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Aleksandr-060686 [28]
2 years ago
5

According to the graph, what is the value of the constant in the equation below?

Mathematics
2 answers:
vazorg [7]2 years ago
7 0
The point (1, 3.2) can be used to advantage, for it tells you
  3.2 = Constant×1

The appropriate selection is
  D.  3.2

GarryVolchara [31]2 years ago
7 0

Answer:  The correct option is (D) 3.2.

Step-by-step explanation:  We are given a graph with respect to which we are to find the value of the constant in the following equation.

Height = constant × Width                       (i)

The graph shows a linear relation between the height and width of some figure.

The width is plotted on the X-axis and the corresponding height is plotted on the Y-axis.

The following four points are plotted n the line:

(x, y) = (0.5, 1.6), (1, 3.2), (2, 6.4) and (3, 9.6).

Since width (x-value) and height (y-value) satisfies equation (i), so we have

1.6=\textup{constant}\times 0.5\\\\\Rightarrow \textup{constant}=\dfrac{1.6}{0.5}=\dfrac{16}{5}=3.2.

Also, we can see that

\dfrac{3.2}{1}=\dfrac{6.4}{2}=\dfrac{9.6}{3}=3.2.

Thus, the value of the constant is 3.2.

Option (D) is correct.

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32.1. It has the least values in the decimal place. Overall, it's 31.00, but the least decimal value is 32.1

Step-by-step explanation:

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2 years ago
You flip a coin and then roll a 6-sided number cube (a die).
expeople1 [14]

Answer:

a)  No, it does not matter whether you roll the die or flip the coin first, as these two events are <u>independent</u> of each other, which means they do not affect each other.

b) Yes.

  • Let event 1 be flipping a coin and event 2 be rolling a die.
  • Let event 1 be rolling a die and event 2 be flipping a coin.

The likelihood that any outcome will occur will not change, as the events are independent.

c) see attached

d)   12 outcomes  (H = head, T = tail, numbers represent the value of the die)

H 1           T 1

H 2          T 2

H 3          T 3

H 4          T 4

H 5          T 5

H 6          T 6

e)  

\sf Probability\:of\:an\:event\:occurring = \dfrac{Number\:of\:ways\:it\:can\:occur}{Total\:number\:of\:possible\:outcomes}

\implies \sf P(even)=\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}=\dfrac{3}{6}=\dfrac{1}{2}

\implies \sf P(head)=\dfrac{1}{2}

\implies \sf P(even)\:and\:P(head)=\dfrac{1}{2} \times \dfrac{1}{2}=\dfrac{1}{4}

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