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3 years ago

The graph shows the absolute value parent function. Which statement best describes the function?

Mathematics

1 answer:

Lubov Fominskaja [6]3 years ago

8 0

Answer:

b. the function is always positive

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Jake has finished pouring 4,865 cubic feet out of a total of 27,000 cubic feet of concrete. What percent of the job has he compl

anzhelika [568]

The total is the 100% and that is 27000, the question is what percentage is 4865
We can use a simple rule of three, direct proportion to solve:
27000 ft^3 --------> 100%
4865 ft^3 ---------> x
x =(4865)(100)/27000
x = 18.02
therefore 4865 ft^3 is 18.02% of 27000 ft^3

6 0

3 years ago

-13 +n + 5 + 6 = 5 + 4n-4

natta225 [31]

Answer:

n=-1

Step-by-step explanation:

-13+n+5+6=5+4n-4

n-2=4n+1

-2=3n+1

-3=3n

-1=n

4 0

3 years ago

Use a table of values with at least 5 values to graph the following function:

34kurt

The given expression :

$y=(\frac{1}{2})^x$

For coordinates:

put x = 0 then :

$\begin{gathered} y=(\frac{1}{2})^0 \\ y=1 \end{gathered}$

Coordinate : (x, y) = (0, 1)

Put x= 1 and simplify :

$\begin{gathered} y=(\frac{1}{2})^1 \\ y=\frac{1}{2} \\ y=0.5 \end{gathered}$

Coordinate : (x, y) = ( 1, 0.5)

Put x = (-2) and simplify :

$\begin{gathered} y=(\frac{1}{2})^{-2} \\ y=\frac{1^{-2}}{2^{-2}} \\ y=\frac{2^2}{1^2} \\ y=2^2 \\ y=4 \end{gathered}$

Coordinate : (x, y) = ( -2, 4)

Put x = (-3) and simplify :

$\begin{gathered} y=(\frac{1}{2})^{-3} \\ y=\frac{1^{-3}}{2^{-3}} \\ y=\frac{2^3}{1^3} \\ y=2^3 \\ y=8 \end{gathered}$

Coordinate : (x, y) = (-3, 8)

Substitute x = (-1) and simplify :

$\begin{gathered} y=(\frac{1}{2})^x \\ y=(\frac{1}{2})^{-1} \\ y=\frac{1^{-1}}{2^{-1}} \\ y=\frac{2}{1} \\ y=2 \end{gathered}$

Coordinate : (x, y) = ( -1, 2)

So, the coordinates are :

The graph is :

4 0

2 years ago

Which statements are true about the interquartile range? Select all that apply. Subtract the lowest and highest values to find t

katen-ka-za [31]

Answer:

The correct options are:

Interquartile ranges are not significantly impacted by outliers.

Lower and upper quartiles are needed to find the interquartile range.

The data values should be listed in order before trying to find the interquartile range.

The option Subtract the lowest and highest values to find the interquartile range is incorrect because the difference between lowest and highest values will give us range.

The option A small interquartile range means the data is spread far away from the median is incorrect because a small interquartile means data is nor spread far away from the median

6 0

3 years ago

Read 2 more answers

CAN SOMEONE PLEASE HELP ME!!!

julia-pushkina [17]

m∠HKF = 87°

Solution:

Given data:

m(ar JF) = 135° and m(ar GH) = 42°

GJ is the diameter of the circle.

Angle in a diameter is 180°.

⇒ m(ar JF) + m(ar FG) = 180°

⇒ 135° + m(ar FG) = 180°

Subtract 135° from both sides of the equation.

⇒ m(ar FG) = 45°

⇒ m(ar FGH) = m(ar FG) + m(ar GH)

⇒ m(ar FGH)= 45° + 42°

⇒ m(ar FGH) = 87°

The measure of the central angle is congruent to the measure of the intercepted arc.

⇒ m∠HKF = m(ar FGH)

⇒ m∠HKF = 87°

7 0

4 years ago