I Need help ASAP if you get right ill name you BRAINLEST

If a number is divided by negative two, the solution is negative sixteen

Sergio [31]

N/-2=-16
Aka
Negative number(n) is divided by negative two equals negative sixteen

3 0

3 years ago

Log3 of 5 <br><br>round to nearest thousandth

Semmy [17]

<span>1.465 is the correct answer I believe.</span>

3 0

3 years ago

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.

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 $z>2.33$

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.

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 $z>1.96 \cup z$

7 0

3 years ago

• Packaging of a product can cost a lot of money. Can you design

podryga [215]

The best arrangement can be

3 rows and 4balls each row

So

Length of one row=1.8(4)=7.2cm
Width=1.8(3)=5.4cm
Height=1.8(2)=3.6cm

So

TSA

2(LB+BH+LH)
2(5.4(7.2)+(7.2)(3.6)+3.6(5.4))
2(84.24)
168.48cm²

Total cost

168.48(0.01)
1.685€

5 0

2 years ago

At last week's football game, 125 tickets were sold. This week 225 tickets were sold. What percent of the

victus00 [196]

You have to do 225 plus 125 divided by 2 times 100 and that is your percent

6 0

3 years ago