Please draw for me when you answer the question above.

What is the minimum value of the graph of y=sin x assumes

Alex

Short answer: - 1
The first thing you should do is draw this or get a program to do it. I use desmos. It is included. The graph shows one complete cycle every 2*pi intervals (or 360 degrees.)

You will find no lower value that y = - 1 (nor any value greater than y = 1)

6 0

3 years ago

Read 2 more answers

Emma measured a restaurant and made a scale drawing. The scale of the drawing was 2 centimeters : 1 meter. If the actual width o

soldi70 [24.7K]

Answer:

20000 meters

Step-by-step explanation:

3 0

3 years ago

Please help me get this right I don’t know how to do it.

velikii [3]

Answer:

tan∠E=1/3

Step-by-step explanation:

Tangent or opposite/adjacent of angle E is segment HF/3 to solve for HF you can use sine of H opposite over hypotenuse or √8/HF=sin45°. Rearranging the equation you get √8/sin45°=HF and sin45°= 2√2 so √8/2√2=1 HF=1 now you know that tan(∠E)=1/3.

7 0

3 years ago

A digital camcorder repair service has set a goal not to exceed an average of 5 working days from the time the unit is brought i

garri49 [273]

Answer:

The degrees of freedom first given by:

$df=n-1=12-1=11$

Then we can find the critical value taking in count the degrees of freedom and the alternative hypothesis and then we need to find a critical value who accumulates 0.05 of the area in the right tail and we got:

$t_{\alpha}= 1.796$

And for this case the rejection region would be:

b) Reject H0 if tcalc >1.7960

Step-by-step explanation:

Information given

5, 7, 4, 6, 7, 5, 5, 6, 4, 4, 7, 5.

System of hypothesis

We want to test if the true mean is higher than 5, the system of hypothesis are :

Null hypothesis: $\mu \leq 5$

Alternative hypothesis: $\mu > 5$

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

The degrees of freedom first given by:

$df=n-1=12-1=11$

Then we can find the critical value taking in count the degrees of freedom and the alternative hypothesis and then we need to find a critical value who accumulates 0.05 of the area in the right tail and we got:

$t_{\alpha}= 1.796$

And for this case the rejection region would be:

b) Reject H0 if tcalc >1.7960

6 0

3 years ago

It takes one week for a crew of workers to pave 3/5 kilometer of a road. At that rate, how long will it take to pave 1 kilometer

Flauer [41]

Answer:

1.67 weeks

Step-by-step explanation:

Let X= time it will take to pave 1 kilometer?

1 week=3/5 kilometer

X week =1 kilometer

Cross multiply

3/5 X= 1

3X= 5

X= 5/3

1.67 weeks

8 0

3 years ago