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harkovskaia [24]
3 years ago
12

Which value is an input of the function

Mathematics
2 answers:
uranmaximum [27]3 years ago
3 0

the answer to this equation is 4

zavuch27 [327]3 years ago
3 0

Answer:

D. 4

Step-by-step explanation:

We have been given graph of a function. We are asked to choose the value, which is input of the given function.

We know that input of a function is value of the independent variable. Upon looking at our given graph, we can see that x is independent variable.

All the values in domain of our a function will be input values.

We can see that domain of our given function is [1,\infty).

Our function is defined for x values that are greater than or equal to 1, so first three options are not correct choice.

Since our function is defined at x=4, therefore, 4 is an input of the given function.

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Distribute and combine like terms.<br> 7(4a+3b-6)-3(3a-2b+4)
andrew11 [14]

Answer:

19a+27b-54

Step-by-step explanation:

First distribute 7 into the first equation. (4a+3b-6) Multiply each coefficient and the constant by 7.

28a+21b-42

Then do the same for -3 and (3a-2b+4).

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4 years ago
A sample of blood pressure measurements is taken for a group of adults, and those values (mm Hg) are listed below. The values ar
Kobotan [32]

Answer:

The coefficient of variation for the systolic measurements is 12.8 %

The coefficient of variation for the diastolic measurements is 16.3 %

Therefore, we can conclude that the systolic measurements are less dispersed as compared to diastolic measurements.

Step-by-step explanation:

The coefficient of variation is the ratio of standard deviation of the data to the mean of the data. It shows the dispersion in the data set.

Coefficient of variation for the systolic measurements:

Mean = μ = (120 + 128 + 156 + 98 + 154 + 122 + 118 + 136 + 128 + 120)/10

Mean = μ = 1280/10

Mean = μ = 128

Standard deviation = σ = \sqrt{\frac{1}{N}\sum_ (x-\mu)^{2} }

Using excel the standard deviation is found to be

Standard deviation = σ = 16.4

Coefficient of variation = σ/μ

Coefficient of variation = 16.4/128

Coefficient of variation = 0.128

Coefficient of variation = 12.8%

Coefficient of variation for the Diastolic measurements:

Mean = μ = (79 + 76 + 75 + 51 + 92 + 87 + 59 + 64 + 72 + 81)/10

Mean = μ = 736/10

Mean = μ = 73.6

Standard deviation = σ = \sqrt{\frac{1}{N}\sum_ (x-\mu)^{2} }

Using excel the standard deviation is found to be

Standard deviation = σ = 12

Coefficient of variation = σ/μ

Coefficient of variation = 12/73.6

Coefficient of variation = 0.163

Coefficient of variation = 16.3%

Conclusion:

We can conclude that the systolic measurements (12.8%) are less dispersed as compared to diastolic measurements (16.3%).

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