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3 years ago

Need help finding the domain and range

Mathematics

1 answer:

pogonyaev3 years ago

3 0

Here are a few things you'll need to know for this question:

Domain: The list of x-values that are possible on a line.
Range: The list of y-values that are possible on a line.
Interval Notation: Shows the domain/range using the endpoints. Brackets mean that the endpoint is included, parentheses mean the endpoint is excluded. Ex: (2,10]. 2 is excluded, 10 is included.
Closed Circles: The endpoint is included.
Open Circles: The endpoint is excluded.

So firstly, let's look at the domain. We see that there is a closed circle at x = -2 and an open circle at x = 5. Using what we know, the interval notation of the domain is [-2,5).

Next, let's look at the range. We see that there is a closed circle at y = -5 and an open circle at y = 2. Using what we know, the interval notation of the range is [-5,2).

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Need help thank you very much

Zarrin [17]

Answer:

a) - 9/2 or -0,45

b) - 3/5 or -0,6

c) -4

d) 1

e) -2

d) 0

Step-by-step explanation:

4 0

3 years ago

I’m stuck and I would really like your help.

Neporo4naja [7]

Answer:

Answer: 1/12

Step-by-step explanation:

2/5x-4=8-3/5x.

We want to tank x out of the denominator by multiplying x on both sides.

2/5-4x=8x-3/5

Now we simplify

-12x=-1

x=1/12

Put x in the equation to check it

6 0

2 years ago

Read 2 more answers

A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.

nika2105 [10]

Answer:

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

$\bar X=2.55$ represent the sample mean for the late time for a flight

$\mu$ population mean

$\sigma=12$ represent the population deviation

n=76 represent the sample size

Confidence interval

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

The confidence interval for the true mean is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The Confidence level given is 0.95 or 95%, th significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ . If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got $z_{\alpha/2}=1.96$