Buying a 30 dollar shirt for $30 and getting a second shirt of equal value for 50%off

Convert this to a decimal then a percent

Ahat [919]

9/27=33.3% repeating 3 = 0.3 repeating 3

18/27= 66.6 repeating 6 = 0.6 repeating 6

4 0

4 years ago

Proof: sin(x) + cos(x) = √2sin(x + π/4)

Oksi-84 [34.3K]

Hello :
sin(x) + cos(x) = √2(1/√2 sin(x) +1/√2 cos(x))
but : 1/√2 = cos(π/4) = sin(<span> π/4)
</span>sin(x) + cos(x) = √2( cos(π/4)sin(x) + sin( π/4) cos(x)) =√2<span>sin(x + π/4)
</span>because : cos(π/4)sin(x) + sin( π/4) cos(x)=sin(x + π/4) by identity :
sin(a+b) = sina cosb +cosa sinb

5 0

3 years ago

Helppppppp unicorn we’re almost done

Inessa05 [86]

Answer:

2 .KL = HF

3 .GIVEN

4 . JKL = GHF , HL

6 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

4 years ago

Math work plz, i just need to get an answer

Neporo4naja [7]

It is 10, and 15 I believe

6 0

3 years ago

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