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sweet-ann [11.9K]
3 years ago
13

A data set of with whole numbers has a low of 20 and a high value of 89

Mathematics
1 answer:
Vikentia [17]3 years ago
7 0

Answer:

The range is 69

Step-by-step explanation:

We have that the least value of the data set is 20 and the highest value is 89.

From the available information, we can only find the range of the data set.

The range is the highest value minus the least value.

This implies that, the range is 89-20=69

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7/10 write the number as hundredths in a fraction form and decimal form
klemol [59]

7/10 as hundredths decimal: .07

7/10 as fraction: 7/100

7 0
3 years ago
3 Latrell made a large pizza. He ate 10 of the
nordsb [41]

Answer:

10 if there is an error in the question (and of the pizza for dinner.)

or 20

Step-by-step explanation:

Just add what he ate for lunch and dinner

It never said how he cut his pizzas. Also it never said if he finished it or not.

7 0
3 years ago
Use the functions and the scenario to solve the problem.
SashulF [63]

The function f(x) - g(x) is an illustration of a composite function

Leanne's end result of f(x) - g(x) = -1/2x + 9 is correct

<h3>How to estimate the student's answer?</h3>

The functions are given as:

f(x) = 1/2x + 6

g(x) = x - 3

The function f(x) - g(x) is calculated using:

f(x) - g(x) = 1/2x + 6 - x + 3

Collect like terms

f(x) - g(x) = 1/2x - x + 6  + 3

Evaluate the like terms

f(x) - g(x) = -1/2x + 9

By comparing the above solution to the students' response, we can see that:

Leanne's end result of f(x) - g(x) = -1/2x + 9 is correct

Read more about composite functions at:

brainly.com/question/13502804

3 0
1 year ago
Find the equation of the line that is perpendicular to the line y = (-1/3)x -1 and passes through the point (1, 5)?
Anit [1.1K]

bearing in mind that perpendicular lines have negative reciprocal slopes, so


\bf \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}~\hspace{10em}\stackrel{slope}{y=\stackrel{\downarrow }{-\cfrac{1}{3}}x-1} \\\\[-0.35em] ~\dotfill


\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-\cfrac{1}{3}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{3}{1}}\qquad \stackrel{negative~reciprocal}{+\cfrac{3}{1}\implies 3}}


so we're really looking for a line whose slope is 3 and runs through (1,5)


\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{5})~\hspace{10em} slope = m\implies 3 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-5=3(x-1) \\\\\\ y-5=3x-3\implies y=3x+2

4 0
3 years ago
Explain why the net shown here cannot be the net of a cube
Schach [20]

Answer:

there is an extra square missing

Step-by-step explanation:

5 0
3 years ago
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