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baherus [9]
4 years ago
14

Adam has 1 2/5 meters of rope that he wants to cut into pieces that are each 1/5 meters long. How many pieces can he cut?

Mathematics
2 answers:
s344n2d4d5 [400]4 years ago
6 0
1. 1 2/5 is 7/5 so he can cut 7
2. Is 8 times three that equals 24 and 24 plus 3 is 27 so 27/8
natita [175]4 years ago
5 0
First one is 7 second one is 8/29 beacause of the eqution
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500 tickets must be sold. 54 students have sold nine tickets each. How many tickets are left to be sold?
Leona [35]

Answer:

14 tickets. Equation form : 500- 486=x.

Step-by-step explanation:

Variables are always the unknown. What you don't know from the given information is the number of tickets left to be sold. 54*9=486. So x is equivalent to 500-486 aka 14

6 0
3 years ago
Read 2 more answers
PLEASE HELP ME WITH THIS<br><br>​
Vladimir [108]

Answer:

A. (0, 60)

B. 15x

Step-by-step explanation:

The initial value is the same as the y-intercept, which is 60 or (0, 60).

The rate of change would be the same as the slope. To find the slope, you do rise/run. You move 15 units up, and 1 to the right. That would be 15x.

Hope this helps!

5 0
3 years ago
Help pleaseeee NO LINKSSSS
Rainbow [258]

Answer:

Step-by-step explanation:

Radius=r. Diameter=d. C=circumference. A=area

Starting from the center of the circle any line drawn to any point on the circumference of the circle is the radius, if the line is drawn from one point on the circumference to another point on the circumference through the centre of the circle that is the diameter. The circumference of the circle is the perimeter of the circle.

Formulas;

r=d÷2

C=2πr

A=πr²

π=\frac{22}{7} or 3.142

4).

a.)r=d÷2

=4.8cm÷2

=2.4cm

b.)d=4.8cm

c.)C=2πr

=2×3.142×2.4cm

=15.08cm

d.) A=πr²

=3.142×(2.4)²

=3.142×5.76

=18.09cm² or 18.1cm²

5.

a.)r=d÷2

=7÷2

=3.5cm

b.)d=7cm

c.)C=2πr

=2×3.142×3.5

=21.99cm or 21.1cm

d.)A=πr²

=3.142×(3.5)²

=3.142×12.25

=38.49cm²

6.)

a.)r=d÷2

=14÷2

=7cm

b.)d=14cm

c.)C=2πr

=2×3.142×7

=43.99cm

d.)A=πr²

=3.142×(7)²

=3.142×49

=153.96cm²

7 0
3 years ago
Write the slope-intercept form of the given line. Include your work in your final answer. Type your answer in the box provided o
Rainbow [258]
Count the rise and run to find the slope, 1. Next, find the y-intercept-(-1). Now make the slope-intercept equation-  y = 1x - 1. Double check, put x = 2 in the equation-  y = 1(2) - 1 (=) y = 2 - 1 (=) y = 1.

So, the slope-intercept equation for this line is y = 1x - 1.
8 0
4 years ago
An article in Technometrics (Vol. 19, 1977, p. 425) presents the following data on the motor fuel octane ratings of several blen
Slav-nsk [51]

Answer:

Median = 90.4

Q_1 = 88.6

Q_3 = 92.2

Step-by-step explanation:

Given

The above data

Required

- A stem and leaf display

- The median

- The quartiles

First, determine the range of the data

Smallest = 83.4

Highest = 100.3

Next, group each dataset base on common whole numbers.

So, we have:

83.4

84.3\ 84.3

85.3

86.7\ 86.7\ 86.7

87.4\ 87.5\ 87.6\ 87.7\ 87.8\ 87.9

88.2\ 88.3\ 88.3\ 88.3\ 88.4\ 88.5\ 88.5\ 88.6\ 88.6\ 88.7\ 88.9

89.0\ 89.2\ 89.3\ 89.3\ 89.6\ 89.7\ 89.8\ 89.8\ 89.9\ 89.9

90.0\ 90.1\ 90.1\ 90.1\ 90.3\ 90.4\ 90.4\ 90.4\ 90.5\ 90.6\ 90.7\ 90.8\ 90.9

91.0\ 91.0\ 91.0\ 91.1\ 91.1\ 91.1\ 91.2\ 91.2\ 91.2\ \ 91.5\  91.6\ 91.6\ 91.8\ 91.8

92.2\ 92.2\ 92.2\ 92.3\ 92.6\ 92.7\ 92.7\ 92.7

93.0\ 93.2\ 93.3\ 93.3\ 93.4\ 93.7

94.2\ 94.2\ 94.4\ 94.7

96.1\ 96.5

98.8

100.3

Next, we construct the stem and leaf plot.

The whole numbers will be the stem while the decimal parts will be the leaf.

So, we have:

\begin{array}{ccc}{Stem} & {} & {Leaf} & {83} & {|} & {.4} & {84} & {|} & {.3\ .3} & {85} & {|} & {.3} & {86} & {|} & {.7\ .7\ .7} & {87} &{|} & {.4\ .5\ .6\ .7\ .8\ .9} & {88} & {|} & {.2\ .3\ .3\ .3\ .4\ .5\ .5\ .6\ .6\ .7\ .9} &{89} & {|} & {.0\ .2\ .3\ .3\ .6\ .7\ .8\ .8\ .9\ .9} & {90} & {|} &{.0\ .1\ .1\ .1\ .3\ .4\ .4\ .4\ .5\ .6\ .7\ .8\ .9} & {91} &{|}&{.0\ .0\ .0\ .1\ .1\ .1\ .2\ .2\ .2\ .5\ .6\ .6\ .8\ .8} &{92} &{|} &{.2\ .2\ .2\ .3\ .6\ .7\ .7\ .7} \ \end{array}

  \begin{array}{ccc} {93} & {|} & {.0\ .2\ .3\ .3\ .4\ .7} & {94} &{|} & {.2\ .2\ .4\ .7} &{96} & {|} & {.1\ .5} & {98} & {|} & {.8} & {100} &{|} &{.3} \ \end{array}

From the above plot,

n = 83

The median is calculated as:

Median = \frac{n+1}{2}th

Median = \frac{83+1}{2}th

Median = \frac{84}{2}th

Median = 42nd

i.e. the median is at the 42nd position.

From the above stem and leaf plot.

The 42nd position is at stem 90 and the leaf  .4

So the median is:

Median = 90.4

The lower quartile (Q) is calculated as:

Q_1 = \frac{n+1}{4}th

Q_1 = \frac{83+1}{4}th

Q_1 = \frac{84}{4}th

Q_1 = 21st

i.e. the lower quartile is at the 21st position.

From the above stem and leaf plot.

The 42nd position is at stem 88 and the leaf .6

So the lower quartile is:

Q_1 = 88.6

The upper quartile (Q3) is calculated as:

Q_3 = 3 * \frac{n+1}{4}th

Q_3 = 3 * \frac{83+1}{4}th

Q_3 = 3 * \frac{84}{4}th

Q_3 = 3 * 21th

Q_3 = 63rd

i.e. the upper quartile is at the 63rd position.

From the above stem and leaf plot.

The 63rd position is at stem 92 and the leaf .2

So the upper quartile is:

Q_3 = 92.2

8 0
3 years ago
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