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ella [17]
2 years ago
15

Square 33 is the first square in Row 5, the first square of the second half of the chessboard. How many pennies are on square 33

?
Mathematics
2 answers:
bulgar [2K]2 years ago
7 0

Answer:

2^{32} is the answer.

Step-by-step explanation:

If we put a penny on the first square of a chess board, 2 on second, six on third and we keep on doing double for the next square then it will form a geometric sequence in the form of T_{n}=2^{n-1}

1, 2, 4, 8, 16.........n terms

Now we have to calculate the pennies on square 33 that will be

So pennies on 33rd square will be

T_{33}=2^{33-1}=2^{32}

Olin [163]2 years ago
4 0
Square 33 is the first square in Row 5, the first square of the second half of the chessboard. How many pennies are on square 33?
Answer : 2^32
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4/9 of students at school are boys if there are 2601 students at the school how many are girls
hram777 [196]

Answer:

1445

Step-by-step explanation:

2601*(9/9-4/9)

4 0
2 years ago
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For the school's sports day, a group of students prepared 12 1/2 litres of lemonade. At the end of the day they had 2 5/8 litres
Hoochie [10]

Given :

For the school's sports day, a group of students prepared 12 1/2 litres of lemonade. At the end of the day they had 2 5/8 litres left over.

To Find :

How many litres of lemonade were sold.

Solution :

Initial amount of lemonade, I = 12 1/2 = 25/2 litres.

Final amount of lemonade, F = 2 5/8 = 21/8 litres.

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2 years ago
Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoc
Bingel [31]

Answer:

We need a sample size of at least 719

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.025 = 0.975, so z = 1.96

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

How large a sample size is required to vary population mean within 0.30 seat of the sample mean with 95% confidence interval?

This is at least n, in which n is found when M = 0.3, \sigma = 4.103. So

M = z*\frac{\sigma}{\sqrt{n}}

0.3 = 1.96*\frac{4.103}{\sqrt{n}}

0.3\sqrt{n} = 1.96*4.103

\sqrt{n} = \frac{1.96*4.103}{0.3}

(\sqrt{n})^{2} = (\frac{1.96*4.103}{0.3})^{2}

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Rouding up

We need a sample size of at least 719

6 0
2 years ago
Craig has $1850 dollars in a bank account that he uses to make automatic payments of $400.73 on his car loan. If Craig stops mak
Ivenika [448]

Given:

Amount in the bank account = $1850

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jok3333 [9.3K]
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