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2 years ago

Which of the following best describes the figure?

Mathematics

2 answers:

stira [4]2 years ago

8 0

Answer:

a convex nonagon

Step-by-step explanation:

adoni [48]2 years ago

8 0

<h2>Answer:</h2>

The given graph represents--

Concave nonagon

<h2>Step-by-step explanation:</h2>

Nonagon--

A nanagon is a polygon with 9 sides.

Regular nonagon means that all the 9 sides are of equal length and all the interior angles are of equal measure.

Convex polygon means that none of the interior angles of the polygon is greater than 180 degree otherwise it is a concave polygon.

From the figure that is provided to us we observe that the polygon has nine sides.

Hence, it is a nonagon.

Also, one of the angle is greater than 180°.

Hence, the polygon is concave nonagon.

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Imaginá que tenés 125 dados cúbicos del mismo tamaño ¿Cuantos dados de altura tiene el cubo de mayor tamaño que podés armar apil

kumpel [21]

Answer:

(i) Debemos apilar 5 dados para construir el cubo de mayor tamaño.

(ii) Se necesita 121 dados cuadrados para formar el cuadrado con la mayor cantidad de dados posibles, quedando 4 dados sobrantes.

Step-by-step explanation:

(i) Sabemos por la Geometría Euclídea del Espacio que un cubo es un sólido regular con 6 caras cuadradas y longitudes iguales. Cada dado tiene un volumen de 1 dado cúbico y 125 dados dan un volumen total de 125 dados cúbicos.

El volumen de un cubo está dado por la siguiente fórmula:

$V = L^{3}$

Donde:

$L$ - Longitud de la arista, medida en dados.

$V$ - Volumen del cubo, medido en dados cúbicos.

Ahora, necesitamos despejar la longitud de la arista para calcular la altura máxima posible:

$L = \sqrt[3]{V}$

Dado que $V = 125\,dados^{3}$ , encontramos que la altura del cubo de mayor tamaño sería:

$L =\sqrt[3]{125\,dados^{3}}$

$L = 5\,dados$

Debemos apilar 5 dados para construir el cubo de mayor tamaño.

(ii) El área cuadrada formada por cubos está determinada por la siguiente fórmula:

$A = L^{2}$

Donde:

$L$ - Longitud de arista, medida en dados.

$A$ - Área, medida en dados cuadrados.

Puesto que la longitud de arista se basa en un conjunto discreto, esto es, el número de dados disponibles, debemos encontrar el valor máximo de $L$ tal que no supere 125 y de un área entera. Es decir:

$L \leq 125\,dados$

Si cada cubo tiene un área de 1 dado cuadrado, entonces un cuadrado conformado por 125 dados tiene un área total de 125 dados cuadrados. Entonces:

$L^{2}< 125\,dados^{2}$

Esto nos lleva a decir que:

$L < 11.180\,dados$

Entonces, la longitud máxima del cuadrado con la mayor cantidad de cubos posible es de 11 dados. El número total requerido de cubos es el cuadrado de esa cifra, es decir:

$n = (11\,dados)^{2}$

$n = 121\,dados$

Se necesita 121 dados cuadrados para formar el cuadrado con la mayor cantidad de dados posibles, quedando 4 dados sobrantes.

4 0

2 years ago

Mopeds (small motorcycles with an engine capacity below 50 cm3) are very popular in Europe because of their mobility, ease of op

Mashcka [7]

Answer:

a) P(X<50)=0.9827

b) P(X>47)=0.4321

c) P(-1.5<z<1.5)=0.8664

Step-by-step explanation:

We will calculate the probability based on a random sample of one moped out of the population, normally distributed with mean 46.7 and standard deviation 1.75.

a) This means we have to calculate P(x<50).

We will calculate the z-score and then calculate the probability accordign to the standard normal distribution:

$z=\dfrac{X-\mu}{\sigma}=\dfrac{50-46.7}{1.75}=\dfrac{3.7}{1.75}=2.114\\\\\\P(X$

b) We have to calculatee P(x>47).

We will calculate the z-score and then calculate the probability accordign to the standard normal distribution:

$z=\dfrac{X-\mu}{\sigma}=\dfrac{47-46.7}{1.75}=\dfrac{0.3}{1.75}=0.171\\\\\\P(X>47)=P(z>0.171)=0.4321$

c) If the value differs 1.5 standard deviations from the mean value, we have a z-score of z=1.5

$z=\dfrac{(\mu+1.5\sigma)-\mu}{\sigma}=\dfrac{1.5\sigma}{\sigma}=1.5$

So the probability that maximum speed differs from the mean value by at most 1.5 standard deviations is P(-1.5<z<1.5):

$P(-1.5$

3 0

2 years ago

On Monday, the closing value of a share of stock in Company ABC, was 74.01. On Tuesday, it closed at 73.67, and on Wednesday, it

Elenna [48]

Answer:

$957.71$

Step-by-step explanation:

Given: On Monday, the closing value of a share of stock in Company ABC, was $74.01$ . On Tuesday, it closed at $73.67$ , and on Wednesday, it closed at $74.32$ . On Thursday, the change in stock value was $-1.48$ . And, on Friday, Company ABC's stock gained $0.89$ points.

To Find: One shareholder owns $13$ shares of Company ABC's stock. What was the total value of their stock in this company at the close of Tuesday.

Solution:

Closing value of Shares of company ABC on Monday $=74.01$

Closing value of Shares of company ABC on Tuesday $=73.67$

Closing value of Shares of company ABC on Wednesday $=74.32$

on Thursday change in stock value was $-1.48$

Closing value of Shares of company ABC on Thursday $=72.84$

on Friday comapny's stock gained $0.89$ points

Closing value of Shares of company ABC on Friday $=73.73$

As one shareholder owns $13$ shares of company

Total value of shares on Tuesday is

$=\text{number of shares}\times\text{closing stock value on tuesday}$

$=13\times73.67$

$=957.71$

Hence total value of shareholder's share on close of Tuesday is $957.71$

6 0

2 years ago

Read 2 more answers

A preimage includes a line segment with a length of x units and a slope of m units. The preimage is dilated by a scale factor of

lapo4ka [179]

Nx is the first one and m is the second answer

8 0

1 year ago

Read 2 more answers

Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in

Hitman42 [59]

Answer:

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval

(3) A survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

Step-by-step explanation:

We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of Americans who decide to not go to college = 48%

n = sample of American adults = 331

p = population proportion of Americans who decide to not go to

college because they cannot afford it

<em>Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.</em>

<em />

<u>So, 90% confidence interval for the population proportion, p is ;</u>

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < $\hat p-p$ < $1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

<u>90% confidence interval for p</u> = [ $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.48 -1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ , $0.48 +1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ ]

= [0.4348, 0.5252]

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) The interpretation of the above confidence interval is that we can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval.

3) Now, it is given that we wanted the margin of error for the 90% confidence level to be about 1.5%.

So, the margin of error = $Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$