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Andrew [12]
3 years ago
13

PLZZZZZ HELP ASAPPPP

Mathematics
1 answer:
blagie [28]3 years ago
3 0

Answer:

the answer to this is definitely 20

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. The probability that a patient catches flu is 3/8, the probability that he has a headache
igomit [66]

Answer:

\boxed {\frac{343}{360}}

Step-by-step explanation:

<u>Probability Rule</u> :

\boxed {P(A\cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B \cap C)}

<u>Solving</u> :

⇒ 3/5 = 3/8 + 5/15 + 4/9 - P (A ∩ B ∩ C)

⇒ P (A ∩ B ∩ C) = 3/8 + 1/3 + 4/9 - 1/5

⇒ P (A ∩ B ∩ C) = 135/360 + 120/360 + 160/360 - 72/360

⇒ P (A ∩ B ∩ C) = (135 + 120 + 160 - 72) / 360

⇒ P (A ∩ B ∩ C) = 343/360

3 0
2 years ago
Help on math equations!! 13 points!!!!!!
masha68 [24]

Answer:

1. \frac{-73}{350}

2. 0.12

3. 13\frac{23}{45}

4. -9/5

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Lines p and q are parallel.
otez555 [7]

Answer:

60 degrees.

Step-by-step explanation:

8 0
4 years ago
A distribution has the five-number summary shown below. What is the
Irina-Kira [14]

Answer:

The interquartile range(IQR) is <u>31</u>.

Step-by-step explanation:

Given:

A distribution has the five-number summary shown below:

24, 36, 42, 57, 65.

Now, to find the interquartile range (IQR).

So, to get the IQR we need to find the quartile 1 and quartile 3:

As, 42 is the median.

Quartile 1 is the average of first half.

<u>24, 36</u>, 42, 57, 65.

Q1 = \frac{24+36}{2}

Q1 = \frac{60}{2}

Q1 = 30

Quartile 3 is the average of last half.

24, 36, 42, <u>57, 65</u>.

Q3 = \frac{57+65}{2}

Q3 = \frac{122}{2}

Q3 = 61

Thus, by putting the formula we get IQR:

IQR = Q3 - Q1.

IQR=61-30

IQR=31.

Therefore, the interquartile range(IQR) is 31.

4 0
3 years ago
Can someone help me in 5 and 7 please
pav-90 [236]

5

Write -22 in the middle of the graph, number the rest of the graph accordingly and draw an open circle around the line under the -22 and draw a line along the graph to the right


7

Write 15 in the middle of the graph and number the rest of the graph accordingly, draw a colored in circle under the 15 on the graph and draw a line to the to the right


Sorry, I'm on the road and couldn't take a picture of the solution


4 0
3 years ago
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