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docker41 [41]
3 years ago
10

What is the slope of the line which contains the ordered pairs (2,1) and (-3,5)?

Mathematics
1 answer:
Vikki [24]3 years ago
5 0

Answer:

4/5

Step-by-step explanation:

slope = x^2 - x^1 / y^2 - y^1

y = mx + b

y = 4/5x + 2.6

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Mark and Don are planning to sell each of their marble collections at a garage sale. If Don has 2 more than 5 times the number o
ch4aika [34]

Mark and Don have 18  and 92 marbles respectively.

<h3>What is the solution of the equation?</h3>

Equations are mathematical statements with two algebraic expressions flanking the equals (=) sign on either side. It demonstrates the equality of the relationship between the expressions printed on the left and right sides. We have LHS = RHS (left hand side = right hand side) in every mathematical equation. To determine the value of an unknown variable that represents an unknown quantity, equations can be solved.

Let the number of marbles Mark has be x.

According to the question,

Don has (5x+2) marbles.

Total number of marbles=110

Thus, the required equation is:

x+5x+2=110

6x=108

x=108/6

x=18

Mark has 18 marbles.

Don has (5*18+2)=92 marbles.

Learn more about equation here:

brainly.com/question/2972832

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4 0
2 years ago
Let a is in {−3, −1, 0}. Evaluate −a for each element of the set.
12345 [234]
Ok so a∈{-3,-1,0}. then just {-3*-1,-1*-1,0*-1} which comes to  {3,1,0}

7 0
3 years ago
72.6 meters in 11 seconds
ira [324]
This would simplify to 6.60 meters/second.
8 0
3 years ago
Read 2 more answers
HELpp
jenyasd209 [6]

Answer:

65,212

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Find the area under the standard normal probability distribution between the following pairs of​ z-scores. a. z=0 and z=3.00 e.
prohojiy [21]

Answer:

a. P(0 < z < 3.00) =  0.4987

b. P(0 < z < 1.00) =  0.3414

c. P(0 < z < 2.00) = 0.4773

d. P(0 < z < 0.79) = 0.2852

e. P(-3.00 < z < 0) = 0.4987

f. P(-1.00 < z < 0) = 0.3414

g. P(-1.58 < z < 0) = 0.4429

h. P(-0.79 < z < 0) = 0.2852

Step-by-step explanation:

Find the area under the standard normal probability distribution between the following pairs of​ z-scores.

a. z=0 and z=3.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 3.00) = 0.9987

Thus;

P(0 < z < 3.00) = 0.9987 - 0.5

P(0 < z < 3.00) =  0.4987

b. b. z=0 and z=1.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 1.00) = 0.8414

Thus;

P(0 < z < 1.00) = 0.8414 - 0.5

P(0 < z < 1.00) =  0.3414

c. z=0 and z=2.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 2.00) = 0.9773

Thus;

P(0 < z < 2.00) = 0.9773 - 0.5

P(0 < z < 2.00) = 0.4773

d.  z=0 and z=0.79

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 0.79) = 0.7852

Thus;

P(0 < z < 0.79) = 0.7852- 0.5

P(0 < z < 0.79) = 0.2852

e. z=−3.00 and z=0

From the standard normal distribution tables,

P(Z< -3.00) = 0.0014  and P(Z< 0) = 0.5

Thus;

P(-3.00 < z < 0 ) = 0.5 - 0.0013

P(-3.00 < z < 0) = 0.4987

f. z=−1.00 and z=0

From the standard normal distribution tables,

P(Z< -1.00) = 0.1587  and P(Z< 0) = 0.5

Thus;

P(-1.00 < z < 0 ) = 0.5 -  0.1586

P(-1.00 < z < 0) = 0.3414

g. z=−1.58 and z=0

From the standard normal distribution tables,

P(Z< -1.58) = 0.0571  and P(Z< 0) = 0.5

Thus;

P(-1.58 < z < 0 ) = 0.5 -  0.0571

P(-1.58 < z < 0) = 0.4429

h. z=−0.79 and z=0

From the standard normal distribution tables,

P(Z< -0.79) = 0.2148  and P(Z< 0) = 0.5

Thus;

P(-0.79 < z < 0 ) = 0.5 -  0.2148

P(-0.79 < z < 0) = 0.2852

8 0
3 years ago
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