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ELEN [110]
3 years ago
5

Luis starts to do a division problem and notices that there is a pattern in the digits to the right of the decimal point.This nu

mber is
Mathematics
1 answer:
Assoli18 [71]3 years ago
5 0

Answer: Rational

Step-by-step explanation:

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Please answer this correctly
Troyanec [42]

Answer:

Mode

Step-by-step explanation:

Mean:

Before replacing = 464/8 = 58

After replacing 42 with 98 = 520/8 = 65

Mode:

Before replacing = 42

After replacing 42 with 98 = 98

Median:

25, 42 , 42 , 47, 55 , 59, 96,98

Before replacing =  47+55/2 = 51

25, 42 ,  47, 55 , 59, 96,98 , 98

After replacing 42 with 98 =  55+59/2 = 114/2 = 57

8 0
2 years ago
The estimate obtained from a sample of which size is likely to be closest to the actual parameter value of a population?
Leno4ka [110]

Answer:

184

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Decide which calculation shows the correct way to find 0.3 - 0.006 and explain your reasoning.
kirza4 [7]
D : 0.300- 0.006 is correct



A is wrong bec 0.3 doesn’t have the extra 0’s in the end

B is wrong bec the decimal places aren’t aligned

C is wrong bec again, the decimal places aren’t aligned
6 0
2 years ago
Read 2 more answers
7=11t
joja [24]

Steps to solve:

7 = 11t

~Divide 11 to both sides

7/11 = 11t/11

~Simplfy

7/11 = t

The fraction can't be simplified any further.

Best of Luck!

5 0
2 years ago
Read 2 more answers
A survey of athletes at a high school is conducted, and the following facts are discovered: 13% of the athletes are football pla
nekit [7.7K]

Answer:

The probability that an athlete chosen is either a football player or a basketball player is 56%.

Step-by-step explanation:

Let the athletes which are Football player be 'A'

Let the athletes which are Basket ball player be 'B'

Given:

Football players (A) = 13%

Basketball players (B) = 52%

Both football and basket ball players = 9%

We need to find probability that an athlete chosen is either a football player or a basketball player.

Solution:

The probability that athlete is a football player = P(A)= \frac{13}{100}=0.13

The probability that athlete is a basketball player = P(B)= \frac{52}{100}=0.52

The probability that athlete is both basket ball player and  football player = P(A\cap B) = \frac{9}{100}=0.09

We have to find the probability that an athlete chosen is either a football player or a basketball player P(A\cup B).

Now we know that;

P(A\cup B)= P(A) + P(B) - P(A \cap B)\\\\P(A\cup B) = 0.13+0.52-0.09=0.56\\\\P(A\cup B) = \frac{0.56}{100}=56\%

Hence The probability that an athlete chosen is either a football player or a basketball player is 56%.

5 0
3 years ago
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