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3 years ago

Could someone please help on this question? I’m giving 20 points and brainliest but I need it quickly so please help!

Mathematics

1 answer:

AnnZ [28]3 years ago

6 0

<h3>Answer: HF = 5</h3>

=============================================

Explanation:

Refer to the diagram below. I'm adding in points A and B, which are marked in blue. I'm also adding in variables x, y, and z.

The goal is to find HF, so let's make that equal to x. This is also equal to HB because they are radii of the same circle.

Let y be the length of segments JA and JB.

Let z be the length of segments CF and CA.

So in short we have

x = HF = HB
y = JA = JB
z = CF = CA

We know that

CH = 13
HJ = 19
CJ = 22

This must mean

CH = CF+HF = z+x = 13
HJ = HB+JB = x+y = 19
CJ = CA+JA = z+y = 22

Or in short,

z+x = 13
x+y = 19
z+y = 22

Let's solve the first equation for z to get z = -x+13. I subtracted x from both sides.

Now plug this into the third equation to get

z+y = 22

(z) + y = 22

(-x+13) + y = 22 ... replace z with -x+13

-x+13+y = 22

-x+y+13 = 22

-x+y = 22-13 .... subtract 13 from both sides

-x+y = 9

------------------------------------

So we have this reduced system of equations of two variables and two equations

x+y = 19 ... found earlier
-x+y = 9 .... what we just found above

Add the equations straight down. This is the elimination property.

Doing so leads to the x terms canceling out since x + (-x) = 0x = 0

The y terms add to y+y = 2y

The terms on the right hand side add to 19+9 = 28

We are left with 2y = 28 which solves to y = 28/2 = 14.

If y = 14, then,

x+y = 19

x+14 = 19

x = 19-14

x = 5

In which we can then say

z = -x+13

z = -5+13

z = 8

Though we don't need to find z (unless you're curious about it).

Since x = 5, and we set HF equal to x, this means that HF = 5.

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Naval intelligence reports that 4 enemy vessels in a fleet of 17 are carrying nuclear weapons. If 9 vessels are randomly targete

icang [17]

Answer:

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

Step-by-step explanation:

The vessels are destroyed without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

In this question:

Fleet of 17 means that $N = 17$

4 are carrying nucleas weapons, which means that $k = 4$

9 are destroyed, which means that $n = 9$

What is the probability that more than 1 vessel transporting nuclear weapons was destroyed?

This is:

$P(X > 1) = 1 - P(X \leq 1)$

In which

$P(X \leq 1) = P(X = 0) + P(X = 1)$

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,17,9,4) = \frac{C_{4,0}*C_{13,9}}{C_{17,9}} = 0.0294$

$P(X = 1) = h(1,17,9,4) = \frac{C_{4,1}*C_{13,8}}{C_{17,9}} = 0.2118$

Then

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0294 + 0.2118 = 0.2412$

$P(X > 1) = 1 - P(X \leq 1) = 1 - 0.2412 = 0.7588$

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

8 0

2 years ago

f(x)=4 hallar el dominio, codominio, ceros o raices, vertices, asintotas, tabla de valores, graficas, identificar el tipo de fun

nekit [7.7K]

Answer:

Tipo de función: Función Constante o Función Polinómica de Orden Cero

Dominio: $Dom \{f(x)\} = \mathbb{R}$

Codominio: $Ran \{f(x)\} = 4$

Asíntota: $y = 4$

Raíces: No tiene.

Vértices: No tiene.

La tabla de valores y el gráfico se encuentran adjuntos al fondo.

Step-by-step explanation:

Sea $f(x) = 4$ , el cual representa a una Función Constante o Función Polinómica de Orden Cero, cuyos dominio y codominio son $Dom \{f(x)\} = \mathbb{R}$ y $Ran \{f(x)\} = 4$ . No tiene vértices. Esta función tiene una asíntota horizontal coincidente con la función ( $y = 4$ ) y carece de raíces al ser $f(x) \ne 0$ para todo $x\in \mathbb{R}$ .