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dmitriy555 [2]
3 years ago
14

Which polygon is a convex heptagon

Mathematics
1 answer:
Darya [45]3 years ago
8 0

Convex Heptagon: A convex heptagon is a polygon with seven sides in which all of its diagonals lies inside the Heptagon. Concave Heptagon: A Concave Heptagon is a polygon with one or more interior angles greater than 180 degrees and some diagonals will lie outside the polygon.

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What is the area of my lawn if it measures 4m x 4.5m? 18 m2
Nadusha1986 [10]

Answer:

17m

Step-by-step explanation:

Add up 4m and 4.5m. 8.5

Multiply that number by 2. 17

6 0
3 years ago
Read 2 more answers
Need some help with this question!
Veronika [31]
1. solve for x in first to just find nterms of y

x=12-y
sub that for x in other one
first divide that one by 2 to make it easier
2x-y=18
2(12-y)-y=18
24-2y-y=18
24-3y=18
minus 24 both sides
-3y=-6
divide -3
y=2

D is answer




2.
find intersection of x+y=5 and -2x+3y=6
or easiers, just test the intersections
wait, they all intersect in the same place
this is super easy
pick ANY point that is in that reigon and see if it is true

A, if we pick lets say (5,5)
that is false for x+y<5
not A

B
pick (10,0)
false for first one
not B

C
pick (0,0)
true for first
false for 2nd
not C


D. pick (-10,0)
true for first
true for 2nd


answer is D


answer is D for both questions
6 0
3 years ago
Ryan spent 24 months living in San Diego.
sertanlavr [38]

Answer:

nothing

Step-by-step explanation:

nothing

8 0
4 years ago
Read 2 more answers
A​ half-century ago, the mean height of women in a particular country in their 20s was 64.7 inches. Assume that the heights of​
Ainat [17]

Answer:

99.5% of all samples of 21 of​ today's women in their 20's have mean heights of at least 65.86 ​inches.

Step-by-step explanation:

We are given that a half-century ago, the mean height of women in a particular country in their 20's was 64.7 inches. Assume that the heights of​ today's women in their 20's are approximately normally distributed with a standard deviation of 2.07 inches.

Also, a samples of 21 of​ today's women in their 20's have been taken.

<u><em /></u>

<u><em>Let </em></u>\bar X<u><em> = sample mean heights</em></u>

The z-score probability distribution for sample mean is given by;

                          Z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean height of women = 64.7 inches

            \sigma = standard deviation = 2.07 inches

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the sample of 21 of​ today's women in their 20's have mean heights of at least 65.86 ​inches is given by = P(\bar X \geq 65.86 inches)

  P(\bar X \geq 65.86 inches) = P( \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } } \geq \frac{65.86-64.7}{\frac{2.07}{\sqrt{21} } } ) = P(Z \geq -2.57) = P(Z \leq 2.57)

                                                                        = <u>0.99492  or  99.5%</u>

<em>The above probability is calculated by looking at the value of x = 2.57 in the z table which has an area of 0.99492.</em>

<em />

Therefore, 99.5% of all samples of 21 of​ today's women in their 20's have mean heights of at least 65.86 ​inches.

3 0
3 years ago
Find the area of each parallelogram plz help
xxTIMURxx [149]
Answer:  The area of the parallelogram is:
_______________________________________________________
\frac{9393}{8}  in² = 1174 ⅛  in² = 1174.125 in² .
_______________________________________________________
Explanation:
_______________________________________________________
Area of a parallelogram:
_______________________________________________________
 A = base * height = b * h ;

From the figure (from the actual "question"):
_______________________________________
         b = 50.5 in.

         h = 23.25 in.
____________________________________________________________
Method 1)  A = b * h =
                 
                     =  (50.5 in) * (23.25 in) = 1174.125 in² ; or, write as:  1174 <span>⅛ .
</span>____________________________________________________________
Method 2)  A = b * h =
             
                      =  (50 ½ in) * (23 <span>¼ in) =
   
                      =    (</span>\frac{101}{2} in) * (\frac{93}{4}<span> in) ;
</span>___________________________________________________________
Note:  "50 ½ " = [(50*2) + 1 ] / 2  = \frac{101}{2} ;

Note:  "23 ¼ " = [(23*4) + 1 ] / 4  = \frac{93}{4} ;
____________________________________________________________
→  A =  (\frac{101}{2} in) * (\frac{93}{4} in) ;
 
→  A = \frac{(101*93)}{(2*4)}  in² = \frac{9393}{8}  in² ; 
          
→  A = (9393/8) in² =  
         
→  A  = \frac{9393}{8}  in² = 1174 ⅛  in² = 1174.125 in² .
________________________________________________________
5 0
3 years ago
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