Answer:
The answer to your question is below
Step-by-step explanation:
14) sin Ф = 32/33
sin Ф = 0.9697
Ф = 75.86°
15) sin Ф = 16/36
sin Ф = 0.444
Ф = 26.39°
16) sin Ф = 29/36
sin Ф = 0.806
Ф = 53.66°
17) cos Ф = 11/34
cos Ф = 0.324
Ф = 71.12°
18) sin Ф = 21/40
sin Ф = 0.525
Ф = 31.67°
19) cos Ф = 6/14
cos Ф = 0.429
Ф = 64.62°
20) cos Ф = 38/39
cos Ф = 0.974
Ф = 13°
The correct statement about the data collected by Ms. Pearson is that there is no association between a student's absences and the final average grades.
<h3>When do variables have a linear relationship?</h3>
The equation that represents a linear relationship is: a + bx
Where x represents the rate of increase. Thus, for linear equations, the functiion increases by a constant term.
Looking at the table, the average final grade does not increase by a constant term.
To learn more about linear functions, please check: brainly.com/question/26434260
I'm doing 3
For 3, using a table is very similar to a double number because the numbers are matching up in both ways.
On the bottom of a double number line we have like, for example,-- the bottom of the double number line would have batches. And its 1, 2, 3 ,4 , etc.
And on a table, it would be the same, the numbers on both diagrams have the same methods, have same way of lining things up but they're just drawn differently.
Hope this helped!
Given:
The two functions are:
To find:
The type of transformation from f(x) to g(x) in the problem above and including its distance moved.
Solution:
The transformation is defined as
.... (i)
Where, a is horizontal shift and b is vertical shift.
- If a>0, then the graph shifts a units left.
- If a<0, then the graph shifts a units right.
- If b>0, then the graph shifts b units up.
- If b<0, then the graph shifts b units down.
We have,
The function g(x) can be written as
...(ii)
On comparing (i) and (ii), we get
Therefore, the type of transformation is translation and the graph of f(x) shifts 2 units up to get the graph of g(x).
If a 1L IV contains 60meq of kcal and the IV was discontinued after 400ml has infused, then the amount of IV received by the patient is 24meq of kcal.
Calculation for the Amount of IV
It is given that,
1L IV contains 60meq of kcal
⇒ 1000 mL of IV contains 60meq of kcal
⇒ 1 mL of IV will contain 60 / 1000 meq of kcal
As per the question,
The amount of IV infused in the patient = 400 mL
⇒ The amount of IV received by the patient in meq of kcal = 400 × (60 / 1000)
= 4 × 6
= 24 meq of kcal
Hence, the patient receives 24meq of kcal amount of IV.
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