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Tresset [83]
3 years ago
9

Find the possibability of 0 heads when flipping 3 coins

Mathematics
2 answers:
weqwewe [10]3 years ago
7 0
The possibility of 0 heads when flipping 3 coins is 1/4
aleksandrvk [35]3 years ago
4 0
There are 3 coins, right and if you had 1 coin the probability of getting heads would be 1/2, so if you have three coins the probability of not flipping heads at all is 3/6
Hope i helped give brainliest!
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Write an explicit formula for an, the nth term of the sequence 4, 20,100. WILL GET BRAINLIEST!!!!!​
anygoal [31]

Answer:

aₙ = 4(5)ⁿ⁻¹

Explanation:

To identify a geometric series: second term = √(first term * third term)

Here given:

  • 1st term: 4
  • 2nd term: 20
  • 3rd term: 100

check ============

20 = √(100*4)

20 = √400

20 = 20  Hence its a geometric term.

Find common difference (r):

20/4 = <u>5</u>       or     100/20 = <u>5</u>

Geometric term formula:
aₙ = a₁(r)ⁿ⁻¹               [where a1 is the 1st term, r is common difference]

<u>insert values found</u>

aₙ = 4(5)ⁿ⁻¹ ......this is the nth term.

5 0
3 years ago
Read 2 more answers
HELP LOOK AT PICTURE
mixas84 [53]

Answer:

It increases.

Step-by-step explanation:

Usually putting a variable in front of a number is multiplying, and as the number your multiplying with increases, the value increases.

6 0
4 years ago
A tank containing 64 gallons lost 6 1/4% of this through leakage how much is remaining is the tank
jolli1 [7]

if we take 64 to be the 100%, how much is 6¼% off of it?

\bf \begin{array}{ccll} amount&\%\\ \cline{1-2} 64&100\\ x&6\frac{1}{4} \end{array}\implies \cfrac{64}{x}=\cfrac{100}{6\frac{1}{4}}\implies \cfrac{64}{x}=\cfrac{\frac{100}{1}}{\frac{25}{4}}\implies \cfrac{64}{x}=\cfrac{100}{1}\cdot \cfrac{4}{25} \\\\\\ \cfrac{64}{x}=16\implies 64=16x\implies \cfrac{64}{16}=x\implies 4=x \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{it had}}{64}-\stackrel{\textit{leakage}}{4}\implies \stackrel{\textit{remaining}}{60}

4 0
3 years ago
Read 2 more answers
If the square root of p2 is an integer greater than 1, which of the following must be true? I. p2 has an odd number of positive
Grace [21]

Answer:

Option | and Option || is True

Step-by-step explanation:

Given:

If the square root of p^{2} is an integer greater than 1,

Lets p = 2, 3, 4, 5, 6, 7..........

Solution:

Now we check all option for p^{2}

Option |.

p^{2} has an odd number of positive factors.

Let p=2

The positive factor of 2^{2}=4=1,2,4

Number of factor is 3

Let p=3

The positive factor of 3^{2}=9=1,3,9

Number of factor is 3

So, p^{2} has an odd number of positive factors.

Therefore, 1st option is true.

Option ||.

p^{2} can be expressed as the product of an even number of positive prime factors

Let p=2

The positive factor of 2^{2}=4=1,2,4

4=2\times 2

Let p=3

The positive factor of 3^{2}=9=1,3,9

9=3\times 3

So, it is expressed as the product of an even number of positive prime factors,

Therefore, 2nd option is true.

Option |||.

p has an even number of positive factors

Let p=2

Positive factor of 2=1,2

Number of factor is 2.

Let p=4

Positive factor of 4=1,2,4

Number of factor is 3 that is odd

So, p has also odd number of positive factor.

Therefore, it is false.

Therefore, Option | and Option || is True.

Option ||| is false.

4 0
3 years ago
Match the following reasons with the statement givem
ANEK [815]

Answer:

AAS(Angle-Angle-Side) postulate states that if two angles and the non-included side one triangle are congruent to two angles and the non-included side of another triangle, then the two triangles are congruent

In triangle RAS and triangle QAT

\angle R =\angle Q   [Angle]

AS =AT    [Side]                                       [Given]

By Base Angle Theorem states that in an isosceles triangle(i.e, AST), the angles opposite the congruent sides(AS =AT) are congruent.

⇒ \angle 5= \angle 6      [By base ∠'s of isosceles triangle are equal]

By definition of supplementary angles, if two Angles are Supplementary when they add up to 180 degrees.

\angle 4, \angle 5 are supplementary and \angle 6, \angle 7 are supplementary.

⇒\angle 4+ \angle 5 =180^{\circ} and

\angle 6+ \angle 7 =180^{\circ}  

Two \angle 's supplementary to equal \angle 's

\angle 4+ \angle 5 =\angle 6+ \angle 7

Since,  \angle 5 =\angle 6  

then, we get;

\angle 4 =\angle 7  [Angle]

then, by AAS postulates,

\triangle RAS \cong \triangle QAT

By CPCT[Corresponding Part of Congruent Triangles are equal]

RS = QT              Hence Proved!

7 0
3 years ago
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