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3 years ago

If (4,y) is a solution to x-2y = 6, what is the value of y?

Mathematics

1 answer:

frutty [35]3 years ago

5 0

Answer:

(4, -1)

Step-by-step explanation:

(4, y) is a solution to x - 2y = 6.

Find the value of y:

$4-2y=6\\\\4-4-2y=6-4\\\\-2y=2\\\\\frac{-2y=2}{-2}\\\\\boxed{y=-1}$

The value of y is '-1'.

(4, -1) is a solution to the equation.

Hope this helps.

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Simply 8×4³+2×4⁴ giving you answer in the form of 4^b

hram777 [196]

Answer:

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Step-by-step explanation:

8×4³+2×4⁴

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4^4×2+2×4^4

4^4(2+2)

4^4×4

4^5

So answer is 4^5

5 0

3 years ago

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statuscvo [17]

Answer:

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Step-by-step explanation:

17000

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3 years ago

I need help on this one

Dominik [7]

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3 years ago

In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to 6977subjects randomly selec

kirza4 [7]

Answer:

Given :An Internet survey was e-mailed to 6977 subjects randomly selected from an online group involved with ears. There were 1337 surveys returned.

To Find : Use a 0.01 significance level to test the claim that the return rate is less than 20%.

Solution:

n = 6977

x = 1337

We will use one sample proportion test

$\widehat{p}=\frac{x}{n}$

$\widehat{p}=\frac{1334}{6977}$

$\widehat{p}=0.1911$

We are given that the claim is the return rate is less than 20%.

$H_0:p=0.2\\H_a:p$

Formula of test statistic = $\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

= $\frac{0.1911-0.2}{\sqrt{\frac{0.2(1-0.2)}{6977}}}$

= $−1.858$

Refer the z table

P(z<-1.85)=0.0332

Significance level = 0.01=α

Since p value > α

So, we accept the null hypothesis .

So,the claim that the return rate is less than 20% is false.

7 0

3 years ago

Two college instructors are interested in whether or not there is any variation in the way they grade math exams. They each grad

AlekseyPX

Answer:

Since the calculated value of F = 1.4397 is less than the critical value of

F (9,9)= 2.4403 we conclude that the first instructor's variance is smaller and reject H0.

Step-by-step explanation:

1)Formulate the hypothesis that first variance is equal or greater than the second variance

H0: σ₁²≥σ₂² against the claim that the first instructor's variance is smaller

Ha: σ₁²< σ₂²

2) Test Statistic F= s₂²/s₁²

F= 84.8/ 58.9= 1.4397

3)Degrees of Freedom = n1-1= 10-1= 9 and n2 = 10-1= 9

4)Critical value at 10 % significance level= F(9,9)= 2.4403

5)Since the calculated value of F = 1.4397 is less than the critical value of

F (9,9)= 2.4403 we conclude that the first instructor's variance is smaller and reject H0.

4 0

3 years ago