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Westkost [7]
3 years ago
14

25 dollars as an interger

Mathematics
2 answers:
strojnjashka [21]3 years ago
5 0

Answer:

-25

Step-by-step explanation:

Hope this helps! :)

shutvik [7]3 years ago
4 0

Answer:

25

Step-by-step explanation:

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Here is your answer.

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2 years ago
One of my factors is equal to 5-2. I am the product 24. What are my other factors?
blsea [12.9K]
5 - 2 = 3

3 x ? = 24
(24 ÷ 3 = 8)
3 x 8 = 24
7 0
2 years ago
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7 0
2 years ago
The number of girl child per family was recorded for 1000 families with 3 children. The data obtained was as follows: [2] Number
ira [324]

Answer:

a.\ Probability = 0.426

b.\ Probability = 0.574

Step-by-step explanation:

Given

Total\ Families = 1000

Number of Girls :      0 ---->  1 -----> 2   ------->3

Number of Families: 112 -> 314 --> 382 ---> 192

Solving (a): At most one girl

From the table, the number of families with at most one girl is: 112 + 314

i.e.

Number of Girls :      0 ---->  1

Number of Families: 112 -> 314

So, we have:

At\ Most\ One\ Girl = 112 + 314

At\ Most\ One\ Girl = 426

The probability is then calculated as:

Probability = \frac{At\ Most\ One\ Girl}{Total}

Probability = \frac{426}{1000}

Probability = 0.426

Solving (b): More girls

From the table, the number of families with more girl is: 382 + 192

i.e.

Number of Girls :      -----> 2   ------->3

Number of Families: --> 382 ---> 192

So, we have:

More\ Girls = 382 + 192

More\ Girls = 574

The probability is then calculated as:

Probability = \frac{More\ Girls}{Total}

Probability = \frac{574}{1000}

Probability = 0.574

4 0
2 years ago
Can you help me solve an infinite geometric sequence
zaharov [31]

A geometric series like

\displaystyle \sum_{n=1}^\infty \dfrac{1}{\alpha^n}

Converges if and only if |\alpha|>1. If this is the case, the sum equals

\displaystyle \sum_{n=1}^\infty \dfrac{1}{\alpha^n} = \dfrac{1}{\alpha-1}

So, in your case, you have convergence if and only if

|2+a|>1 \iff 2+a>1 \lor 2+a-1 \lor a

And if this is the case, the sum equals

\displaystyle \sum_{n=1}^\infty \dfrac{1}{(2+a)^n} = \dfrac{1}{2+a-1} = \dfrac{1}{a+1}

8 0
3 years ago
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