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dybincka [34]
2 years ago
11

Help plssssssssssssss

Mathematics
1 answer:
katen-ka-za [31]2 years ago
7 0

Answer: AB 90 IB 150 IC 135 AC 90

Step-by-step explanation:

All triangles have to equal 180, if B is 60 and I is 90, 60+90=150, 180-150=30 then A=30 if a equals 30 than 60+30=90 A+B=AB (same with the others.)

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One number is 3 less than another number if their sum is 49
Musya8 [376]
One number is 26 the other is 23


x+y=49

Y=x+3,X (now substitute it in the equation)

x+3+x=49

2x+3=49

2x=46. X=23


Y=x+3. Y=23+3 =26 for checking 23+26=49
6 0
2 years ago
Find the mode of the data.<br> 1, 2, 1, 3, 3, 4, 1<br> The mode is
drek231 [11]

Answer:

1 is the answer.

Step-by-step explanation:

The mode of a data set is the number that occurs most frequently in the set.

5 0
2 years ago
Read 2 more answers
Choose the model that shows the correct answer to this problem:<br> 6/8<br> 9/12<br> 3/4<br> 3/12
Veronika [31]

Answer:

1/3 x 3/4

Step-by-step explanation:

in this case, 3/4 is correct because it is 3 of 4 shaded.

hope this helps!

3 0
2 years ago
Read 2 more answers
Mathematics mh hhhhh
crimeas [40]

Answer:

A. x≥4 ( red on 4)

B. x≤4 ( red on 4)

C. x>4 (blue on 4)

D. x<4 (blue on 4)

Step-by-step explanation:

illustrated

A. x≥4 ( red on 4)

B. x≤4 ( red on 4)

C. x>4 (blue on 4)

D. x<4 (blue on 4)

4 0
3 years ago
Read 2 more answers
A box contains 5 red balls, 6 white balls and 9 black balls. Two balls are drawn at
valina [46]

Answer:

P(Same)=\frac{61}{190}

Step-by-step explanation:

Given

Red = 5

White = 6

Black = 9

Required

The probability of selecting 2 same colors when the first is not replaced

The total number of ball is:

Total = 5 + 6 + 9

Total = 20

This is calculated as:

P(Same)=P(Red\ and\ Red) + P(White\ and\ White) + P(Black\ and\ Black)

So, we have:

P(Same)=\frac{n(Red)}{Total} * \frac{n(Red) - 1}{Total - 1} + \frac{n(White)}{Total} * \frac{n(White) - 1}{Total - 1}  + \frac{n(Black)}{Total} * \frac{n(Black) - 1}{Total - 1}

<em>Note that: 1 is subtracted because it is a probability without replacement</em>

P(Same)=\frac{5}{20} * \frac{5 - 1}{20- 1} + \frac{6}{20} * \frac{6 - 1}{20- 1}  + \frac{9}{20} * \frac{9- 1}{20- 1}

P(Same)=\frac{5}{20} * \frac{4}{19} + \frac{6}{20} * \frac{5}{19}  + \frac{9}{20} * \frac{8}{19}

P(Same)=\frac{20}{380} + \frac{30}{380}  + \frac{72}{380}

P(Same)=\frac{20+30+72}{380}

P(Same)=\frac{122}{380}

P(Same)=\frac{61}{190}

4 0
2 years ago
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