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Airida [17]
3 years ago
15

-32 / (-4) -40 / (-8) 348 / 26 x 29

Mathematics
1 answer:
Jlenok [28]3 years ago
5 0

Answer:

Your answer is: ↓

A) 8

B) 5

C) 6 / 13

Reduce the expression, if possible, by cancelling the common factors.

Step-by-step explanation:

Hope this helped : )

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Please help me with this
Helga [31]

Answer:

3/5

Step-by-step explanation:

I just looked up how to do this since I've never done it before, so if I'm wrong I'm sorry, but from what I found all you have to do is put the opposite(o) over the hypotenuse(h) o/h which would be 9/15 and simplified it would be 3/5, and again if I'm wrong I'm sorry.

I hope this helps, and have a great day!

8 0
3 years ago
What two terms are required to write a geometric sequence??
Tanya [424]

You need the first term and the common ratio.

8 0
4 years ago
Quarter past 7 , what time is it ?
Phoenix [80]
A quarter of an hour is 15 minutes, so a quarter past 7 is 7+15 mins. This gives you:

7:15

Hope this helps and good luck!
3 0
3 years ago
Read 2 more answers
HELP ASAP Indicate the equation of the line through (2, -4) and having slope of 3/5. =
DerKrebs [107]

Answer: No solution

Step-by-step explanation:

Since putting it in by y=mx+b and there is no solution

4 0
4 years ago
In a class of students, the following data table summarizes how many students have a cat or a dog. What is the probability that
leonid [27]

Given:

Number of students who has a cat and a dog = 5

Number of students who has a cat but do not have a dog = 11

Number of students who has a dog but do not have a cat = 3

Number of students who neither have a cat nor a dog = 2

To find:

The probability that a student has a cat given that they do not have a dog.

Solution:

Let the following events:

A = Student has a cat

B = Do not have a dog

Total number of outcomes is:

5+3+11+2=21

The probability that a student has a cat but do not have a dog is:

P(A\cap B)=\dfrac{11}{21}

The probability that a student do not have a dog is:

P(B)=\dfrac{11+2}{21}

P(B)=\dfrac{13}{21}

The conditional probability is:

P\left(\dfrac{A}{B}\right)=\dfrac{P(A\cap B)}{P(B)}

P\left(\dfrac{A}{B}\right)=\dfrac{\dfrac{11}{21}}{\dfrac{13}{21}}

P\left(\dfrac{A}{B}\right)=\dfrac{11}{13}

Therefore, the probability that a student has a cat given that they do not have a dog is \dfrac{11}{13}.

5 0
3 years ago
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