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shutvik [7]
2 years ago
9

En una muestra de 50 restaurantes de comida rápida, la venta media fue de $3.000, y la desviación estándar, $200. El intervalo d

e confianza del 95% es
Mathematics
1 answer:
Step2247 [10]2 years ago
8 0

Respuesta:

(2945.411; 3054.589)

Explicación paso a paso:

Dado ;

Tamaño de la muestra, n = 50

Media, xbar = 3000

Desviación estándar, s = 200

Nivel de confianza, Zcrítico al 95% = 1,96

El intervalo de confianza se define como:

Xbar ± margen de error

Margen de error = Zcrítico * s / sqrt (n)

Margen de error = 1,96 * 200 / sqrt (50)

Margen de error = 54.589

Límite inferior = (3000 - 54.589) = 2945.411

Límite superior = (3000 + 54.589) = 3054.589

(2945.411; 3054.589)

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drek231 [11]

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52 minutes because if you write the problem out on a peice of papaer it will make sense

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3 years ago
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The perimeter of the base of a right square pyramid is 32 cm.
Serga [27]

Answer:

  170.67 cm^3

Step-by-step explanation:

The volume is given by ...

  V = 1/3s²h

We are given that 4s=32, so s=8. We are also given that h=8. (All linear dimensions are in centimeters.) Then the volume is ...

  V = (1/3)(8²)(8) = 512/3 = 170.67 . . . . cm³

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3 0
3 years ago
If 11250 = m(600) + b and 7650 = m(400) + b, how do I find the values of b and m? I forgot how to compare when I have two differ
Lelu [443]

Answer:

b=450 and m=18

Step-by-step explanation:

This exercise is an example of <em>linear system equations</em>. A <em>system of linear equations</em> is a set of (linear) equations that have more than one unknown. The unknowns appear in several of the equations, but not necessarily in all of them. What these equations do is relate the unknowns to each other.

The easy way to solve this problem is by using the<em> reduction method</em>.The <em>reduction method</em> consists of operating between the equations, such as adding or subtracting both equations, so that one of the unknowns disappears. Thus, we obtain an equation with a single unknown.

Ordering the equations as a system equations.

\left \{ {{600m+b=11250} \atop {400m+b=7650}} \right.

Using the reduction method, let's substract the second equation with the first equation, in order to clear m.

600m + b = 11250

<u>-400m -  b = -7650</u>

200m       = 3600    -------->  m = 3600/200 -----> m = 18

We can substitute the value of m in any of the two equations to obtain the unknow b.

Let's substitute the value of m in the second equation.

400(18) + b = 7650

7200 + b = 7650 -----> b = 7650 - 7200 -------> b = 450

We can check this values in both equations.

600m + b = 11250 -----> 600(18) + 450 = 11250

400m + b = 7650 ----->  400(18) + 450 = 7650

Satisfying the result of both equations.

6 0
3 years ago
Which linear function represents the table?
lara31 [8.8K]

Answer:

A) y= 5x - 3

Step-by-step explanation:

(5 • x) - 3 = y

Problem Solving:

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5 0
3 years ago
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NO SCAM PLEASEEEE
damaskus [11]
The answer is 11/36

2/12 chance of rolling fours

because there are 2 sides containing a four on both dice combined and 12 sides in total.

Doubles mean you have to roll the same number simultaneously so let’s say we want to calculate the probability for double ones: then it’s 1/6 on the first dice for a one, and 1/6 on the second dice to land on a one as well.

I personally like to imagine a box like this:
_ _ _ _ _ _
|
|
|
|
|
|

If you have one dice then it’s just a random segment on one of the lines. If you want the specific result from two dice then you want two specific segments which is also the 1 specific tile out of 36 (6 width times 6 height). So you multiply.

1/6 * 1/6 = 1/36 chance to roll double of ones

And 1/36 chance to roll double twos, threes, fours, fives, and sixes. But we don’t count the double fours because any four will do. So:

1/36 * 5 = 5/36

So for the probability of either doubles or containing a four is the probability of doubles of either number plus the probability of either dice being a four:

5/36 + 2/12 =

5/36 + 6/36 =

11/36
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2 years ago
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