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Mademuasel [1]
3 years ago
13

1. A line contains the points (−12,−60) and (60,− 42). What is the slope of the line in simplified form? PLEASE HELP QUICKLY!!!!

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! 20 points
Mathematics
1 answer:
Galina-37 [17]3 years ago
8 0

Answer:

the answer should be 0.25

formula for slope is m=(y2-y1)/(x2-x1)

therefore m=(-42-(-60))/(60-(-12))

(-42+60)/(60+12)

18/72

simplify

1/4

further simplify

0.25 should be the final answer

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0.00000000828860 in scientific notation
azamat

0.00000000828860 upto which decimal?


6 0
3 years ago
I need help for 7 and 8. i also need the solution on how you get the answer. thanks in advance!
ycow [4]
7) Rectangular pyramid:
Length = 8m ; width = 4.6m ; Volume = 88m³

Volume of a rectangular pyramid = (Length * Width * Height)/3
88m³ = (8m * 4.6m * height)/3
88m³ * 3 = 36.8m² * height
264m³ = 36.8m² * height
264m³ / 36.8m² = height
7.2 m = height

8) Cone:
r = 5 in ; volume = 487 in
³

Volume of a cone = π r² h/3
487 in³ = 3.14 * (5in)² * h/3
487 in³ * 3 = 3.14 * 25in² * h
1,461 in³ = 78.5 in² * h
1,461 in³ / 78.5 in² = height
18.6 in = height
8 0
3 years ago
2x – 4y = 16<br> -2x - 3y = 7
noname [10]
Bro use the app Socratic
5 0
2 years ago
Write an inequality and use substitution to solve the following problem:
nekit [7.7K]

They already spent $2000, and they only had $5100

Subtract

5100 - 2000 = 3100

$3,050 is your answer, because it is the only one that is less than 3100

hope this helps

7 0
2 years ago
Consider two independent tosses of a fair coin. Let A be the event that the first toss results in heads, let B be the event that
aliina [53]

Answer with Step-by-step explanation:

We are given that two independent tosses of a fair coin.

Sample space={HH,HT,TH,TT}

We have to find that A, B and C are pairwise independent.

According to question

A={HH,HT}

B={HH,TH}

C={TT,HH}

A\cap B={HH}

B\cap C={HH}

A\cap C={HH}

P(E)=\frac{number\;of\;favorable\;cases}{total\;number\;of\;cases}

Using the formula

Then, we get

Total number of cases=4

Number of favorable cases to event A=2

P(A)=\frac{2}{4}=\frac{1}{2}

Number of favorable cases to event B=2

Number of favorable cases to event C=2

P(B)=\frac{2}{4}=\frac{1}{2}

P(C)=\frac{2}{4}=\frac{1}{2}

If the two events A and B are independent then

P(A)\cdot P(B)=P(A\cap B)

P(A\cap)=\frac{1}{4}

P(B\cap C)=\frac{1}{4}

P(A\cap C)=\frac{1}{4}

P(A)\cdot P(B)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}

P(B)\cdot P(C)=\frac{1}{4}

P(A)\cdot P(C)=\frac{1}{4}

P(A)\cdot P(B)=P(A\cap B)

Therefore, A and B are independent

P(B)\cdot P(C)=P(B\cap C)

Therefore, B and C are independent

P(A\cap C)=P(A)\cdot P(C)

Therefore, A and C are independent.

Hence, A, B and C are pairwise independent.

6 0
3 years ago
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