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nexus9112 [7]
3 years ago
11

An 80 confidence interval for is found to be (200, 240). what is the margin of error

Mathematics
1 answer:
Bogdan [553]3 years ago
8 0

Answer:

It’s 20

Step-by-step explanation:

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The length of the hypotenuse of 30 -60 -90 triangle is 4 find the longer leg
11Alexandr11 [23.1K]
The longer leg is the leg adjacent to the larger angle

so 60 

length of longer leg = 4 sin60 = 2sqrt(3) 

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What is the vertex of function f(x)=(x-3)(x+5)?
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Answer:

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Step-by-step explanation:

6 0
2 years ago
April worked 1 1/2 times as long on her math project as did Carl. Debbie worked 1 1/4 times as long as Sonia. Richard worked 1 3
vlada-n [284]

Answer:

        Student                                                            Hours worked

             April.                                                                  7\frac{7}{8} \ hrs

        Debbie.                                                                   8\frac{1}{8}\ hrs

        Richard.                                                                   7\frac{19}{24}\ hrs

Step-by-step explanation:

Some data's were missing so we have attached the complete information in the attachment.

Given:

Number of Hours Carl worked on Math project = 5\frac{1}{4}\ hrs

5\frac{1}{4}\ hrs can be Rewritten as \frac{21}{4}\ hrs

Number of Hours Carl worked on Math project = \frac{21}{4}\ hrs

Number of Hours Sonia worked on Math project = 6\frac{1}{2}\ hrs

6\frac{1}{2}\ hrs can be rewritten as \frac{13}{2}\ hrs

Number of Hours Sonia worked on Math project = \frac{13}{2}\ hrs

Number of Hours Tony worked on Math project = 5\frac{2}{3}\ hrs

5\frac{2}{3}\ hrs can be rewritten as \frac{17}{3}\ hrs.

Number of Hours Tony worked on Math project = \frac{17}{3}\ hrs.

Now Given:

April worked 1\frac{1}{2} times as long on her math project as did Carl.

1\frac{1}{2}  can be Rewritten as \frac{3}{2}

Number of Hours April worked on math project = \frac{3}{2} \times Number of Hours Carl worked on Math project

Number of Hours April worked on math project = \frac{3}{2}\times \frac{21}{4} = \frac{63}{8}\ hrs \ \ Or \ \ 7\frac{7}{8} \ hrs

Also Given:

Debbie worked 1\frac{1}{4} times as long as Sonia.

1\frac{1}{4}  can be Rewritten as \frac{5}{4}.

Number of Hours Debbie worked on math project = \frac{5}{4} \times Number of Hours Sonia worked on Math project

Number of Hours Debbie worked on math project = \frac{5}{4}\times \frac{13}{2}= \frac{65}{8}\ hrs \ \ Or \ \ 8\frac{1}{8}\ hrs

Also Given:

Richard worked 1\frac{3}{8} times as long as tony.

1\frac{3}{8} can be Rewritten as \frac{11}{8}

Number of Hours Richard worked on math project = \frac{11}{8} \times Number of Hours Tony worked on Math project

Number of Hours Debbie worked on math project = \frac{11}{8}\times \frac{17}{3}= \frac{187}{24}\ hrs \ \ Or \ \ 7\frac{19}{24}\ hrs

Hence We will match each student with number of hours she worked.

        Student                                                            Hours worked

             April.                                                                  7\frac{7}{8} \ hrs

        Debbie.                                                                   8\frac{1}{8}\ hrs

        Richard.                                                                   7\frac{19}{24}\ hrs

5 0
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√245c^2 simplify :
raketka [301]
My boiii, tiger algebra helped me out a little on this one, I got 7c * sqrt(5)
8 0
3 years ago
3x2 - 2x = 0<br> Factorise and solve
malfutka [58]

 Solve  :    3x-2 = 0  

Add  2  to both sides of the equation :  

                     3x = 2

Divide both sides of the equation by 3:

                    x = 2/3 = 0.667

Two solutions were found :

x = 2/3 = 0.667

x = 0

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6 0
3 years ago
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