Is the relationship shown by the data linear? If so, model the data with an equation. X:1,5,9,13 Y:5,10,15,20
1 answer:
Let
A (x1,y1)-------> (1,5)
B (x2,y2)------> (5,10)
C (x3,y3)------> (9,15)
D (x4,y4)------> (13,20)
find the slope
AB
m=(10-5)/(5-1)-----> m=5/4
find the slope
CB
m=(15-10)/(9-5)-----> m=5/4
find the slope
DC
m=(20-15)/(13-9)-----> m=5/4
so
<span>the relationship shown is linear
</span>
<span>to model the data with an equation find the equation of a line
</span>with m=5/4 and point A (1,5)
y-y1=m*(x-x1)-----> y-5=(5/4)*(x-1)----> y=1.25x-1.25+5-----> y=1.25x+3.75
the equation to model data is
y=1.25x+3.75
see the attached figure
A (x1,y1)-------> (1,5)
B (x2,y2)------> (5,10)
C (x3,y3)------> (9,15)
D (x4,y4)------> (13,20)
find the slope
AB
m=(10-5)/(5-1)-----> m=5/4
find the slope
CB
m=(15-10)/(9-5)-----> m=5/4
find the slope
DC
m=(20-15)/(13-9)-----> m=5/4
so
<span>the relationship shown is linear
</span>
<span>to model the data with an equation find the equation of a line
</span>with m=5/4 and point A (1,5)
y-y1=m*(x-x1)-----> y-5=(5/4)*(x-1)----> y=1.25x-1.25+5-----> y=1.25x+3.75
the equation to model data is
y=1.25x+3.75
see the attached figure
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Answer:
0.9115 is the proportion of paint cans that contain less than 5.54 ml of the dye.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 5 ml
Standard Deviation, σ = 0.4 ml
We are given that the distribution of amount of dye dispensed is a bell shaped distribution that is a normal distribution.
Formula:
P(paint cans contain less than 5.54 ml of the dye)
Calculation the value from standard normal z table, we have,
0.9115 is the proportion of paint cans that contain less than 5.54 ml of the dye.