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3 years ago

When solving a system by multiplying and then adding or subtracting, how do you decide whether to add or subtract?

1 answer:

8 0

i don’t know if your school does PEMDAS but it stands for parentheses, exponents, multiplication, division, addition, and then subtraction. so you would multiply, then add, and then subtract.

You might be interested in

Given the coordinates for the function below, who of the following are coordinates for its inverse?

lapo4ka [179]

Answer:

If we have a function f(x) that has an inverse g(x), we have the relations:

f( g(x)) = x

g( f(x)) = x

Then if the coordinates for the function f(x) are points like (x₁, y₁)

The coordinates for the inverse, g(x), will be (y₁, x₁)

Now let's look at the table, here we can see that the coordinates for f(x) are:

(0, 650)

(100, 550)

(200, 450)

(340, 310)

(650, 0)

Then the coordinates of the inverse function will be:

(650, 0)

(550, 100)

(450, 200)

(340, 310)

(0, 650)

5 0

2 years ago

Write an equation of the line that passes through the given points.<br> (-2,8) and (1, -1)

KIM [24]

/Answer:

Short Answer Type:

P1 - (-2,8)

P2 - (1,1)

Slope: (1 - 8) / (1 + 2)

=> -7/3

Therefore: 1 = -7/3 + c

=> 10/3 = c

<h2><u>Therefore equation: y = -7x/3 + 10/3</u> </h2>

6 0

3 years ago

What is the slope of the line through (-4,2) and (3, 3)? "/ yo Choose 1 answer: 19 21 42 142 A А 7 5 B 7 5 5 7 5 7 e slope an

NNADVOKAT [17]

Answer:

1/7

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(3-2)/(3-(-4))

m=1/(3+4)

m=1/7

5 0

3 years ago

You are testing the claim that the mean GPA of night students is different from the mean GPA of day students. You sample 30 nigh

kirill [66]

Answer:

Step-by-step explanation:

This is a test of 2 independent groups. The population standard deviations are not known. Let μ1 be the mean GPA of night students and μ2 be the mean GPA of day students.

The random variable is μ1 - μ2 = difference in the mean GPA of night students and the mean GPA of day students.

We would set up the hypothesis.

The null hypothesis is

H0 : μ1 = μ2 H0 : μ1 - μ2 = 0

The alternative hypothesis is

H1 : μ1 ≠ μ2 H1 : μ1 - μ2 ≠ 0

This is a two tailed test.

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(x1 - x2)/√(s1²/n1 + s2²/n2)

From the information given,

μ1 = 2.35

μ2 = 2.58

s1 = 0.46

s2 = 0.47

n1 = 30

n2 = 25

t = (2.35 - 2.58)/√(0.46²/30 + 0.47²/25)

t = - 1.8246

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [0.46²/30 + 0.47²/25]²/[(1/30 - 1)(0.46²/30)² + (1/25 - 1)(0.47²/25)²] = 0.00025247091/0.00000496862

df = 51

We would determine the probability value from the t test calculator. It becomes

p value = 0.0746

Since alpha, 0.05 < than the p value, 0.0746, then we would fail to reject the null hypothesis.

3 0

3 years ago

Two random variables x and y are independent if the value of x does not affect the value of y. If the variables are not indepen

jeyben [28]

Answer: $\mu_{x+y} =$ 3026

$\mu_{x-y}=$ 30

Step-by-step explanation: Average sum of the female and male's test score is the sum of expected value of each gender:

$\mu_{x+y}=\mu_{x}+\mu_{y}$

Assuming x represents the random male selected and y represents the random female selected:

$\mu_{x+y}=1528+1498$

$\mu_{x+y}=$ 3026

The average sum of their scores is 3026.

Average difference is the difference between the expected value (mean) of each gender:

$\mu_{x-y}=\mu_{x}-\mu_{y}$

$\mu_{x-y}=$ 1528 - 1498

$\mu_{x-y}=$ 30

The average difference of their scores is 30.

6 0

3 years ago