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2 years ago

This word is used to describe an event which is almost, but not quite, certain to occur.

Mathematics

2 answers:

elixir [45]2 years ago

8 0

The answer is Probabilty. Hope it helps!

babunello [35]2 years ago

4 0

I think the answer is "probably"

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A new baby is born in France the newspaper reports the baby's mass to be 1.5×10³ kg

Ira Lisetskai [31]

The question only supplied us with the mass of the new baby which is:

$1.5\cdot10^3\operatorname{kg}\Rightarrow1500\operatorname{kg}$

This mass is too bizzarre & unrealistic for

4 0

1 year ago

Please help me!! You must be good in math.

ivolga24 [154]

It is SAS because
1) it is given that sides are equal (AB=CD)
2)it is also given that angles B and C are equal
3) BC is a common side so it is equal in both triangles.

answer: SAS

8 0

3 years ago

Use complete sentences to describe the relationship between sets A and B if A ⊆ B.

Alex17521 [72]

Answer:

Step-by-step explanation:

A is a subset of B while B is the universal set.

AB = A U B

A is a subset of B. That is A is contained in B.

5 0

3 years ago

Read 2 more answers

Write out the first four terms of the series to show how the series starts. Then find the sum of the series or show that it dive

Nostrana [21]

Answer:

The first four terms of the series are

$(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})$

$\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$ = 14.25

Step-by-step explanation:

We know that

Sum of convergent series is also a convergent series.

We know that,

$\sum_{k=0}^\infty a(r)^k$

If the common ratio of a sequence |r| <1 then it is a convergent series.

The sum of the series is $\sum_{k=0}^\infty a(r)^k=\frac{a}{1-r}$

Given series,

$\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$

$=(9+3)+(\frac97+\frac35)+(\frac9{7^2}+\frac3{5^2})+(\frac9{7^3}+\frac3{5^3})+.......$

The first four terms of the series are

$(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})$

Let

$S_n=\sum_{n=0}^\infty \frac{9}{7^n}$ and $t_n=\sum_{n=0}^\infty \frac{3}{5^n}$

Now for $S_n$ ,

$S_n=9+\frac97+\frac{9}{7^2}+\frac9{7^3}+.......$

$=\sum_{n=0}^\infty9(\frac 17)^n$

It is a geometric series.

The common ratio of $S_n$ is $\frac17$

The sum of the series

$S_n=\sum_{n=0}^\infty \frac{9}{7^n}$

$=\frac{9}{1-\frac17}$

$=\frac{9}{\frac67}$

$=\frac{9\times 7}{6}$

=10.5

Now for $t_n$

$t_n= 3+\frac35+\frac{3}{5^2}+\frac3{5^3}+.......$

$=\sum_{n=0}^\infty3(\frac 15)^n$

It is a geometric series.

The common ratio of $t_n$ is $\frac15$

The sum of the series

$t_n=\sum_{n=0}^\infty \frac{3}{5^n}$

$=\frac{3}{1-\frac15}$

$=\frac{3}{\frac45}$

$=\frac{3\times 5}{4}$

=3.75

The sum of the series is $\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$

= $S_n+t_n$

=10.5+3.75

=14.25

4 0

3 years ago

Can someone help me with this two column proofs? -this one is confusing to me

Shalnov [3]

Perpendicular bisectors have a particular property: if AB is a perpendicular bisector of CD, then every point lying on AB has the same distance from C and D.

In your case, we have that every point lying on AC has the same distance from B and D.

So, in particular, we have EB=ED, because E lies on AC.

Moreover, since AC is a perpendicular bisector, it is the height of the triangle (if we choose BD as base), and it bisects BD: this means that the triangle is isosceles, so AD=AB.

This means that triangles ABE and ADE have:

AD=AB because ABD is isosceles
EB=ED because AC is the perpendicular bisector of BD
AE in common

So, their sides are all equal, and thus they are congruent.

6 0

3 years ago