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charle [14.2K]
3 years ago
12

2) Solving an equation - using the distributive property and inverse operations to solve 4/5

Mathematics
1 answer:
ella [17]3 years ago
6 0
What is there to solve
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20 POINTS <br><br> Please only have to answer question 9
KengaRu [80]

Answer:

l = -16/3   m = 80/3

Step-by-step explanation:

see attached for step by step

8 0
2 years ago
The product of 3 and 1 less than a number is
MAVERICK [17]

Answer: 4 is lesser than the all of the numbers above it.

Step-by-step explanation: I don’t really understand what you are trying to ask, Maybe re-word what you were trying to say?

8 0
3 years ago
Use the following ordered pairs
Lynna [10]

The ordered pair for Y is (-4,-2)

Step-by-step explanation:

Given

X(-2,3)

Z(-6,-7)

As Y is the mid-point of XZ, it will divide Xz in two equal segments. Y is the mid-point of the segment.

Let (x_Y, y_Y) be the coordinates of the mid-point

Then

(x_Y, y_Y) = (\frac{x_1+_2}{2} , \frac{y_1+y_2}{2})

Putting the values

=(frac{-2-6}{2} , \frac{3-7}{2})\\=(\frac{-8}{2} , \frac{-4}{2})\\=(-4, -2)

Hence,

The ordered pair for Y is (-4,-2)

Keywords: Coordinate geometry, Mid-point

Learn more about coordinate geometry at:

  • brainly.com/question/2116906
  • brainly.com/question/2131336

#LearnwithBrainly

3 0
3 years ago
Any1 know the answers for this?
Bezzdna [24]
X=4 because you combine 2+4 and subtract -6 then divide
7 0
2 years ago
The owner of a fish market has an assistant who has determined that the weights of catfish are normally​ distributed, with mean
LiRa [457]

Answer:

P(X\geq 3.4)=0.0228

Step-by-step explanation:

Given the mean is 3.2, standard deviation is 0.8 and the sample size is 64.

-We calculate the  probability of a mean of 3.4 as follows:

#First determine the z-value:

z=\frac{\bar X -\mu}{\sigma/\sqrt{n}}\\\\=\frac{3.4-3.2}{0.8/\sqrt{64}}\\\\=2.000

#We then determine the corresponding probability on the z tables:

Z(X\geq 3.4)=1-P(X

Hence, the probability of obtaining a sample mean this large or​ larger is 0.0228

8 0
2 years ago
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