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anygoal [31]
3 years ago
7

Help please. I don’t understand and my teacher didn’t bother to teach us any of this and my brain hurts!!!!

Mathematics
1 answer:
const2013 [10]3 years ago
8 0

Answer:

  23.  42 m^2

  24.  144 ft^2

  25.  68 cm^2

Step-by-step explanation:

The teacher should not need to spend any time teaching this. The area of the total figure is the sum of the areas of its parts. You already know that. And, you already know how to compute the areas of triangles and rectangles.

__

You may recall that the area of a triangle is ...

  A = 1/2bh

and the area of a rectangle with the same base and height is ...

  A = bh

So, a triangle with base b and height h has the same area as a rectangle with the same base (b) and half the height (h/2). Recognizing this makes these problems really simple.

__

23. The base of the rectangle and triangle is 7 m. The height of the triangle is 4 m, so its area is equivalent to adding 4/2 = 2 m to the 4 m height of the rectangle. Then the total area is ...

  A = (7 m)(4 m +(1/2)(4 m)) = (7 m)(6 m) = 42 m^2

__

24. Since there are two identical triangles of height 5, their total area is equal to that of a rectangle of height 5.

  A = (12 ft)(7 ft +5 ft) = 144 ft^2

__

25. The 5 cm height of the triangle makes it have an area equivalent to a rectangle 5/2 cm high.

  A = (8 cm)(6 cm +5/2 cm) = 68 cm^2

Note: we have read the fuzzy numbers as 8 cm wide, 6 cm high for the rectangle, and 5 cm high for the triangle. If your figure is different, please use the appropriate numbers in the calculation.

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Two surfers and statistics students collected data on the number of days on which surfers surfed in the last month for 30 longbo
Alisiya [41]

Answer:

Do not reject H0. The mean days surfed for longboarders is significantly larger than the mean days surfed for all shortboarders

Step-by-step explanation:

The null hypothesis is that  the mean days surfed for all long boarders is larger than the mean days surfed for all short boarders

H0:  μL > μs      against the claim Ha:  μL≤ μs

the alternate hypothesis is the mean days surfed for all long boarders isless or equal to  the mean days surfed for all short boarders (because long boards can go out in many different surfing conditions)

The test statistic is

t= x1- x2/  √s1/n1+ s2/n2

1) Calculations

Longboards

Mean

ˉx=∑x/n=4+8+9+4+9+7+9+6+6+11+15+13+16+12+10+12+18+20+15+10+15+19+21+9+22+19+23+13+12+10/30

=377/30

=12.5667

Longboard Variance S2=[∑dx²-(∑dx)²/n]/n-1

=[831-(-13)²/30]/29

=831-5.6333/29

=825.3667/29

=28.4609

Shortboard Mean

ˉx=∑x/n=6+4+6+6+7+7+7+10+4+6+7+5+8+9+4+15+13+9+12+11+12+13+9+11+13+15+9+19+20+11/30

=288/30

=9.6

Shortboard Variance S2=[∑x²-(∑x)²/n]/n-1

=[ 3270-(288)2/30]/29

=3270-2764.8/29

=505.2/29

=17.4207

2) Putting values in the test statistic

t=|x1-x2|/√S²1/n1+S²2/n2

t =|12.5667-9.6|/√28.4609/30+17.4207/30

t =|2.9667|/√0.9487+0.5807

t=|2.9667|/√1.5294

t=|2.9667|/1.2367

t=2.3989

3) Degree of freedom =n1+n2-2=30+30-2=58

4) The critical region is t ≤ t(0.05) (58) =1.6716

5) Since the calculated t= 2.4 does not fall in the critical region t(0.05) (58)  ≤ 1.6716 we do not reject H0.

The p-value is 0.008969. The result is significant at p <0 .05.

6 0
3 years ago
An ounce is 1/16 of a pound. How many ounces are in 8 3/4 pounds?
ivolga24 [154]
Multiply 8 3/4 lb by the conversion factor   16 oz
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                                                                     1  lb

Then 35 lb       16 oz            560 oz
         -------- * -----------  =  ------------- =  140 oz (answer)
            4            1 lb                4
5 0
3 years ago
PLEASE HELP WILL GIVE BRAINLIEST!!! Type the correct answer in each box. If necessary, use / for the fraction bar(s).
eduard

Answer:

NEED EXAMPLE PLZ TO ANSWERS!?

5 0
3 years ago
He table shows the solution to the equation |2x − 5| − 2 = 3:
ozzi
The solution is completely correct, as putting 5 or 0 in the formula again will make it true.

|2(0)-5|=5
|-5|=5
5=5

|2(5)-5|=5
|10-5|=5
5=5
6 0
4 years ago
<img src="https://tex.z-dn.net/?f=%5Ctext%7BI%20was%20eating%20cookies%20and%20had%20some%20thoughts.%20If%20I%20wanted%20to%20c
andrew-mc [135]

In the attachement, there is what I came up with so far. I think that finding 'a' is non-trivial, if possible at all.

A_c - the area of a circle

A_{cs} - the area of a circular segment

5 0
3 years ago
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