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Hatshy [7]
3 years ago
15

Witch of the following is a true statement about a circle inscribed in a regular polygon

Mathematics
1 answer:
ss7ja [257]3 years ago
4 0
The area of the circle is equal to the area of the polygon
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Evaluate the expression when a = 4.
Rus_ich [418]


2(4)^3

2(64)

=128 Your final answer would be C

6 0
3 years ago
Read 2 more answers
I WILL GIVE BRAINLIEST.
den301095 [7]

Answer:

(-a)n is negative  number, and since you did not provide a clear first option, I will assume the answer is none of the above. However, the accurate answer is C

Step-by-step explanation:

4 0
3 years ago
A training field is formed by joining a rectangle and two semicircles, as shown below. The rectangle is 96m long and 74m wide.
Oduvanchick [21]

Answer:

11402.66 m^{2}

Step-by-step explanation:

The width of rectangle is the diameter of the semi-circle part

Area of one semicircle is given by \frac {0.5\pi d^{2}}{4}

Total area of semi circle will be 2\times\frac {0.5\pi d^{2}}{4}

Substituting 74 m for d and \pi as 3.14 we obtain

Total area semi-circle=2\times\frac {0.5*3.14\times 74^{2}}{4}=4298.66 m^{2}

Area of rectangle is given by the product of length and width

Rectangular area=96 m*74 m=7104 m^{2}

Total area of rectangular and semi-circles will be

4298.66 m^{2}+7104 m^{2}=11402.66 m^{2}

Therefore, area of training field is 11402.66 m^{2}

6 0
3 years ago
a nationwide study of american homeowners revealed that​ 65% own at least one lawn mower. a lawn equipment​ manufacturer, locate
svetoff [14.1K]

Answer:

z=\frac{0.6841 -0.65}{\sqrt{\frac{0.65(1-0.65)}{497}}}=1.5938  

We have a right tailed test then the p value would be:  

p_v =P(z>1.5938)=0.0555  

Step-by-step explanation:

Information provided

n=497 represent the random sample of homes selected

X=340 represent the number of homes with one or more lawn

\hat p=\frac{340}{497}=0.6841 estimated proportion of homes with one or more lawn

p_o=0.65 is the value that we want to test

z would represent the statistic

p_v represent the p value (variable of interest)  

System of hypothesis

We want to check if proportion of homeowners owning lawn mowers in charlotte is higher than​ 65%m so then the correct hypothesis are .:  

Null hypothesis:p\leq 0.65  

Alternative hypothesis:p > 0.65  

The statistic is given by:

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

Replacing the info given we got:

z=\frac{0.6841 -0.65}{\sqrt{\frac{0.65(1-0.65)}{497}}}=1.5938  

We have a right tailed test then the p value would be:  

p_v =P(z>1.5938)=0.0555  

4 0
3 years ago
Help guys soal integral​
Afina-wow [57]

Answer:

E. \purple { \bold{ \frac{2}{3} ( \frac{x - 1}{x} ) \sqrt{\frac{x - 1}{x} }  + c }}

Step-by-step explanation:

\int \sqrt{ \frac{x - 1}{ {x}^{5} } } dx \\  \\  =  \int \sqrt{ \frac{x - 1}{ {x}^{4} .x} } dx \\  \\ =  \int \frac{1}{ {x}^{2}}\sqrt{   \frac{x - 1}{ x} } dx \\  \\ =  \int \frac{1}{ {x}^{2}}\sqrt{  1 -  \frac{1}{ x} } dx \\  \\ let \: 1 -  \frac{1}{ x} = t \\  \\  \implies \:  \frac{1}{ {x}^{2} } dx = dt \\  \\ \implies \int \frac{1}{ {x}^{2}}\sqrt{  1 -  \frac{1}{ x} } dx =  \int \sqrt{t} dt \\  \\ =  \int  {t}^{ \frac{1}{2} } dt \\  \\ =  \frac{t ^{ \frac{3}{2} } }{ \frac{3}{2} }  + c \\  \\  =  \frac{2}{3} t ^{ \frac{3}{2} }  + c \\  \\ =  \frac{2}{3}  \sqrt{ {t}^{3} }   + c \\  \\ =  \frac{2}{3}  t\sqrt{ {t} }   + c \\  \\  =  \frac{2}{3} (1 -  \frac{1}{x} ) \sqrt{1 -  \frac{1}{x} }  + c \\  \\  \red{ \bold{=  \frac{2}{3} ( \frac{x - 1}{x} ) \sqrt{\frac{x - 1}{x} }  + c }}\\  \\

5 0
3 years ago
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