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Annette [7]
3 years ago
14

The probability of a chance event is 0.5, it is __________ to happen when compared to a chance event with a probability of 1.

Mathematics
2 answers:
AysviL [449]3 years ago
6 0

Answer:

less likely

Step-by-step explanation:

0.5 is less then one there for the probability of 0.5 is less then 1.

nignag [31]3 years ago
3 0

Answer:

less likely

Step-by-step explanation:

Probability is between 0 and 1 so 1 is guaranteed to happen and 0 is guaranteed to not happen

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The diameter of a sphere with a volume of 12,348 cubic ft is ___ ft.
aniked [119]

Answer:

108.62ft

Step-by-step explanation:

12348=⁴/3×3.14×r²

r²=12348×¾÷3.14

r²=2,949.36

r=squareroot of 2,949.36

r=54.308

d=2r

d=54.308×2

d=108.616

6 0
2 years ago
What is a common factor for the two fractional terms StartFraction 5 over 8 EndFraction x andStartFraction 11 over 8 EndFraction
Dafna1 [17]

Answer:

\frac{1}{8x}

Step-by-step explanation:

Given two fractional terms \frac{5}{8x}\ and\ \frac{11}{8xy}. Their common factor is a value or function that can go in both fractional terms. The terms can be written as shown.

\frac{5}{8x} = 5 *\frac{1}{8x}

\frac{11}{8xy} = 11 * \frac{1}{8x} * \frac{1}{y}

It can be seen from the both equations that they both have \frac{1}{8x} as one of their factors i.e <em>1/8x is common to both fractional terms</em>. This gives the common factor for the two fractional terms as \frac{1}{8x}

4 0
3 years ago
Read 2 more answers
Hi guys, can anyone help me with this triple integral? Many thanks:)
Crank

Another triple integral.  We're integrating over the interior of the sphere

x^2+y^2+z^2=2^2

Let's do the outer integral over z.   z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

y^2=4-x^2-z^2

x is the inner integral so at this point we conservatively say its zero.  That means y goes from -\sqrt{4-z^2} and +\sqrt{4-z^2}

Similarly the inner integral x goes between \pm-\sqrt{4-y^2-z^2}

So we rewrite the integral

\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz

Let's work on the inner one,

\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz

There's no z in the integrand, so we treat it as a constant.

=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}

So the middle integral is

\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy  

I gotta go so I'll stop here, sorry.

7 0
3 years ago
Read 2 more answers
Dau coroana <br> daca il faceti
Andre45 [30]

Answer:

I give the crown

if you do

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
11) Given that the points (-1, 6), (3, 6), (3, 1), and (-1, 1) are vertices of a rectangle, how much shorter is the width than t
quester [9]

-1 unit is the correct answer.

5 0
3 years ago
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