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Step2247 [10]
2 years ago
11

In a recent survey of 655 working Americans ages 25-34, the average weekly amount spent on lunch was 43.76 with standard deviati

on 2.67. The weekly amounts are approximately bell-shaped. (c) Between what two values will approximately 95% of the amounts be?
Mathematics
1 answer:
klemol [59]2 years ago
8 0

Answer:

Between 38.42 and 49.1.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean of 43.76, standard deviation of 2.67.

Between what two values will approximately 95% of the amounts be?

By the Empirical Rule, within 2 standard deviations of the mean. So

43.76 - 2*2.67 = 38.42

43.76 + 2*2.67 = 49.1

Between 38.42 and 49.1.

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A bar model can be used to a subtraction problem by showing how much of it your actually losing or winning so subtract the totals and you have gotten your answer.
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3 years ago
Sellus
igor_vitrenko [27]

Answer:

3.9

Step-by-step explanation:

Given the data:

Payout ($) (x) : 0 2 4 8 10

Probability p(x) : 0.35 0.2 0.1 0.2 0.15

The expected winning ; E(X) = Σ(x * p(x))

Σ(x * p(x)) = (0*0.35)+(2*0.2)+(4*0.1)+(8*0.2)+(10*0.15)

= 0 + 0.4 + 0.4 + 1.6 + 1.5

= 3.9

8 0
3 years ago
The domain of discourse is sets. Suppose some predicate Q is true for all elemenst of A: ∀a ∈ A Q(a). Recall the definition of s
Sloan [31]

Answer:

See the step-by-step explanation

Step-by-step explanation:

Let c be any element of C. (I'm not sure wether you have to assume that C is non-empt or not)

C is a subset of B. That means that as c is  in C, it is also in B. (c \in C \Rightarrow  c \in B)

Now, B is a subset of A. It follows that as c \in B \Rightarrow c \in A.

That means c is an element of A. The predicate Q is true for all elements of A, including c.

Because we let c be any element of C, we have proven that the predicate Q is true for all elements in C.

5 0
3 years ago
PLEASEEE ANSWERR
Vinvika [58]

Answer:

b

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Mi padre compró un radio por cierta suma de dinero, un televisor que costó 10 veces más que el radio y un reproductor cuyo preci
Maru [420]

Answer:

For radio = $40.43

Television =  $444.73

Player =  $364.73

Step-by-step explanation:

The computation of the each device cost is shown below:

let us assume the following things

radio be x

television x + 10x

player x + 10x - $80

Now the equation is

x + x + 10x + x + 10x - $80 = $850

3x + 20x - $80 = $850

23x = $930

x = $40.43

For radio = $40.43

Television = $40.43 + 10(40.43)

= $40.43 + $404.3

= $444.73

and, the player would be

= $444.73 - $80

= $364.73

7 0
3 years ago
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