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

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