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Elza [17]
3 years ago
10

The percentage, P, of U.S. voters who use punch cards or lever machines in national elections can be modeled by the formula: P =

−2.5x + 63.1 where x is the number of years after 1994. In which years will fewer than 38.1% of U.S. voters use punch cards or lever machines?
Mathematics
1 answer:
Ne4ueva [31]3 years ago
7 0

Answer:

From 2004 onwards, fewer than 38.1% of U.S. voters use punch cards or lever machines

Step-by-step explanation:

The percentage of U.S. voters who use punch cards or lever machines in national elections, in x years after 1994, is given by the following equation:

P(x) = -2.5x + 63.1

In which years will fewer than 38.1% of U.S. voters use punch cards or lever machines?

In x years after 1994 onwards.

x is found when P(x) = 38.1.

So

P(x) = -2.5x + 63.1

38.1 = -2.5x + 63.1

2.5x = 25

x = \frac{25}{2.5}

x = 10

From 2004 onwards, fewer than 38.1% of U.S. voters use punch cards or lever machines

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Is 0.2971 irrational?
Vladimir [108]

Answer:

Yes

Step-by-step explanation:

An irrational number can be written as a decimal, but not as a fraction. An irrational number has endless non-repeating digits to the right of the decimal point.

7 0
3 years ago
You cut square corners from a piece of cardboard that has dimensions 32 cm by 40 cm. You then fold the cardboard to create a box
jenyasd209 [6]

Answer:

h = 6 cm  

l = (32-2(6)) =  20 cm

w = (40-2(6)) = 28 cm

Step-by-step explanation:

Given in the question a piece of cardboard having dimensions 32 by 40

Suppose we cut square of length x

When you cut square corners from a piece of cardboard, two sides are 32-2x, and other two sides are 40-2x lengths

<h3>Step 1</h3>

Formula for the volume of box

v = height * length  * width

v = x  (32-2x)  (40-2x)

v = 4x³ - 144x² + 1280x

<h3>Step 2</h3>

Find derivative and equal it to 0

dv/dx = 4(3)x² - (2)144x + 1280

      0 = 12x² - 288x + 1280

x1 = 18.110

x2 =5.88

<h3>Step 3</h3>

Select x and plug value of x to find dimension

(32-2x) > 0

2x < 32

 x < 16

so domain say that x should be less than 16 which means x = 5.88 ≈ 6 cm

h = 6 cm  

l = (32-2(6)) =  20 cm

w = (40-2(6)) = 28 cm

6 0
4 years ago
Divide using a common factor of 6 to find an equivalent fraction for
Savatey [412]

5: six twelfths is one half

6: eight tenths is four fifths

7: four eighths is one half

8: ten tenths is two halves

9: two sixths is one third

10: one quarter, because you divide the top and bottom number by 3

method:

take the first one - four eighths, which is 4/8

the question gives you what to divide them by, 4, so u divide both the top number and bottom number to get 1/2, which is one half :)

7 0
3 years ago
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They want to know different types of areas
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4 years ago
Orange M&amp;M’s: The M&amp;M’s web site says that 20% of milk chocolate M&amp;M’s are orange. Let’s assume this is true and set
SOVA2 [1]

Answer:

The correct option is (A).

Step-by-step explanation:

Let <em>X</em> = number of orange  milk chocolate M&M’s.

The proportion of orange milk chocolate M&M’s is, <em>p</em> = 0.20.

The number of candies in a small bag of milk chocolate M&M’s is, <em>n</em> = 55.

The event of an milk chocolate M&M being orange is independent of the other candies.

The random variable <em>X</em> follows a Binomial distribution with parameter <em>n</em> = 55 and <em>p</em> = 0.20.

The expected value of a Binomial random variable is:

E(X)=np

Compute the expected number of orange  milk chocolate M&M’s in a bag of 55 candies as follows:

E(X)=np

         =55\times 0.20\\=11

It is provided that in a randomly selected bag of milk chocolate M&M's there were 14 orange ones, i.e. the proportion of orange milk chocolate M&M's in a random bag was 25.5%.

This proportion is not surprising.

This is because the average number of orange milk chocolate M&M’s in a bag of 55 candies is expected to be 11. So, if a bag has 14 orange milk chocolate M&M’s it is not unusual at all.

All unusual events have a very low probability, i.e. less than 0.05.

Compute the probability of P (X ≥ 14) as follows:

P(X\geq 14)=\sum\limits^{55}_{x=14}{{55\choose x}0.20^{x}(1-0.20)^{55-x}}

                 =0.1968

The probability of having 14 or more orange candies in a bag of milk chocolate M&M’s is 0.1968.

This probability is quite larger than 0.05.

Thus, the correct option is (A).

4 0
3 years ago
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