The following set of coordinates represent which figure (1,1) (5,3) (7,7) (3,5)

The length of a rectangular field is 24 meters this is 3 meters less than twice the width find the width

stepladder [879]

Answer:

The width is 5 meters

Step-by-step explanation:

The rectangular field is 24 meaning adding just one width and length of the rectangle would be 12.

x = width

2x - 3 + x = 12

3x - 3 = 12

3x - 3 + 3 = 12 + 3

3x = 15

x = 5

5 0

3 years ago

What is the value of x?<br> X=32<br> X=36<br> X=37<br> X=40

kumpel [21]

A line is 180 degrees so write x+(4x-20)=180

5 0

3 years ago

Read 2 more answers

Si el lado de un cuadrado mide 10cm cuanto mide él área del círculo que se encuentra adentro de ese cuadrado

Alex17521 [72]

Answer:

El área del círculo que se encuentra en el cuadrado es de 78.5cm²

Step-by-step explanation:

Para resolver este ejercicio tenemos que pensar que un cuadrado tiene sus 4 lados iguales, por lo que todos sus lados medirán 10cm.

Ahora nos fijamos que necesitamos saber para calcular el área de un circulo

a = área

r = radio

π = 3.14

a = π * r²

como podemos ver no sabemos el valor del radio

como el circulo toca con los 4 lados del cuadrado sabemos que su radio sera la distancia del centro del cuadrado a cualquiera de los lados.

Entonces tenemos que dividir un lado por 2

10cm/2 = 5cm

El radio del circulo sera 5cm

Ahora que tenemos todos los datos podemos calcular el valor del área

a = 3.14 * (5cm)²

a = 3.14 * 25cm²

a = 78.5cm²

El área del círculo que se encuentra en el cuadrado es de 78.5cm²

8 0

3 years ago

YES OR NOOO PLEASEE ANSWER

dlinn [17]

Answer:

Not a Function.... hope you pass!

7 0

2 years ago

1. A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur at n

Nuetrik [128]

Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

For each fatality, there are only two possible outcomes, either it involved an intoxicated driver, or it did not. The probability of a fatality involving an intoxicated driver is independent of any other fatality, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

70% of fatalities involve an intoxicated driver, hence $p = 0.7$ .

A sample of 15 fatalities is taken, hence $n = 15$ .

The probability is:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)$

Hence

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{15,10}.(0.7)^{10}.(0.3)^{5} = 0.2061$

$P(X = 11) = C_{15,11}.(0.7)^{11}.(0.3)^{4} = 0.2186$

$P(X = 12) = C_{15,12}.(0.7)^{12}.(0.3)^{3} = 0.1700$

$P(X = 13) = C_{15,13}.(0.7)^{13}.(0.3)^{2} = 0.0916$

$P(X = 14) = C_{15,14}.(0.7)^{14}.(0.3)^{1} = 0.0305$

$P(X = 15) = C_{15,15}.(0.7)^{15}.(0.3)^{0} = 0.0047$

Then:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.2061 + 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047 = 0.7215$

0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

A similar problem is given at brainly.com/question/24863377

5 0

3 years ago