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LiRa [457]
4 years ago
7

A line plot has a range of 4, from 1 to 5, with 5 modes. How would you describe the graph?

Mathematics
2 answers:
weqwewe [10]4 years ago
4 0

Answer:

Each column will be the same height.

Step-by-step explanation:

Mode refers to the most occurring number, and there are five within a data set that is 4 wide, so there will be 5 columns of equal length.

My name is Ann [436]4 years ago
4 0
Each colum willl be the same heights
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It takes a hospital employee 24 minutes to drive from home to the hospital, a distance of 16 miles. What is the employee’s avera
zlopas [31]

Answer:

r = 40 miles / hour

Step-by-step explanation:

<em><u>Formula</u></em>

d = r * t

<em><u>Givens</u></em>

d = 16 miles

t = 24 minutes = 24/ 60 minutes in an hour = 0.4 hours

r = ???

<em><u>Substitute and solve</u></em>

16 miles = r * 0.4 hours  Divide by 0.4

16/0.4 = r   Switch sides

r = 16/0.4

r = 40 miles / hour


6 0
3 years ago
Which expression is equivalent to 9(-3f - 1) - 2f?​
Zigmanuir [339]

Answer:

-29f-9

Step-by-step explanation:

7 0
3 years ago
The half-life of a certain radioactive substance is 45 days. There are 6.2 grams present initially. On what day
kvasek [131]

Answer:

There will be less than 1 gram of the radioactive substance remaining by the elapsing of 118 days

Step-by-step explanation:

The given parameters are;

The half life of the radioactive substance = 45 days

The mass of the substance initially present = 6.2 grams

The expression for evaluating the half life is given as follows;

N(t) = N_0 \left (\dfrac{1}{2} \right )^{\dfrac{t}{t_{1/2}}

Where;

N(t) = The amount of the substance left after a given time period = 1 gram

N₀ = The initial amount of the radioactive substance = 6.2 grams

t_{1/2} = The half life of the radioactive substance = 45 days

Substituting the values gives;

1 = 6.2 \left (\dfrac{1}{2} \right )^{\dfrac{t}{45}

\dfrac{1}{6.2}  =  \left (\dfrac{1}{2} \right )^{\dfrac{t}{45}

ln\left (\dfrac{1}{6.2} \right )  =  {\dfrac{t}{45} \times ln \left (\dfrac{1}{2} \right )

t = 45 \times \dfrac{ln\left (\dfrac{1}{6.2} \right ) }{ln \left (\dfrac{1}{2} \right )} \approx 118.45 \ days

The time that it takes for the mass of the radioactive substance to remain 1 g ≈ 118.45 days

Therefore, there will be less than 1 gram of the radioactive substance remaining by the elapsing of 118 days.

3 0
3 years ago
PLEASE PLEASE HELP!!
HACTEHA [7]
The answer is 1, 4, and 5 because to be linear, they have to make up a straight line together.
4 0
3 years ago
Can someone help me with this
SOVA2 [1]
X is how much it costs for one hour, so when you multiply it by 3 1/2, you get the amount of money charged for 3 and a half hours.
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4 years ago
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