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4 years ago

What is 475.53 minus 67.37

Mathematics

2 answers:

lozanna [386]4 years ago

8 0

The answer is 408.16

Ivanshal [37]4 years ago

3 0

The answer would be 408.16

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1. Find the equation of a line passing through the points (1,-2) and (-3,10).

liraira [26]

Answer:

Step-by-step explanation:

(1 , -2) x₁ =1 & y₁ = -2

(-3,10) x₂= -3 & y₂ = 10

First we need to find the slope.

$\boxed{Slope = $\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$}$ $\boxed{Slope=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}}$

$=\dfrac{10-[-2]}{-3-1}\\\\\\=\dfrac{10+2}{-4}\\\\\\=\dfrac{12}{-4}\\\\\\= - 3$

m = -3

Equation of the line: y = mx + b

y = -3x + b

To find the value of b, take one of the point and substitue in the above equaiton.

(1 , -2)

-2 = -3*1 + b

-2 = -3 + b

-2 + 3 = b

b = 1

Equation of the line:

y = -3x + 1

4 0

3 years ago

PLEASE HELP ME PLEASE!

Mamont248 [21]

Answer:

-4 and d

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Please give brainliest, I really want to rank up, thank you!

5 0

3 years ago

Read 2 more answers

What is (x) in this problem 3/5=(-x)+1

Maurinko [17]

3/5= -x+1
-2/5= -x

x = 2/5

3 0

4 years ago

Last week Andrew bought 3 lbs. of zucchini and 2 lbs. of tomatoes for $7.05 at a farm stand. This week he bought 4 lbs. of zucch

olga nikolaevna [1]

$3x+2y=7.05\\4x+3y=9.83\\x+y=2.78$

$2.78

3 0

3 years ago

In a large school, it was found that 77% of students are taking a math class, 74% of student are taking an English class, and 70

Iteru [2.4K]

Answer:

0.81 = 81% probability that a randomly selected student is taking a math class or an English class.

0.19 = 19% probability that a randomly selected student is taking neither a math class nor an English class

Step-by-step explanation:

We solve this question working with the probabilities as Venn sets.

I am going to say that:

Event A: Taking a math class.

Event B: Taking an English class.

77% of students are taking a math class

This means that $P(A) = 0.77$

74% of student are taking an English class

This means that $P(B) = 0.74$

70% of students are taking both

This means that $P(A \cap B) = 0.7$

Find the probability that a randomly selected student is taking a math class or an English class.

This is $P(A \cup B)$ , which is given by:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

$P(A \cup B) = 0.77 + 0.74 - 0.7 = 0.81$

0.81 = 81% probability that a randomly selected student is taking a math class or an English class.

Find the probability that a randomly selected student is taking neither a math class nor an English class.

This is

$1 - P(A \cup B) = 1 - 0.81 = 0.19$

0.19 = 19% probability that a randomly selected student is taking neither a math class nor an English class

6 0

3 years ago