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3 years ago

What is the largest odd number you can make with 4536.

Mathematics

1 answer:

muminat3 years ago

5 0

Answer:

6543

Step-by-step explanation:

The largest number you can make with those digits is ...

6543

It happens to be odd.

<em>Additional comment</em>

This is done by putting the largest digit in the number place with the greatest place value, repeating until you run out of digits. Here, the smallest digit is odd, so the number ends up being odd.

If the smallest digit were even, you'd have reserve the smallest odd digit to put in the ones place. For example, if your digits were 2, 4, 5, 6, then the largest odd number you could build would be 6425.

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The diagram below shows a circular pizza with a diameter of 12 inches. The pizza is in a square box with side lengths of 14 inch

saw5 [17]

Answer:

Step-by-step explanation:

Area of box = 14^2 = 196 in^2

Radius r of pizza = 12/2 = 6

Area of pizza = πr² = 36π ≈ 113.04 in^2

5 0

3 years ago

Read 2 more answers

A survey of 1000 high school students found that 781 students own a cell phone and 133 of those students who own a cell phone al

kirza4 [7]

This question is not complete. Find attached the complete question and the figure showing the two way table.

Answer:

a. P(owns a cellphone) = 781/1000

b. P(does not own a PS4) = 421/500

c. P(owns a PS4 U does not own a cellphone) = 377/1000

d. P(owns a PS4 n does not own a cellphone) = 25/158

e. P(owns a PS4 | own a cell phone) =

158/1000

f. P(owns a cellphone | own a PS4) = 781/1000

Step-by-step explanation:

Total number of students = 1000 students

We have two variables,

1. The students who own cellphones

2. The students who own PS4

a. P(owns a cellphone)

From the question and the table, we are told that 781 people own cellphones.

Hence, P(owns a cellphone) = 781/1000

b. P(does not own a PS4)

From the attached table, we can see that the number of students that does not own a PS4 = 842

Hen P(does not own a PS4) = 842/1000 = 421/500

c. P(owns a PS4 U does not own a cellphone) =

First we find the Probability of student that owns a PS4

Number of students that own a PS4 = 158

P(owns a PS4 ) = 158/1000

Number of students that does not own a cell phone = 219

P(does not own a cell phone) = 219/1000

Therefore, P(owns a PS4 U does not own a cellphone) = P(owns a PS4 ) + P(does not own a cell phone)

= 158/1000 +219/1000

= 377/1000

d. P(owns a PS4 n does not own a cellphone) =

From the table, we obtain that as =

25/158

e. P(owns a PS4 | own a cell phone) =

P(owns a PS4 n own a cell phone) ÷ P(own a cell phone)

= P(owns a PS4)

From the table that would give us = 158/1000

f. P(owns a cellphone | own a PS4) =

P(owns a cellphone n own a PS4) ÷ P(own a PS4)

= P(owns a cellphone

From the table, this is given as :

781/1000

4 0

3 years ago

I need help on eighteen

Damm [24]

I don't get it a all I tried fr

5 0

3 years ago

Arrange the summation expressions in increasing order of their values.

nalin [4]

We'll solve for each one of them:

$\sum_{i=1}^{4}4(5)^{i-1} = (4(5)^{1-1}) + (4(5)^{2-1}) + (4(5)^{3-1}) + (4(5)^{4-1})$
$=4+20+100+500 = 624$

$\sum_{i=1}^{5}3(4)^{i-1} = (3(4)^{1-1}) + (3(4)^{2-1}) + (3(4)^{3-1}) + (3(4)^{4-1}) + (3(4)^{5-1})$
$=3+12+48+192+768=1023$

$\sum_{i=1}^{2}5(6)^{i-1} = (5(6)^{1-1}) + (5(6)^{2-1})$
$=5+30=35$

$\sum_{i=1}^{4}5^{i-1} = (5^{1-1})+(5^{2-1})+(5^{3-1})+(5^{4-1})$
$=1 + 5+ 25+125 = 156$

So, our totals are: 624, 1023, 35, and 156. So clearly, 1023 > 624 > 156 > 35. W can write it as (your answer):

$\sum_{i=1}^{2}5(6)^{i-1}$

Hope this helps! If anything is confusing, you can always DM me.

3 0

3 years ago

Read 2 more answers

Is it less or greater or negative or positive

eimsori [14]

Answer: A negative number is less than a positive number.

Think about it this way: a negative number is less than 0. A

positive number is greater than 0. Therefore, a negative number

must be less than a positive number.

3 0

4 years ago