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IrinaK [193]
3 years ago
7

The mean of a soccer team goals is 6 goals after 4 games. If the team scored 5 goals in the first game, 6 goals in the second ga

me and 6 goals in the third game, how many goals did they score at the 4th game?
Mathematics
2 answers:
Sedaia [141]3 years ago
8 0
7


Mark brainliest please


Hope this helps you
Hitman42 [59]3 years ago
8 0
<h3>Answer:  7 goals</h3>

=====================================================

Work Shown:

We have this set of data

\begin{tabular}{|l|l|}\textbf{game} & \textbf{goals}\\        1 & 5 \\        2 & 6 \\        3 & 6\\         4 & x    \end{tabular}

where x is some positive whole number. Add up everything in the second column:

5+6+6+x = x+17

This represents the total number of goals made over the four games.

Divide this over the number of games played (4) and set that equal to the mean of 6. Solve for x.

(x+17)/4 = 6

x+17 = 4*6

x+17 = 24

x = 24-17

x = 7

7 goals were scored in the fourth game.

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4 years ago
In an arithmetic sequence, a17 = -40 anda28 = -73. Explain how to use this information to write a recursive formula for this seq
ohaa [14]

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Step-by-step explanation:

4 0
3 years ago
Help. I wil give credit
IrinaK [193]

Answer:

Step-by-step explanation:

1) Angle 1 and angle 2 are complementary angles. If angle 1 measures (3x + 2), what is the measure of angle 2? 3x+2 + y = 90

2) Angle A and angle b are supplementary angles. Angle A measures (2m – 10) degrees and angle b measures (m + 25) degrees. Find the measure of angle A and angle b.

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7 0
2 years ago
Which is equivalent to RootIndex 5 StartRoot 1,215 EndRoot Superscript x?
SOVA2 [1]

Answer:

1,215 Superscript one-fifth x

Step-by-step explanation:

Given:

The expression to simplify is given as:

(\sqrt[5]{1215})^x

We know that,

\sqrt[n]{a} = a^{\frac{1}{n}}

Here, n=5, a=1215

So, \sqrt[5]{1215} = 1215^{\frac{1}{5}}

So, the above expression becomes:

(\sqrt[5]{1215})^x = (1215^{\frac{1}{5}})^x

Now, using the law of indices (a^m)^n=a^{(m\times n)}

Here, a=1215,m=\frac{1}{5},n=x

So, the expression is finally simplified to;

=(1215)^{({\frac{1}{5}}\times x)}\\\\=(1215)^{\frac{1}{5}x}

Therefore, the second option is the correct one.

(\sqrt[5]{1215})^x = 1,215 Superscript one-fifth x

6 0
3 years ago
Read 2 more answers
Use mathematical induction to prove that for each integer n ≥ 4, 5^n ≥ 2 2^n+1 + 100
anastassius [24]

The given Statement which we have to prove using mathematical induction is

   5^n\geq 2*2^{n+1}+100

for , n≥4.

⇒For, n=4

LHS

=5^4\\\\5*5*5*5\\\\=625\\\\\text{RHS}=2.2^{4+1}+100\\\\=64+100\\\\=164

 LHS >RHS

Hence this statement is true for, n=4.

⇒Suppose this statement is true for, n=k.

 5^k\geq 2*2^{k+1}+100

                      -------------------------------------------(1)

Now, we will prove that , this statement is true for, n=k+1.

5^{k+1}\geq 2*2^{k+1+1}+100\\\\5^{k+1}\geq 2^{k+3}+100

LHS

5^{k+1}=5^k*5\\\\5^k*5\geq 5 \times(2*2^{k+1}+100)----\text{Using 1}\\\\5^k*5\geq (3+2) \times(2*2^{k+1}+100)\\\\ 5^k*5\geq 3\times (2^{k+2}+100)+2 \times(2*2^{k+1}+100)\\\\5^k*5\geq 3\times(2^{k+2}+100)+(2^{k+3}+200)\\\\5^{k+1}\geq (2^{k+3}+100)+3\times2^{k+2}+400\\\\5^{k+1}\geq (2^{k+3}+100)+\text{Any number}\\\\5^{k+1}\geq (2^{k+3}+100)

Hence this Statement is true for , n=k+1, whenever it is true for, n=k.

Hence Proved.

4 0
3 years ago
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