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lyudmila [28]
3 years ago
14

A random sample of adults were asked whether they prefer reading an e-book over a printed book. The survey resulted in a sample

proportion of p′=0.14, with a sampling standard deviation of σp′=0.02, who preferred reading an e-book. Use the empirical rule to construct a 95% confidence interval for the true proportion of adults who prefer e-books.
Mathematics
1 answer:
djyliett [7]3 years ago
8 0

Answer: (0.10\ ,\ 0.18)

Step-by-step explanation:

According to the empirical rule, the 95% confidence interval for the true population proportion is given by :-

(p'-2\sigma_{p'}\ ,\ p'+2\sigma_{p'})

Given : . The survey resulted in a sample proportion of p′=0.14, with a sampling standard deviation of σp′=0.02, who preferred reading an e-book.

Then, the 95% confidence interval for the true proportion of adults who prefer e-books will be :-

(0.14-2(0.02)\ ,\ 0.14+2(0.02))\\\\=(0.14-0.04\ ,\ 0.14+0.04)\\\\=(0.10\ ,\ 0.18)

Hence, the 95% confidence interval for the true proportion of adults who prefer e-books= (0.10\ ,\ 0.18)

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75

Step-by-step explanation:

Q1=55 Q2=67 Q3=75

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Solve question in image (yr 8 math, 40 pt)
Nady [450]

Answer:

y = 3

Step-by-step explanation:

(5y - 3)/4 + 6 = 3y

5y - 3 + 24 = 12y

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7 0
2 years ago
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Suppose it is found that a certain model of car sells y= -1.6(x-2)^2 + 4.8(x-2) + 16 cars per week, where x is the number of sho
Vilka [71]

Answer:

Approximately 4 shops would be the best number of shops which the car is sold, and the number of cars is 20.

Step-by-step explanation:

This is a quadratic function. Therefore, the first thing you need to do is assume some values for x, and then, calculate the value of y, (which would be the number of cars sold). In this way, you can do the graph.

Watch the following attachment, which is a document of excel, showing the graph and some values of x, that I assume

Now, even if you don't have the graph you can estimate the number of shops with the following expression of the summit:

X = -b / 2a

Where:

a: number that goes along with the x elevated.

b: number that goes along with the x, but without being elevated.

In this equation, the first thing we need to do, is rearrange it, that's because we have a (x-2) in the equation, and this could be really annoying.

First let's solve the (x-2)^2

(x-2)^2 = (x-2)(x-2) = x^2 - 2*2x + 2^2 = x^2 - 4x + 4

This is now multiplied by -1.6:

-1.6x^2 + 6.4x - 6.4

Now, multiply 4.8 by x-2:

4.8x - 9.6

Finally, let's arrange this:

-1.6x^2 + 6.4x - 6.4 + 4.8x - 9.6 + 16

-1.6x^2 + 11.2x - 6.4 - 9.6 + 16 = -1.6x^2 + 11.2x

This means that the graph do not have a "y" intercept, and the values of a and b are -1.6 and 11.2, therefore, we can estimate the best number of shops, calculating the summit of the graph (This is because s a quadratic function, and the graph is a parable, and the minimum or maximum point of the graph is reached in the summit)

Summit has X and Y values, the x value is (see formula above):

X = -11.2 / 2 * -1.6 = 3.5

and for the y value, just replace the X value in the equation:

Y = -1.6(3.5)^2 + 11.2(3.5) = -19.6 + 39.2 = 19.6

With this we can conclude that the best number of shops that sells this model of car, is 4 (rounded) and they all sell 20 cars (also rounded). See the graph below in the attachment.

Download xlsx
4 0
3 years ago
Read 2 more answers
1. Consider the following hypotheses:
Andrej [43]

Answer:

See deductions below

Step-by-step explanation:

1)

a) p(y)∧q(y) for some y (Existencial instantiation to H1)

b) q(y) for some y (Simplification of a))

c) q(y) → r(y) for all y (Universal instatiation to H2)

d) r(y) for some y (Modus Ponens using b and c)

e) p(y) for some y (Simplification of a)

f) p(y)∧r(y) for some y (Conjunction of d) and e))

g) ∃x (p(x) ∧ r(x)) (Existencial generalization of f)

2)

a) ¬C(x) → ¬A(x) for all x (Universal instatiation of H1)

b) A(x) for some x (Existencial instatiation of H3)

c) ¬(¬C(x)) for some x (Modus Tollens using a and b)

d) C(x) for some x (Double negation of c)

e) A(x) → ∀y B(y) for all x (Universal instantiation of H2)

f)  ∀y B(y) (Modus ponens using b and e)

g) B(y) for all y (Universal instantiation of f)

h) B(x)∧C(x) for some x (Conjunction of g and d, selecting y=x on g)

i) ∃x (B(x) ∧ C(x)) (Existencial generalization of h)

3) We will prove that this formula leads to a contradiction.

a) ∀y (P (x, y) ↔ ¬P (y, y)) for some x (Existencial instatiation of hypothesis)

b) P (x, y) ↔ ¬P (y, y) for some x, and for all y (Universal instantiation of a)

c) P (x, x) ↔ ¬P (x, x) (Take y=x in b)

But c) is a contradiction (for example, using truth tables). Hence the formula is not satisfiable.

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