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AysviL [449]
3 years ago
6

Write the equation of a line that is perpendicular to x=3x=3x, equals, 3 and that passes through the point (0,-4)(0,−4)left pare

nthesis, 0, comma, minus, 4, right parenthesis.
Mathematics
1 answer:
leva [86]3 years ago
8 0

Answer: x+3y=12

Step-by-step explanation:

Slope of two lines that are perpendicular to each other is 1.

If one line is y=3x, then its slope = 3   [by comparing to the linear equation y= mx+c, here m=3]

Let n be the slope of the required line, then

n\times3=-1\\\\\rightarrow\ n=\dfrac{-1}{3}

Equation of line with slope n and passers through (a,b) is

(y-b)=n(x-a)

Equation of line with slope n= \dfrac{-1}{3} and passes through point ( 0,-4) :

(y-(-4))=\dfrac{-1}{3}(x-0)\\\\\Rightarrow\ y+4=\dfrac{-1}{3}x\\\\\Rightarrow\ -3(y+4)=x\\\\\Rightarrow-3y-12=x\\\\\Rightarrow x+3y=12

Hence, Required equation : x+3y=12

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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
Caroline is considering two video game rental plans. Plan A can be modeled with the equation C = 2n, and Plan B can be modeled w
meriva

Options

A. Caroline rents exactly 7 games each month.

B. Caroline rents exactly 6 games each month.

C. Caroline rents 6 or more games each month.

D. Caroline rents from 1 to 5 games each month.

Answer:

D. Caroline rents from 1 to 5 games each month.

Step-by-step explanation:

Given

Plan A:

C = 2n

Plan B:

C = n + 6

Required

Which options justifies A over B

The solution to this question is option (d).

In option d, n = 1,2,3,4,5

When any of the values of n is substituted in plan A and B, respectively; the cost of plan A is cheaper than plan B.

This is not so, for other options (A - C)

To show:

Substitute 1 for n in A and B

Plan A:

C = 2n  = 2 * 1 = 2

Plan B:

C = n + 6 = 1 + 6 = 7

Substitute 5 for n in A and B

Plan A:

C = 2n  = 2 * 5 = 10

Plan B:

C = n + 6 = 5 + 6 = 11

<em>See that A < B</em>

8 0
3 years ago
Write the equation of a line that passes through the point (1,4) and has a slope of 9.
Kryger [21]

Answer:

y=9x-5

Step-by-step explanation:

You want to find the equation for a line that passes through the point (1,4) and has a slope of 9.

First of all, remember what the equation of a line is:

y = mx+b

Where:

m is the slope, and

b is the y-intercept

To start, you know what m is; it's just the slope, which you said was 9. So you can right away fill in the equation for a line somewhat to read:

y=9x+b.

Now, what about b, the y-intercept?

To find b, think about what your (x,y) point means:

(1,4). When x of the line is 1, y of the line must be 4.

Because you said the line passes through this point, right?

Now, look at our line's equation so far: b is what we want, the 9 is already set and x and y are just two "free variables" sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specifically passes through the point (1,4).

So, why not plug in for x the number 1 and y the number 4? This will allow us to solve for b for the particular line that passes through the point you gave!.

(1,4). y=mx+b or 4=9 × 1+b, or solving for b: b=4-(9)(1). b=-5.The equation of the line that passes through the point (1,4) with a slope of 9

is

y=9x-5

4 0
3 years ago
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