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sineoko [7]
3 years ago
11

What is the slope of the line whose equation is y−4=5/2(x−2)?

Mathematics
2 answers:
postnew [5]3 years ago
7 0

Answer:

m=\dfrac{5}{2}

Step-by-step explanation:

If the equation of the line is

y=mx+b,

then m represents the slope of the line and b represents the y-intercept of the line. This equation is called the equation of the line in the slope form.

Rewrite the equation of the line in the slope form

y-4=\dfrac{5}{2}(x-2)\\ \\y-4=\dfrac{5}{2}x-\dfrac{5}{2}\cdot 2\\ \\y-4=\dfrac{5}{2}x-5\\ \\y=\dfrac{5}{2}x-1

Thus, the slope of the line is

m=\dfrac{5}{2}

Debora [2.8K]3 years ago
6 0

The slope of a line whose equation is y-4 = \frac{5}{2}(x-2) is \frac{5}{2}

<h2>Further Explanation</h2><h3>Slope/gradient</h3>
  • Slope or the gradient of a line refers to the change along the y-axis divided by the change along the x-axis.
  • The slope of the line can be calculated from two co-ordinates of the line in question or obtained from the equation of a line
<h3>Equation of a straight line  </h3>
  • Equation of a straight line is written in the form y=mx+ c, where m and c are numbers.
  • m is the slope or gradient of the line while c is the y-intercept.
<h3>Equation of a straight line can be found when given:</h3>
  1. A slope of the line and one point where the line is passing through  
  2. Two points where the line is passing through
  3. A slope of the line and the y-intercept  

In this case;

The equation in question is;

y-4 = \frac{5}{2}(x-2)

Combining like terms;

y= \frac{5}{2}x-5+4

The equation of the line is

y= \frac{5}{2}x-1

From the equation the slope of the line is \frac{5}{2}, while

The y-intercept is -1

Keywords: Slope, Equation of a straight line, y-intercept,  

<h3>Learn more about:  </h3>
  • Equations of a straight line: brainly.com/question/4932386
  • Slope of a straight line: brainly.com/question/4932386
  • Double intercept: brainly.com/question/4932386

Level: High school  

Subject: Mathematics  

Topic: Equation of a straight line  

Sub-topic: Slope/gradient of a line

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An article includes the accompanying data on compression strength (lb) for a sample of 12-oz aluminum cans filled with strawberr
Sliva [168]

Answer:

A) Option B is correct.

H₀: μ₁ = μ₂

Hₐ: μ₁ - μ₂ < 0

B) t = -2.502

p-value = 0.0112

C) Option A is correct.

Reject H₀. The data suggests that cola has a higher average compression strength than the strawberry drink.

D) Option A is correct.

The distributions of compression strengths are approximately normal.

Step-by-step explanation:

The complete Question is presented in the two attached images to this answer.

A) To perform this test we first define the null and alternative hypothesis.

The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.

While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.

For this question, we want to test if the extra carbonation of cola results in a higher average compression strength. That is, that cola has a higher average compression strength than the strawberry drink.

Hence, the null hypothesis would be that there isn't significant evidence to suggest that the extra carbonation of cola results in a higher average compression strength, that is, cola has a higher average compression strength than the strawberry drink.

The alternative hypothesis is that there is significant evidence to suggest that the extra carbonation of cola results in a higher average compression strength, that is, cola has a higher average compression strength than the strawberry drink.

Mathematically, if the average compression strength of strawberry drink is μ₁, the average compression strength of cola is μ₂ and the difference in compression strengths is μ = μ₁ - μ₂

The null hypothesis is represented as

H₀: μ = 0 or μ₁ = μ₂

The alternative hypothesis is represented as

Hₐ: μ < 0 or μ₁ - μ₂ < 0

B) So, to perform this test, we need to compute the test statistic

Test statistic for 2 sample mean data is given as

Test statistic = (μ₁ - μ₂))/σ

σ = √[(s₂²/n₂) + (s₁²/n₁)]

μ₁ = average compression strength of strawberry drink = 537

n₁ = sample size of the sample of strawberry drink in cans surveyed = 10

s₁ = standard deviation of the compression strength of strawberry drink in cans surveyed= 22

μ₂ = average compression strength of cola = 559

n₂ = sample size of the sample of cola in cans surveyed = 10

s₂ = standard deviation of the compression strength of strawberry drink in cans surveyed = 17

σ = [(17²/10) + (22²/10)] = 77.5903160379 = 8.792

We will use the t-distribution as no information on population standard deviation is provided

t = (537 - 559) ÷ 8.792

= -2.502 = -2.50

checking the tables for the p-value of this t-statistic

Degree of freedom = df = n₁ + n₂ - 2 = 10 + 10- 2 = 18

Significance level = 0.05

The hypothesis test uses a one-tailed condition because we're testing in only one direction (whether compression strength of cola in can is greater).

p-value (for t = -2.50, at 0.05 significance level, df = 18, with a one tailed condition) = 0.011154 = 0.0112 to 4 d.p.

C) The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 0.05

p-value = 0.0112

0.0112 < 0.05

Hence,

p-value < significance level

This means that we reject the null hypothesis accept the alternative hypothesis & say that the extra carbonation of cola results in a higher average compression strength, that is, cola has a higher average compression strength than the strawberry drink.

D) The necessary conditions required before a t-test is deemed valid include.

- The samples used must be a random sample of the population distribution with each variable in the sample independent of other one.

- The distribution of the population where the samples were extracted from must be normal or approximately normal to ensure some degree of normality for the samples.

Hence, the necessary assumption for this t-test among the options is that the distributions of compression strengths are approximately normal.

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