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Alja [10]
3 years ago
11

There is a 20% chance of thunderstorms tomorrow. Describe the likelihood of the event.

Mathematics
2 answers:
Pachacha [2.7K]3 years ago
8 0

Answer:

Step-by-step explanation:

20% = 1/5 or 0.2

There is a 1/5 chance of a thunderstorm.

There really is no clear answer. I would try 0.2

labwork [276]3 years ago
7 0

Answer: The vent is unlikely to happen.

Step-by-step explanation:

Given : There is a 20% chance of thunderstorms tomorrow.

i.e. the probability of thunderstorms tomorrow =0.2

We know that when the probability of any event lies between 0 and 0.5 then the event is said to be unlikely to happen.

Since , the probability of thunderstorms tomorrow is between 0 and 0.5.

Therefore, it is unlikely to have  thunderstorms tomorrow.

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Which of the following Statements are true?
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Answer: Only Statement 1 is True.

Step-by-step explanation:

A box is essentially a rectangular prism, which has 8 corners. Tubes, pipes, and cans are cylinders because the corss-section of them are circular and the connection between the bases are parallel.

8 0
2 years ago
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crimeas [40]
B is the correct answer
8 0
3 years ago
In a diagram, X is the midpoint of Segment VZ. VW =5, and VY =20. Find the coordinates of W,X, and Y.
guapka [62]

solution:

x(a_{x},b_{x})is midpoint of vz\\
a_{x}=\frac{a_{v}+a_{z}}{2}=\frac{-12+22}{2}=\frac{10}{2}=5\\
b_{x}=\frac{b_{v}+b_{z}}{2}=\frac{0+0}{2}=0\\
and,\\
x(5,0)\\
w(a_{w},b_{w}),v_{w} and w is on the right of v.\\
a_{w}-a_{v}=5\\
a_{w}-(-12)=5\\
a_{w}+12=5\\
a_{w}=5-12=-7\\
w(-7,0)\\
y(a_{y},b_{y})=20 and y is on the right of v.\\
a_{y}-a_{v}=20\\
a_{y}-(-12)=20\\
a_{y}+12=20\\
a_{y}=20-12=8\\
y=(8,0)

7 0
3 years ago
Help me on that one please
hoa [83]
The answer is 363:716
8 0
3 years ago
Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth. −6y2 − 9y = −1
Tomtit [17]
-6y^2 - 9y = -1

y= \dfrac{-9+ \sqrt{9^2-4*-6} }{2*6}, \dfrac{-9-\sqrt{9^2-4*-6}}{2*6}


y=  \dfrac{-9+ \sqrt{105} }{12}, \dfrac{-9- \sqrt{105} }{12}
8 0
3 years ago
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