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3 years ago

Beyonce is solving a system of equations: 2x – 3y = -2

Mathematics

1 answer:

nlexa [21]3 years ago

7 0

Answer:

Step-by-step explanation:

A system of equations has no solutions if you end up with a contradiction. For example, the system of two equations

y = 2x + 4

y = 2x

If you subtract the second equation from the first, you get

0 = 4

an obvious contradiction. If you were to graph the equations, you would see that the lines are parallel. They never intersect, so there is no point that satisfies both equations.

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The value of 7y–2 is 10 more than the value of 2y

const2013 [10]

Answer:

y=12/5 (in decimal notation y=2.4)

Step-by-step explanation:

the value of 7y-2 = 10 more than 2y

7y-2 = 10+2y

7y-2y = 10+2

5y = 12

y = 12/5

y = 2.4

3 0

4 years ago

Describe how the graph of y= x2 can be transformed to the graph of the given equation. y = x2 + 3 (5 points)

kari74 [83]

Answer:

x=-1 1/2,x=-1.5

Step-by-step explanation: x=-3/2

6 0

3 years ago

What is the rate of change of y with respect to x for this function?

lilavasa [31]

Answer:

Given the function: y=f(x) = 3x+2

when x=-2 at the beginning of the interval [-2, 5],

then;

y = 3x+2 begins at

y= 3(-2)+2 = -6+2= -4.

and

when x=5 at the end of the interval [-2, 5],

y = 3x+2 ends up at

y= 3(5)+2 = 15+2= 17.

So,

y has changed -4 to 17, which is a change of 17-(-4)= 17+4 = 21

and x has changed from -2 to 5, which is a change of 5-(-2)=5+2=7

So, the average rate of change of y with respect to x over the interval

[-2, 5] is given by ;

Therefore, the average rate of change y with respect to x over the interval is, 3

Step-by-step explanation:

4 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:

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