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Arte-miy333 [17]
3 years ago
12

Tell whether each number is rational or irrational. 3.525252... Rational Irrational

Mathematics
1 answer:
Serjik [45]3 years ago
5 0

Answer:

rational

Step-by-step explanation:

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Lynn asked six friends how many cars their parents own. She recorded 1, 1, 2, 2, 2, and 4 cars.
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D

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4 years ago
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State the domain and range of the following relation<br> x^2+y^2=16
Vlada [557]

The domain and range is  [-4, 4] and [0, 4]

<h3 /><h3>What is Domain and range?</h3>

The domain of a function is the set of values that we are allowed to plug into our function.

The range of a function is the set of values that the function assumes.

x² + y² = 16

y = √16 - x²

For domain under root should not be negative quantity,

16 - x²≥0

16≥x²

So, x≤4 or x≥4

Thus, the domain is  [-4, 4]

Range:

y is maximum at x=0, y=4

y is minimum at x=4, y=0

Thus, range = [0, 4]

Learn more about domain and range here:

brainly.com/question/12751831

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8 0
2 years ago
A study finds that 80% of the local high school students text while doing homework. Ten students are selected at random from the
AysviL [449]

Answer:

Step-by-step explanation:

6 0
3 years ago
Mickey keeps picking playing cards out of a standard deck of 52 cards, hoping that she will
andrew-mc [135]

Answer:

03%

Step-by-step explanation:

In this situation Mickey should have drew first 7 cards that are not diamond and then on the 8th attempt she should draw an ace.

Let A be the event of drawing 7 cards which are not diamond and B the event of drawing an ace card. therefore required probability is P(A∩B)=P(A)*P(B)

Now, there are 13 diamonds and 39 other cards. Hence,

P(A)=(\frac{39}{52})⁷=0.1335

After drawing 7 non-diamond cards, the 8th card must be a diamond.Hence,

P(B)=\frac{13}{52}=0.25

Hence, P(A∩B)=P(A)*P(B)=0.1335*0.25=0.033(approximately 0.03,i.e. 3%)

Therefore probability that Mickey will draw her first diamond on the 8th

attempt is 3%

7 0
3 years ago
Simplify the question
zimovet [89]
Answer: 50.4537849152
Hope this helped!
8 0
3 years ago
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