Write the set of points from -6 to 0 but excluding -4 and 0 as a union of intervals
First we take the interval -6 to 0. In that -4 and 0 are excluded.
So we split the interval -6 to 0.
Start with -6 and go up to -4. -4 is excluded so we break at -4. Also we use parenthesis for -4.
Interval becomes [-6,-4) . It says -6 included but -4 excluded.
Next interval starts at -4 and ends at 0. -4 and 0 are excluded so we use parenthesis not square brackets
(-4,0)
Now we take union of both intervals
[-6,-4) U (-4,0) --- Interval from -6 to 0 but excluding -4 and 0
Answer:
YUGOYOTFOTFTYFIYT
Step-by-step explanation:
Answer:
Inverse of g(x)=x+15/3
Step-by-step explanation:
g(x)=3x-15
Let, g(x) be y
y=3x-15
Interchange x and y
x=3y-15
x+15=3y
y=x+15/3
Answer:
Step-by-step explanation:
Both 115 and 145 mph are above the mean. Draw a normal curve and mark these speeds. 115 mph is 1 standard deviation above the mean; 130 would be 2 standard deviations above the mean; and 145 would be 3 s. d. above it.
We need to find the area under the standard normal curve between 115 and 145. This is equivalent to the area under the standard normal curve between z = 1 and z = 3.
I used my TI-83 Plus calculator's DISTR function "normalcdf(" to calculate this area: normalcdf(1, 3) = 0.1573.
The area between z = 1 and z = 3 is 0.1573. In other words, the percentage of serves that were between 115 and 145 mph was 15.73%.
Answer:
( J - 6 ) × (J + 6 )
Step-by-step explanation:
j^2 -6^2
( J - 6 ) × ( j + 6 )
