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r-ruslan [8.4K]
3 years ago
15

Find the domain of y= 2/3 square root of x+5-4

Mathematics
1 answer:
3241004551 [841]3 years ago
4 0
Set the radicand in
√
x
−
5
x
-
5
greater than or equal to
0
0
to find where the expression is defined.
x
−
5
≥
0
x
-
5
≥
0
Add
5
5
to both sides of the inequality.
x
≥
5
x
≥
5
The domain is all values of
x
x
that make the expression defined.
Interval Notation:
[
5
,
∞
)
[
5
,
∞
)
Set-Builder Notation:
{
x
|
x
≥
5
}
{
x
|
x
≥
5
}
The range of an even indexed radical starts at its starting point
(
5
,
0
)
(
5
,
0
)
and extends to infinity.
Interval Notation:
[
0
,
∞
)
[
0
,
∞
)
Set-Builder Notation:
{
y
|
y
≥
0
}
{
y
|
y
≥
0
}
Determine the domain and range.
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How many 2-letter combinations can be created from the distinct letters in the word “mathematician”? Assume that the order of th
kupik [55]
Hello!

this problem comes with a slightly complex math equation that I had to ask my math teacher about, so I hope it helps.

<span>The expression n! means "the product of the integers from 1 to n"
</span>this can be said for any variable

say n is the total and r is the 2-letter combo

C(n,r<span>)=13!/</span>(2!(13−2)!)

If you use this equation, you will get a final answer of: 78

I hope this helps, and have a nice day.
3 0
3 years ago
Read 2 more answers
The true average diameter of ball bearings of a certain type is supposed to be 0.5 in. A one-sample t test will be carried out t
yulyashka [42]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

   C

b

    C

c

    C

d  

     A

Step-by-step explanation:

From the question we are told that

    The population mean is  \mu  =  0.5 \  in

     

Generally the Null hypothesis is  H_o  :  \mu = 0. 5 \ in

                The Alternative hypothesis is  H_a  :  \mu  \ne  0.5 \ in

Considering the parameter given for part a  

       The sample size is  n =  15  

        The  test statistics is  t =  1.66

        The level of significance \alpha  =  0.05

The degree of freedom is evaluated as

            df =  n-  1

           df =  15-  1

           df =  14

Using the critical value calculator at (social science statistics web site )  

           t_{\frac{\alpha}{2} ,df } =  t_{\frac{0.05 }{2} ,14} =  2.145

We are making use of this  t_{\frac{\alpha }{2} } because it is a one-tail test

Looking at the value of  t and t_{\frac{\alpha }{2} } the we see that  t < t_{\frac{\alpha }{2}  } so the null hypothesis would not be rejected

Considering the parameter given for part b  

       The sample size is  n =  15  

        The  test statistics is  t =  -1.66

        The level of significance \alpha  =  0.05

The degree of freedom is evaluated as

            df =  n-  1

           df =  15-  1

           df =  14

Using the critical value calculator at (social science statistics web site )  

           t_{\frac{\alpha}{2} ,df } =  t_{\frac{0.05 }{2} ,14} =  -2.145

Looking at the value of  t and t_{\frac{\alpha}{2} ,df } the we see that t does not lie in the area covered by  t_{\frac{\alpha}{2}  , df } (i.e the area from -2.145 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

 

Considering the parameter given for part  c

       The sample size is  n =  26  

        The  test statistics is  t =  -2.55

        The level of significance \alpha  =  0.01

The degree of freedom is evaluated as

            df =  n-  1

           df =  26-  1

           df =  25

Using the critical value calculator at (social science statistics web site )  

           t_{\frac{\alpha}{2} ,df } =  t_{\frac{0.01 }{2} ,25} =  2.787

Looking at the value of  t and t_{\frac{\alpha }{2} } the we see that t does not lie in the area covered by  t_{\alpha , df } (i.e the area from -2.787 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

Considering the parameter given for part  d

       The sample size is  n =  26  

        The  test statistics is  t =  -3.95

        The level of significance \alpha  =  0.01

The degree of freedom is evaluated as

            df =  n-  1

           df =  26-  1

           df =  25

Using the critical value calculator at (social science statistics web site )  

           t_{\frac{\alpha}{2} ,df } =  t_{\frac{0.01 }{2} ,25} =  -2.787

Looking at the value of  t and t_{\frac{\alpha}{2}  } the we see that  t  lies in the area covered by  t_{\alpha , df } (i.e the area from -2.787 downwards on the normal distribution curve ) hence we  reject the null hypothesis

6 0
3 years ago
Which of the following is less than 527 m?
Andru [333]
1m = 100cm = 1000mm = 1,000,000,000nm
1Mm = 1,000,000m

A. 0.000699 Mm = 0.000699 · 1,000,000 m = 699 m

B. 914,879,710 nm = (914,879,710 : 1,000,000,000) m = 0.914879710 m

C. 51,723 cm = (51,723 : 100) m = 517.23 m

Answer: C. 51,723 cm = 517.23 m ≈ 527 m
8 0
3 years ago
Write an equation to determine the number(t) that Raymond was on the trampoline.
nata0808 [166]
An equation you can write is 1.25t+7=43.25.  When you solve this equation you will get 29 as t.  So, he was on the trampoline for 29 minutes.

Hope this helped!
5 0
4 years ago
Need help ASAP!!!!!!!
Temka [501]

D

Step-by-step explanation:

2/6 is litterally right there cause it has seven lines and tge 0 doesnt count

3 0
3 years ago
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