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Ann [662]
3 years ago
9

Draw a square and shade about 1/8 of it. How did you decide how much to shade?

Mathematics
2 answers:
timama [110]3 years ago
8 0
I decided how much to shade because when you divide the square into 8 sections, you know to shade in 1 because you want to shade 1/8 of it.
Hope this helps.
(Please mark this brainliest, I would really appreciate it) Thanks!

jeka943 years ago
4 0
I drew a square with each side being 8 inches long. I then divided the square into 8 sections. Each section was 1 inch wide and 8 inches long. Then I shaded in one section of the square.
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What is 4/5 11/25 4/50 as a percent?
svp [43]
I believe the 11/25 one would be 44% and 4/50 would be 8%, 4/5 would be 80%. Hope it helps!
6 0
3 years ago
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Which equations show inverse operations of each other? Choose exactly two answers that are correct. 16×1/4  =4     1/4 ÷4 =116  
qaws [65]

Answer:

6 × 1/4 = 4

4 × 1/4 = 1

4 ÷ 1/4 = 16

All three of these are correct

Step-by-step explanation:

16 × 1/4 = 4

16/4 = 4

4 = 4


1/4 ÷ 4 = 116

1/4 × 1/4 = 116

1/16 ≠ 116


4 x 1/4 = 1

4/4 = 1

1 = 1


4 ÷ 1/4 = 16

4 × 4 = 16

16 = 16

8 0
3 years ago
How will adding the value 1000 affect the mean and median of the data set 5, 10, 17, 19, 20?
WITCHER [35]
The answer should be C
3 0
3 years ago
Find the probability of the following events , when a dice is thrown once:
Fudgin [204]

Answer:

Step-by-step explanation:

s a die is rolled once, therefore there are six possible outcomes, i.e., 1,2,3,4,5,6.

(a) Let A be an event ''getting a prime number''.

Favourable cases for a prime number are 2,3,5,

i.e., n(A)=3

Hence P(A)=n(A)n(S)=36=12

(b) Let A be an event ''getting a number between 3 and 6''.

Favourable cases for events A are 4 or 5.

i.e., n(A)=2

P(A)=n(A)n(S)=26=13

(c) Let A be an event ''a number greater than 4''.

Favourable cases of events A are 5, 6.

i.e., n(A)=2

P(A)=n(A)n(S)=26=13

(d) Let A be the event of getting a number at most 4.

∴ A={1,2,3} ⇒ n(A)=4,n(S)=6

∴ Required probability =n(A)n(S)=42=23

(e) Let A be the event of getting a factor of 6.

∴ A={1,2,36} ⇒ n(A)=4,n(A)=6

∴ Required probability =46=23

(ii) Since, a pair of dice is thrown once, so there are 36 possible outcomes. i.e.,

(a) Let A be an event ''a total 6''. Favourable cases for a total of 6 are (2,4), (4,2), (3,3), (5,1), (1,5).

i.e., n(A)=5

Hence P(A)=n(A)n(S)=536

(b) Let A be an event ''a total of 10n. Favourable cases for total of 10 are (6,4), (4,6), (5,5).

i.e., n(A)=5

P(A)=n(A)n(S)=336=112

(c) Let A be an event ''the same number of the both the dice''. Favourable cases for same number on both dice are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).

i.e., n(A)=6

P(A)=n(A)n(S)=636=16

(d) Let A be an event ''of getting a total of 9''. Favourable cases for a total of 9 are (3,6), (6,3), (4,5), (5,4).

i.e., n(A)=4

P(A)=n(A)n(S)=436=19

(iii) We have, n(S) = 36

(a) Let A be an event ''a sum less than 7'' i.e., 2,3,4,5,6.

Favourable cases for a sum less than 7 ar

7 0
3 years ago
Read 2 more answers
Hope researches the impact of one variable on another. She graphs the data from her research and sees that the data forms a shap
zaharov [31]

Answer:

She can determine that while there is an association between the variables, there is no correlation, and she cannot determine causation.

Step-by-step explanation:

Since there is an arch shape to her graph, we know that as one variable changes, the other changes in the same manner. This means there is an association between the variables.

However, since the graph is not linear, there is no correlation between the variables. Since there is no correlation, we cannot determine causation.

Mark me as brainiest

6 0
3 years ago
Read 2 more answers
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