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3 years ago

Does this graph represent a function?

Mathematics

1 answer:

Novosadov [1.4K]3 years ago

6 0

Answer:

d or c hope you get it right

Step-by-step explanation:

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75+12.55 = 100 +105. <br><br> I need the Asnwer

kolezko [41]

Step-by-step explanation:

Add 75 to 12.55. the answer would be 87.55.

100+105=205

205-87.55=117.45

5 0

2 years ago

Which relation is a function?

Ksivusya [100]

B. To see if it’s a function, it has to pass the vertical line test. That means draw a vertical line (going up and down) through any point on the graph and it can only go through one point. If it goes through more than 1, it’s not a function. Like in graph A, if you draw a vertical line at x=-2, it will go through two points, same at x=2. For graph b, no matter where you move a vertical line, it only passes through one point at a time, so it passes the vertical line test.

8 0

2 years ago

What is -3y = 12+3x?<br>m=<br>y=

lana66690 [7]

Step-by-step explanation:

-3y=12+3x

-y=4+x

y=-4-x and m which is the gradient is the coefficient of x which is -1, therefore m=-1

4 0

3 years ago

Read 2 more answers

Health insurers are beginning to offer telemedicine services online that replace the common office visit. A company provides a v

Veronika [31]

Answer:

$\text {CI} = (60.54, \: 81.46)\\\\$

Therefore, we are 95% confident that actual mean savings for a televisit to the doctor is within the interval of ($60.54 to $81.46)

Step-by-step explanation:

Let us find out the mean savings for a televisit to the doctor from the given data.

Using Excel,

=AVERAGE(number1, number2,....)

The mean is found to be

$\bar{x} = \$71$

Let us find out the standard deviation of savings for a televisit to the doctor from the given data.

Using Excel,

=STDEV(number1, number2,....)

The standard deviation is found to be

$s = \$ 22.35$

The confidence interval is given by

$\text {confidence interval} = \bar{x} \pm MoE\\\\$

Where the margin of error is given by

$$ MoE = t_{\alpha/2}(\frac{s}{\sqrt{n} } ) $ \\\\$

Where n is the sample of 20 online doctor visits, s is the sample standard deviation and $t_{\alpha/2}$ is the t-score corresponding to a 95% confidence level.