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3 years ago

What is the y-intercept: 7x + y = 10

Mathematics

2 answers:

scoray [572]3 years ago

8 0

Answer:

Step-by-step explanation:

7(0)+y=10

y=10

Karo-lina-s [1.5K]3 years ago

4 0

Answer: (0, 10)

Step-by-step explanation:

The given expression is in the standard form of linear equations: Ax + By = C, where A and B are constants.

The definition of y-intercept is the point that intersects with the y-axis, meaning that the [x] value of the point will be zero.

To find the y-intercept of an equation, substitute the [x] variable with 0, as mentioned above, the [x] value of a y-intercept is 0.

$7x+y=10\\$

$7(0)+y=10$

$0+y=10$

$y=10$

As we find the [y] value is 10, while the given [x] value is 0, therefore, the y-intercept of $7x+y=10$ is $\boxed{(0,10)}$ .

Hope this helps!! :)

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Find the appropriate rejection regions for the large-sample test statistic z in these cases. (Round your answers to two decimal

Usimov [2.4K]

Answer:

a) We have that the significance is given by $\alpha =0.01$ and we know that we have a right tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 1% of the area on the right and 99% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.01,0,1)"

And we got for this case $z_{crit}=2.33$

So then the rejection region would be $z>2.33$

b) We have that the significance is given by $\alpha =0.05$ , $\alpha/2 =0.025$ and we know that we have a two tailed test.

So for this case we need to look in the normal standard dsitribution a critical value that accumulates 2.5% of the area on the right and 97.5% of the area on the left. This value can be founded with the following excel code:

"=NORM.INV(1-0.025,0,1)"

And we got for this case $z_{crit}=\pm 1.96$

So then the rejection region would be $z>1.96 \cup z$

Step-by-step explanation:

Part a

We have that the significance is given by $\alpha =0.01$ and we know that we have a right tailed test.

"=NORM.INV(1-0.01,0,1)"

And we got for this case $z_{crit}=2.33$

So then the rejection region would be $z>2.33$

Part b

We have that the significance is given by $\alpha =0.05$ , $\alpha/2 =0.025$ and we know that we have a two tailed test.

"=NORM.INV(1-0.025,0,1)"

And we got for this case $z_{crit}=\pm 1.96$

So then the rejection region would be $z>1.96 \cup z$

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3 years ago

Is is true that the coefficient of the expression 18z-7(-3+2z) of z is 13

Sever21 [200]

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2 years ago

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Zina [86]

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