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choli [55]
3 years ago
10

The average of 10 numbers is -5. A new number is added to the list making the average -6. What is the new number?

Mathematics
1 answer:
Shtirlitz [24]3 years ago
4 0

Answer:

- 16

Step-by-step explanation:

Average is calculated as

average = \frac{total}{count}

Thus for the given 10 numbers we have

\frac{total}{10} = - 5 ( multiply both sides by 10 )

total = - 50

let the number added be x, then

\frac{total+x}{11} = - 6 ( multiply both sides by 11 )

total + x = - 66, that is

- 50 + x = - 66 ( add 50 to both sides )

x = - 16

The new number is - 16

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A pendulum is set swinging. its first oscillation is through 30 and each succeeding oscillation is through 95% of the angle of t
Marat540 [252]

<u>Answer-</u>

<em>After 76 swings</em><em> the angle through which it swings less than 1°</em>

<u>Solution-</u>

From the question,

Angle of the first of swing = 30° and then each succeeding oscillation is through 95% of the angle of the one before it.

So the angle of the second swing = (30\times \frac{95}{100})^{\circ}

Then the angle of third swing = (30\times (\frac{95}{100})^2)^{\circ}

So, this follows a Geometric Progression.

(30,\ 30\cdot \frac{95}{100},\ 30\cdot (\frac{95}{100})^2............,0)

a = The initial term = 30

r = Common ratio = \frac{95}{100}

As we have to find the number swings when the angle swept by the pendulum is less than 1°.

So we have the nth number is the series as 1, applying the formula

T_n=ar^{n-1}

Putting the values,

\Rightarrow 1=30(\frac{95}{100})^{n-1}

\Rightarrow \frac{1}{30} =(\frac{95}{100})^{n-1}

Taking logarithm of both sides,

\Rightarrow \log \frac{1}{30} =\log (\frac{95}{100})^{n-1}

\Rightarrow \log \frac{1}{30} =(n-1)\log (\frac{95}{100})

\Rightarrow -1.5=(n-1)(-0.02)

\Rightarrow 1.5=(n-1)(0.02)

\Rightarrow n-1=\dfrac{1.5}{0.02}

\Rightarrow n-1=75

\Rightarrow n=76

Therefore, after 76 swings the angle through which it swings less than 1°

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velikii [3]
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<h3><u><em>Hope this helps!!!</em></u></h3><h3><u><em>Please mark as brainliest!!!</em></u></h3>
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