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3 years ago

Write the equation 2y − 4x = 5 in the form y = mx +

2 answers:

7 0

Starting with <span>2y − 4x = 5, solve for y. To do this, add 4x to both sides, obtaining 2y = 4x+5. Divide both sides by 2, obtaining

y = 2x + 5/2 (answer)</span>

tekilochka [14]3 years ago

5 0

Answer:

$y=2x+\frac{5}{2}$ .

Step-by-step explanation:

We have been equation of a line in standard form of equation $2y-4x=5$ . We are asked to write our given equation in slope-intercept form of equation $y=mx+b$ .

To convert our given equation in slope-intercept form, we need to separate y on one side of equation using opposite operations.

Add $4x$ on both sides:

$2y-4x+4x=5+4x$

$2y=4x+5$

Now, we will divide both sides of our equation by 2.

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

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

$y=2x+\frac{5}{2}$

Therefore, our required equation would be $y=2x+\frac{5}{2}$ .

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Can someone please help me.. very quickly .. thank you

Alisiya [41]

X/8 + 12 = 16
start by subtracting twelve from both sides
x/8 + 12 = 16
- 12 -12
You're left with x/8 = 4
Multiply both sides by 8
8× x/8 = 4 ×8
Your answer is x = 32
If you need to simplify, the answer is 2.

3 0

3 years ago

Read 2 more answers

Gas station X charges $4.00 per gallon of gas. Gas station Y charges $4.12 per gallon of gas. What is the percent increase from

kirill [66]

First, find the increase in charge between the two station. That is:

$4.12-$4.00=$0.12

Then, divide the increase by the price charged by Gas Station X.

$0.12/$4.00=0.03

Lastly, since it is a decimal, transform it to percentage by multiplying 100%

0.03x100%=3%

Hope this is the brainliest answer.

3 0

3 years ago

Can someone help please

elena-s [515]

Answer:

I'm sorry I can help you maybe a little bit if that's okay. Btw what grade are you in?

Step-by-step explanation:

3 0

3 years ago

A market analyst is hired to provide information on the type of customers who shop at a particular store. A random survey is tak

sveta [45]

Complete question:

A market analyst is hired to provide information on the type of customers who shop at a particular store. A random survey is taken of 100 shoppers at this store. Of these 100, 73 are women. The shoppers The data is summarized below. we grouped in three age categories, under 30, 30 up to 50 and 50 and over. Women under 30 are 30, Men under 30 are 8 Women 30 to 50 are 25 and Men are 14 Women 50 and over are 18 and Men are 5.

Let W be the event that a randomly selected shopper is a woman. Let A be the event that a randomly selected shopper is under 30. a.) Find the probability of W. W (women Shoppers) = Find the probability of A. b.) Find the probability of A and W. c.) P (A and W) (Shoppers) d.) Find the probability of A or W. P(A or W) (Shoppers) e.) Find the probability of A given W

Answer:

A) P(W) = 73/100; B) P(A) = 38/ 100 C) 30 / 100

D) 81/100 E) 30/73

Step-by-step explanation:

- - - - - - - - Women Men Total

Under 30 - - - 30 - - - 8 - - 38

30 to 50 - - - - 25 - - 14 - - 39

50 & over - - - 18 - - - 5 - - -23

Totals - - - - - - 73 - - -27 - - 100

Recall:

Probability : P= (required outcome / Total possible outcomes)

A) probability of women shoppers: P(W)

Number of women shoppers = 73; total shoppers = 100

P(W) = 73/100

B) Probability of under 30: P(A)

= number of shoppers under 30 = 38

Tital number of shoppers = 100

P(A) = 38/ 100

C) probability of A and W; This is the probability that the selected person is a woman and under 30.

(W n A) = 30

= 30 / 100

D) probability of A or W:

P(A) + P(W) - P(A n W) :

(38 / 100 + 73 / 100 - 30 / 100) = (38+73-30) /100

= 81/100

E) probability of A given W:

P(A n W) / P(W) = 30/73

5 0

3 years ago

A survey collected data on annual credit card charges in seven different categories of expenditures: transportation, groceries,

Andru [333]

Answer:

A. Null and alternative hypothesis:

$H_0: \mu_d=0\\\\H_a:\mu_d\neq 0$

B. Yes. At a significance level of 0.05, there is enough evidence to support the claim that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

P-value = 0.00002

C. As the difference is calculated as (population 1 − population 2), being population 1: groceries and population 2: dinning out, and knowing there is evidence that the true mean difference is positive, we can say that the groceries annual credit card charge is higher than dinning out annual credit card charge.

The point estimate is the sample mean difference d=$840.

The 95% confidence interval for the mean difference between the population means is (490, 1190).

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

Then, the null and alternative hypothesis are:

$H_0: \mu_d=0\\\\H_a:\mu_d\neq 0$

The significance level is 0.05.

The sample has a size n=42.

The sample mean is M=840.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=1123.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{1123}{\sqrt{42}}=173.2827$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{840-0}{173.2827}=\dfrac{840}{173.2827}=4.848$

The degrees of freedom for this sample size are:

$df=n-1=42-1=41$

This test is a two-tailed test, with 41 degrees of freedom and t=4.848, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=2\cdot P(t>4.848)=0.00002$

As the P-value (0.00002) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

We have to calculate a 95% confidence interval for the mean difference.

The t-value for a 95% confidence interval and 41 degrees of freedom is t=2.02.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=2.02 \cdot 173.283=350$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 840-350=490\\\\UL=M+t \cdot s_M = 840+350=1190$

The 95% confidence interval for the mean difference is (490, 1190).

7 0

3 years ago