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3 years ago

X +3y =6? use the x and y intercepts to graph each linear equation. please show all work.

1 answer:

8 0

X and y intercepts
get a graphing paper
when x=0 then the line intercepts or passes through the y axis
when y=0 then the line intecepts or passes through the x axis
so subsitute 0 for x and solve for the y intercept
0+3y=6
3y=6
diivded by 3
y=2
put a point on the y axis 2 units up from thhe origin (0,0) on the y axis

subsitute 0 for y and solve for x intercept
x+3(0)=6
x=6
pput a point on the x axis 6 units to the right of the origin on the x axis

now that you have 2 points, draw a straight line connecting the two points

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Im so confused please help with 5 and 6

Drupady [299]

Answer:

5)102,78

6)56,56

See THE IMAGE FOR SOLUTION 

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3 years ago

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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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4 years ago

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katen-ka-za [31]

The answer is

B) 2n - 5

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3 years ago

Can anyone answer this question please 6[y-2]/7-12=2[y-7]/3

Kaylis [27]

Answer:

y =47.5.

Step-by-step explanation:

First eliminate the fractions by multiplying through by the LCM of 7 and 3 which is 21:

21* 6[y-2]/7-21*12 = 21*2[y-7]/3

18(y - 2) - 252 = 14(y - 7)

18y -36 - 252 = 14y - 98

18y - 14y = -98 + 36 + 252

4y = 190

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y = 47.5.

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2 years ago

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kifflom [539]

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So y represents the initial deposit of Zen, which is 1000.

ANSWER: Answer choice D (Amount of money Zen deposited in the savings account)

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3 years ago

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