MATHEMATICS COLLEGE

Simplify the following expression by using the distributive property 6(3a-4)

Answers

Answer 1

Answer:

$\huge\boxed{\boxed{Hello\:there!}}$

The Distributive property states that

a(b+c)=ab+ac

Now, let's simplify:

6(3a-4)

18a-24

$\huge\bold{Hope\:it\:helps\;and\;good\;luck.}$

$\huge\rm{LoveLastsAllEternity :)$

Answer 2

Answer:

Answer:

first you multiply 6 and 3a which equals 18a. Then you multiply 6 and 4 which equals 24. So in the end it should look like: 18a-24

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Need Help Fast!!!!!!!!!!!!!!!!!!

Answers

7.2, 3, 8.09, 2.22, 5.06, 2.5

Consider the following system of equations:Equation 1: 82 + 2y = 30
Equation 2: 70 + 2y = 24
Which variable pair should we try to eliminate?
The x's because the coefficients are the same.
The y's because the coefficients are the same.
The x's because the coefficients are different?
The y's because the coefficients are different?

Answers

The correct answer is: "The y's because the coefficients are the same."

Further explanation:

we always try to eliminate the variable from the system of equations which involves lesser calculations and complexity.

Assuming that Given equations are:

Equation 1: 82x + 2y = 30

Equation 2: 70x + 2y = 24

we can see that the coefficients of x are very large and different. While the coefficients of y are same. We can only subtract the equations to eliminate y.

Hence,

The correct answer is: "The y's because the coefficients are the same."

Keywords: Linear equations, variables

Learn more about linear equations at:

#LearnwithBrainly

The linear equation of the initial (red ) incline would be y = __x +__

Answers

Answer:

$y=\underline {1.5}x+\underline 0$

Step-by-step explanation:

From the graph,

The initial red incline is a straight line passing through the origin.

So, a straight line passing through the origin is of the form:

$y=mx+0$

Where, 'm' is the slope.

Now, the slope of a given as the change in y value to change in x value.

Consider the two end points of the red line (0, 0) and (4, 6).

The slope with two points $(x_1,y_1)\ and\ (x_2,y_2)$ on the line is given as:

$m=((y_2-y_1))/((x_2-x_1))$

Plug in $x_1=0,y_1=0,x_2=4,y_2=6$ . Therefore, slope is:

$m=(6-0)/(4-0)\nm=(6)/(4)=1.5$

Hence, the equation of the initial red incline is:

$y=\underline {1.5}x+\underline 0$

11/5 divided by 51/4 equals

Answers

When 11/5 is divided by 51/4 then the quotient will be equal to 44/255.

To divide two fractions, multiply the first fraction by the reciprocal of the second fraction.

The reciprocal of a fraction is obtained by swapping the numerator and the denominator.

So, to calculate (11/5) ÷ (51/4), we can do the following:

(11/5) ÷ (51/4)

= (11/5) × (4/51)

Now, multiply the numerators and denominators:

Numerator: 11 × 4 = 44

Denominator: 5 × 51 = 255

Therefore, (11/5) ÷ (51/4) = 44/255.

Hence, when 11/5 divided by 51/4 equals to 44/255.

To learn more on Division click:

brainly.com/question/21416852

#SPJ6

Answer:

44/255

Step-by-step explanation:

Since you you are dividing fraction you have to multiply by the reciprocal.

11/5 x 4/51

11 x 4 = 44

51 x 5 = 255

= 44/255

Troy and Lisa were shopping for school supplies. Each purchased different quantities of the same notebook and thumb drive. Troy bought five notebooks and five thumb drives for $110. Lisa bought three notebooks and five thumb drives for $102. Find the cost of each notebook and each thumb drive.

Answers

Answer:

Notebooks cost $4

Thumb drives cost $18

Step-by-step explanation:

x=$ for notebooks

y=$ for thumb drives

5x + 5y = 110 ----> for Troy

3x + 5y = 102 ----> for Lisa

Use subtraction to find x.

(5x + 5y = 110)

-(3x + 5y = 102)

_____________

2x + 0y = 8

2x = 8

x=4

Sub 4 for x into 5x + 5y = 110

20 + 5y = 110

-20 -20

___________

5y = 90

y=18

Notebooks cost $4

Thumb drives cost $18

Data taken from a random sample of 60 students chosen from the student population of a large urban high school indicated that 36 of them planned to pursue post-secondary education. An independent random sample of 50 students taken at a neighboring large suburban high school resulted in data that indicated that 31 of those students planned to pursue post-secondary education. Do these data provide sufficient evidence at the 5% level to reject the hypothesis that these population proportions are equal

Answers

Answer:

No, these data do not provide sufficient evidence at the 5% level to reject the hypothesis that these population proportions are equal.

Step-by-step explanation:

We are given that data taken from a random sample of 60 students chosen from the student population of a large urban high school indicated that 36 of them planned to pursue post-secondary education.

An independent random sample of 50 students taken at a neighboring large suburban high school resulted in data that indicated that 31 of those students planned to pursue post-secondary education.

Let $p_1$ = population proportion of students of a large urban high school who pursue post-secondary education.

$p_2$ = population proportion of students of a large suburban high school who pursue post-secondary education.

So, Null Hypothesis, $H_0$ : $p_1-p_2$ = 0 {means that these population proportions are equal}

Alternate Hypothesis, $H_A$ : $p_1-p_2\neq$ 0 {means that these population proportions are not equal}

The test statistics that would be used here Two-sample z proportionstatistics;

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{(\hat p_1(1-\hat p_1))/(n_1)+(\hat p_2(1-\hat p_2))/(n_2) } }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of students of a large urban high school who pursue post-secondary education = $(36)/(60)$ = 0.60

$\hat p_2$ = sample proportion of students of a large urban high school who pursue post-secondary education = $(31)/(50)$ = 0.62

$n_1$ = sample of students of a large urban high school = 60

$n_2$ = sample of students of a large suburban high school = 50

So, the test statistics = $\frac{(0.60-0.62)-(0)}{\sqrt{(0.60(1-0.60))/(60)+(0.62(1-0.62))/(50) } }$

= -0.214

The value of z test statistics is -0.214.

Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that these population proportions are equal.

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