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4 years ago

I MARK BRAINLEIST! WHEN CONVERTING STANDARD FORM TO SCIENTIFIC FORM WITH A POSITIVE EXPONENT

Mathematics

1 answer:

Grace [21]4 years ago

3 0

Answer:

You move the 0 to the right

Step-by-step explanation:

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Find the least common multiple of 18, 24 *

emmainna [20.7K]

Answer:

Step-by-step explanation:

Find the prime factorization of 18

18 = 2 × 3 × 3

Find the prime factorization of 24

24 = 2 × 2 × 2 × 3

Multiply each factor the greater number of times it occurs in steps above to find the LCM:

LCM = 2 × 2 × 2 × 3 × 3

LCM = 72

i hope this helps :)

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4 years ago

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What is the square root of 600?

BaLLatris [955]

24. 49489743 or 24 and 1/2 approximate
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4 years ago

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Please Help, i am so confused

Brut [27]

Answer: d

Step-by-step explanation:

5 0

3 years ago

A store sells 8 colors of balloons with at least 28 of each color. How many different combinations of 28 balloons can be chosen?

Len [333]

Answer:

(a) $Selection = 6724520$

(b) $At\ most\ 12 = 6553976$

(c) $At\ most\ 8 = 6066720$

(d) $At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 5896638$

Step-by-step explanation:

Given

$Colors = 8$

$Balloons = 28$ --- at least

Solving (a): 28 combinations

From the question, we understand that; a combination of 28 is to be selected. Because the order is not important, we make use of combination.

Also, because repetition is allowed; different balloons of the same kind can be selected over and over again.

So:

$n => 28 + 8-1$ $= 35$

$r = 28$

$Selection = ^{35}^C_{28$

$Selection = \frac{35!}{(35 - 28)!28!}$

$Selection = \frac{35!}{7!28!}$

$Selection = \frac{35*34*33*32*31*30*29*28!}{7!28!}$

$Selection = \frac{35*34*33*32*31*30*29}{7!}$

$Selection = \frac{35*34*33*32*31*30*29}{7*6*5*4*3*2*1}$

$Selection = \frac{33891580800}{5040}$

$Selection = 6724520$

Solving (b): At most 12 red balloons

First, we calculate the ways of selecting at least 13 balloons

Out of the 28 balloons, there are 15 balloons remaining (i.e. 28 - 13)

So:

$n => 15 + 8 -1 = 22$

$r = 15$

Selection of at least 13 =

$At\ least\ 13 = ^{22}C_{15}$

$At\ least\ 13 = \frac{22!}{(22-15)!15!}$

$At\ least\ 13 = \frac{22!}{7!15!}$

$At\ least\ 13 = 170544$

Ways of selecting at most 12 =

$At\ most\ 12 = Total - At\ least\ 13$ --- Complement rule

$At\ most\ 12 = 6724520- 170544$

$At\ most\ 12 = 6553976$

Solving (c): At most 8 blue balloons

First, we calculate the ways of selecting at least 9 balloons

Out of the 28 balloons, there are 19 balloons remaining (i.e. 28 - 9)

So:

$n => 19+ 8 -1 = 26$

$r = 19$

Selection of at least 9 =

$At\ least\ 9 = ^{26}C_{19}$

$At\ least\ 9 = \frac{26!}{(26-19)!19!}$

$At\ least\ 9 = \frac{26!}{7!19!}$

$At\ least\ 9 = 657800$

Ways of selecting at most 8 =

$At\ most\ 8 = Total - At\ least\ 9$ --- Complement rule

$At\ most\ 8 = 6724520- 657800$

$At\ most\ 8 = 6066720$

Solving (d): 12 red and 8 blue balloons

First, we calculate the ways for selecting 13 red balloons and 9 blue balloons

Out of the 28 balloons, there are 6 balloons remaining (i.e. 28 - 13 - 9)

So:

$n =6+6-1 = 11$

$r = 6$

Selection =

$^{11}C_6 = \frac{11!}{(11-6)!6!}$

$^{11}C_6 = \frac{11!}{5!6!}$

$^{11}C_6 = 462$

Using inclusion/exclusion rule of two sets:

$Selection = At\ most\ 12 + At\ most\ 8 - (12\ red\ and\ 8\ blue)$

$Only\ 12\ red\ and\ only\ 8\ blue\ = 170544+ 657800- 462$

$Only\ 12\ red\ and\ only\ 8\ blue\ = 827882$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = Total - Only\ 12\ red\ and\ only\ 8\ blue$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 6724520 - 827882$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 5896638$

3 0

3 years ago

Diana has been saving nickels and dimes. She opened up her piggy bank and determined that it contained 48 coins worth $4.50. Det

tino4ka555 [31]

Answer:

6 Nickels

42 Dimes

Step-by-step explanation:

There are 48 coins in total (nickels and dimes). Let the number of nickels be "n" and the number of dimes be "d". So we can write:

n + d = 48

n = 48 - d

In dollars, we know nickels are worth 0.05 and dimes are worth 0.10. In total there is $4.50, so we can write:

0.05n + 0.10d = 4.50

Now replacing first equation [n = 48 - d] into 2nd, we get and equation in "d" and solve for d first:

$0.05n + 0.10d = 4.50\\0.05(48-d) + 0.10d = 4.50\\2.4-0.05d+0.10d=4.50\\0.05d=4.50-2.4\\0.05d=2.1\\d=\frac{2.1}{0.05}=42$

So, there are 42 dimes.

We know from before, n = 48 - d, so

n = 48 - 42 = 6

There are 6 nickels

3 0

3 years ago