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postnew [5]
3 years ago
7

Help! I’m trying to help my friend get caught up who had been ill.

Mathematics
2 answers:
zavuch27 [327]3 years ago
6 0

Answer:

down below

Step-by-step explanation:

i answered so that the other guy can get a brainliest:)

aniked [119]3 years ago
3 0

Answer:

well, r = 79°

s = 63°

Step-by-step explanation:

r = 90° - 11° = 79°

s = 90° - 27° = 63°

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Put each shape through this classification flow chart
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3 years ago
Math help asap timed 24 points!
S_A_V [24]

3x+8.5 =2.5x + 14.5

subtract 2.5x from each side

.5x + 8.5 = 14.5

subtract 8.5 from each side

.5x =6

multiply by 2

x =12

at 12 weeks they will be the same height

Choice B

3 0
4 years ago
Hardest one yet..<br> complete all of these<br> i mean all of em
Nonamiya [84]

Answer:

1st one: 16

2nd one: 16

3rd one: 64

4th one: 27

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Step-by-step explanation:

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3 years ago
Which number is a reasonable estimate for the sum of the equation 1,956 + 1,183 = ________?
jonny [76]

Answer:

3,139

Step-by-step explanation:

1,956 + 1,183 = 3,139

6 0
3 years ago
Read 2 more answers
Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a seco
kari74 [83]

Answer:

C. \frac{1}{18}

Step-by-step explanation:

Given: Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl then a second card is drawn.

To Find: If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5.

Solution:

Sample space for sum of cards when two cards are drawn at random is \{(1,1),(1,2),(1,3)......(6,6)\}

total number of possible cases =36

Sample space when sum of cards is 8 is \{(3,5),(5,3),(6,2),(2,6),(4,4)\}

Total number of possible cases =5

Sample space when one of the cards is 5 is \{(5,3),(3,5)\}

Total number of possible cases =2

Let A be the event that sum of cards is 8

p(\text{A}) =\frac{\text{total cases when sum of cards is 8}}{\text{all possible cases}}

p(\text{A})=\frac{5}{36}

Let B be the event when one of the two cards is 5

probability than one of two cards is 5 when sum of cards is 8

p(\frac{\text{B}}{\text{A}})=\frac{\text{total case when one of the number is 5}}{\text{total case when sum is 8}}

p(\frac{\text{B}}{\text{A}})=\frac{2}{5}

Now,

probability that sum of cards 8 is and one of cards is 5

p(\text{A and B}=p(\text{A})\times p(\frac{\text{B}}{\text{A}})

p(\text{A and B})=\frac{5}{36}\times\frac{2}{5}

p(\text{A and B})=\frac{1}{18}

if sum of cards is 8 then probability that one of the cards is 8 is \frac{1}{18}, option C is correct.

3 0
3 years ago
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