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3 years ago

Are there less than 1 million, exactly 1 million, or greater than 1 million milligrams in 1 kilogram? Explain how you know

Mathematics

2 answers:

NISA [10]3 years ago

5 0

It is exactly one million milligrams

erastovalidia [21]3 years ago

5 0

Okay! Metric system.
K(kilo)
H(hecto)
D(eca)
B(gram, liter, etc.)
d(deci)
C(centi)
M(milli)

Since we have to go from Milli to Kilo, we count the number of spaces between them. 6 spaces. 6 zeroes. Slap a one onto that (since it's one kilo and one milli)
and you've got 1000000. One Million. It's not 0.000001 because it goes up, or if it were horizontal, to the left, meaning the decimal point goes to the right. (yes, confusing. it's hard to explain without visuals) This sounds really confusing, and even reading back at it, it's confusing for me, but I'm not good at explaining. Sorry.

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Seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Find the probability that (a)

UNO [17]

Answer:

a) P=0.226

b) P=0.6

c) P=0.0008

d) P=0.74

Step-by-step explanation:

We know that the seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Therefore, we have 46 balls.

a) We calculate the probability that are 3 red, 2 blue, and 2 green balls.

We calculate the number of possible combinations:

$C_7^{46}=\frac{46!}{7!(46-7)!}=53524680$

We calculate the number of favorable combinations:

$C_3^{12}\cdot C_2^{16}\cdot C_2^{18}=660\cdot 120\cdot 153=12117600$

Therefore, the probability is

$P=\frac{12117600}{53524680}\\\\P=0.226$

b) We calculate the probability that are at least 2 red balls.

We calculate the probability withdrawn of 1 or none of the red balls.

We calculate the number of possible combinations:

$C_7^{46}=\frac{46!}{7!(46-7)!}=53524680$

We calculate the number of favorable combinations: for 1 red balls

$C_1^{12}\cdot C_7^{34}=12\cdot 1344904=16138848$

Therefore, the probability is

$P_1=\frac{16138848}{53524680}\\\\P_1=0.3$

We calculate the number of favorable combinations: for none red balls

$C_7^{34}=5379616$

Therefore, the probability is

$P_0=\frac{5379616}{53524680}\\\\P_0=0.1$

Therefore, the the probability that are at least 2 red balls is

$P=1-P_1-P_0\\\\P=1-0.3-0.1\\\\P=0.6$

c) We calculate the probability that are all withdrawn balls are the same color.

We calculate the number of possible combinations:

$C_7^{46}=\frac{46!}{7!(46-7)!}=53524680$

We calculate the number of favorable combinations:

$C_7^{12}+C_7^{16}+C_7^{18}=792+11440+31824=44056$

Therefore, the probability is

$P=\frac{44056}{53524680}\\\\P=0.0008$

d) We calculate the probability that are either exactly 3 red balls or exactly 3 blue balls are withdrawn.

Let X, event that exactly 3 red balls selected.

$P(X)=\frac{C_3^{12}\cdot C_4^{34}}{53524680}=0.57\\$

Let Y, event that exactly 3 blue balls selected.

$P(Y)=\frac{C_3^{16}\cdot C_4^{30}}{53524680}=0.29\\$

We have

$P(X\cap Y)=\frac{18\cdot C_3^{12} C_3^{16}}{53524680}=0.12$

Therefore, we get

$P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\\\P(X\cup Y)=0.57+0.29-0.12\\\\P(X\cup Y)=0.74$

8 0

4 years ago

What is the value of X

skelet666 [1.2K]

X is 3 your are welcome

8 0

3 years ago

Read 2 more answers

Question in screenshot.

Kitty [74]

Your answer is b. hope this helps :)

4 0

4 years ago

X + 4y = 16 -X + 3y = -2

jeyben [28]

Answer:

(8, 2 )

Step-by-step explanation:

Given the 2 equations

x + 4y = 16 → (1)

- x + 3y = - 2 → (2)

Adding the 2 equations term by term will eliminate the x- term

0 + 7y = 14

7y = 14 ( divide both sides by 7 )

y = 2

Substitute y = 2 into either of the 2 equations and solve for x

Substituting into (1)

x + 4(2) = 16

x + 8 = 16 ( subtract 8 from both sides )

x = 8

solution is (8, 2 )

7 0

3 years ago

4.26 as a mixed number in simplest form

VikaD [51]

The answer is 4 13/50

8 0

3 years ago