells us that the original function has been shifted 3 units downward. So the correct option has been marked in the square red below.

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It is 5:15 pm now. What time will it be in 40 minutes?
zimovet [89]

Answer:

5:55 pm

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Use the empirical rule to solve the problem. The annual precipitation for one city is normally distributed with a mean of 288 in
nadezda [96]

Answer:

z=-1.99

z=1.99

And if we solve for a we got

a=288 -1.99*3.7=214.4

a=288 +1.99*3.7=295.4

And the limits for this case are: (214.4; 295.4)

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the annual precipitation of a population, and for this case we know the distribution for X is given by:

X \sim N(288,3.7)  

Where \mu=288 and \sigma=3.7

The confidence level is 95.44 and the signficance is 1-0.9544=0.0456 and the value of \alpha/2 =0.0228. And the critical value for this case is z = \pm 1.99

Using this condition we can find the limits

z=-1.99

z=1.99

And if we solve for a we got

a=288 -1.99*3.7=214.4

a=288 +1.99*3.7=295.4

And the limits for this case are: (214.4; 295.4)

8 0
3 years ago
Please help, I've been working on this for 30 min now...
MaRussiya [10]

Step 1: Y is equal to \frac{3}{4}x + 2 so where ever you see a y in the equation 3x - 4y = -5 replace it with \frac{3}{4}x + 2

3x - 4(\frac{3}{4}x + 2) = -5

Now all your variables are x and you can solve for x

Step 2: Distribute for to the numbers in the parentheses

3x - 3x - 8 = -5

Step 3: Combine like terms

0x - 8 = -5

0x = 3

Step 4: Isolate x

0x/0 = 3/0

3/0 is undefined

^^^ This means that there is no answer

If you look at the graph below you can see that these lines are parallel and therefore have no intersection point

Hope this helped!

3 0
3 years ago
Let X equal the number of flips of a fair coin that are required to observe
Nataliya [291]

Answer:

(a) I attached a photo with the diagram.

(b) f(x) = \big( \frac{1}{2} \big)^{x-1}

(c) 1/4

(d) 4

(e) k^2/2

Step-by-step explanation:

(a) I attached a photo with the diagram.

(b) The easiest way to think about this part is in terms of combinatorics. Think about it like this.

To begin with, look at the three each level of the three represents a possible outcome of throwing the coin n-times. If you throw the coin 3 times at the end in total there are 8 possible outcomes. But The favorable outcomes are just 2.

 1  - Your first outcome is HEADS and all the others are different except the last one.

 2  - Your first outcome is TAILS and all the others are different except the last one.

Therefore the probability of the event is

f(x) = P(X=x) = \frac{2}{2^x}  = \frac{1}{2^{x-1}} = \big( \frac{1}{2} \big)^{x-1}

(c)

P(X = 0) = 0 because it is not possible to have two consecutive tails or heads.

E(X > 4) = 1 - P(X \leq 3) = 1 - ( P(X = 0 ) + P(X = 1) + P(X = 2) + P(X = 3))\\= 1 - ( \frac{1}{2} + \frac{1}{4} ) = \frac{1}{4}

(d)

Remember that this is a geometric distribution therefore

E[X] = p/(1-p), in this case  p = 1/2 so E[X] = 1 and

E[X+1]^2 = ( E[X] +1 )^2  = (1+1)^2 = 2^2 = 4

Also

(e)

This is a geometric distribution so its variance is

Var(X) = \frac{1-p}{p^2} = 1/2 / 1/4 = 1/2

And using properties of variance

Var(kX - k ) = Var(kX) = k^2 var(X) = k^2 /2

3 0
4 years ago
(6 2/5 + 1 4/5) + 3 1/5
Pie
The answer (im pretty sure) is: 10 2/5
6 0
3 years ago
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