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3 years ago

Write one hundred twenty-one and twenty-one thousandths in decimal notation.

Mathematics

2 answers:

Likurg_2 [28]3 years ago

4 0

*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆**☆*――*☆*――*☆*――*☆

Answer: 121.021

I hope this helped!

<!> Brainliest is appreciated! <!>

- Zack Slocum

*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆**☆*――*☆*――*☆*――*☆

Tomtit [17]3 years ago

4 0

Answer:

121.021

Step-by-step explanation:

You might be interested in

I need help with this

SVETLANKA909090 [29]

Answer:

The answer is 0.675

Step-by-step explanation:

If you were going to ask Brainly, you could have easily just used a calculator lol

4 0

3 years ago

WILL GIVE BRAINLIEST AND 20 POINTS

MrRissso [65]

Answer:

14 yd²

Step-by-step explanation:

4 • 7 = 28

Since this is a triangle you divide the answer by 2

28/2 = 14

14 yd²

Hope this helps dude

4 0

3 years ago

Jill added 450+790+123 and 1163 .is this sum reasonable? Explain.

AlexFokin [52]

Let us manually add the given numbers.

450

790

123

____

1363

Therefore we see that the correct answer 1363 which is far from what Jill got which is only 1163.

Therefore the answer is No, the sum should be 1363 (closer to 1400).

3 0

3 years ago

Read 2 more answers

There are three power plants [X, Y, Z] that at any given time each one either generates electricity or idles. Event A is that pl

insens350 [35]

We're told that

$P(A\cap B)=0.15$

$P(A\cup B)^C=0.06\implies P(A\cup B)=0.94$

$P(B\mid A)=P(B^C\mid A)=0.5$

where the last fact is due to the law of total probability:

$P(A)=P(A\cap B)+P(A\cap B^C)$

$\implies P(A)=P(B\mid A)P(A)+P(B^C\mid A)P(A)$

$\implies 1=P(B\mid A)+P(B^C\mid A)$

so that $B\mid A$ and $B^C\mid A$ are complementary.

By definition of conditional probability, we have

$P(B\mid A)=P(B^C\mid A)$

$\implies\dfrac{P(A\cap B)}{P(A)}=\dfrac{P(A\cap B^C)}{P(A)}$

$\implies P(A\cap B)=P(A\cap B^C)$

We make use of the addition rule and complementary probabilities to rewrite this as

$P(A\cap B)=P(A\cap B^C)$

$\implies P(A)+P(B)-P(A\cup B)=P(A)+P(B^C)-P(A\cup B^C)$

$\implies P(B)-[1-P(A\cup B)^C]=[1-P(B)]-P(A\cup B^C)$

$\implies2P(B)=2-[P(A\cup B)^C+P(A\cup B^C)]$

$\implies2P(B)=[1-P(A\cup B)^C]+[1-P(A\cup B^C)]$

$\implies2P(B)=P(A\cup B)+P(A\cup B^C)^C$

$\implies2P(B)=P(A\cup B)+P(A^C\cap B)\quad(*)$

By the law of total probability,

$P(B)=P(A\cap B)+P(A^C\cap B)$

$\implies P(A^C\cap B)=P(B)-P(A\cap B)$

and substituting this into $(*)$ gives

$2P(B)=P(A\cup B)+[P(B)-P(A\cap B)]$

$\implies P(B)=P(A\cup B)-P(A\cap B)$

$\implies P(B)=0.94-0.15=\boxed{0.79}$

8 0

3 years ago

Solve for y.

MariettaO [177]

5y = 135....divide both sides by 5
y = 135 / 5
y = 27 <==

7 0

3 years ago

Read 2 more answers