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IceJOKER [234]
3 years ago
6

PLEASE HELP! The length of time a full length movie runs from opening to credits is normally distributed with a mean of 1.9 hour

s and standard deviation of 0.3 hours. Calculate the following: A random movie is between 1.8 and 2.0 hours. A movie is longer than 2.3 hours. The length of movie that is shorter than 94% of the movies.
Mathematics
1 answer:
Jobisdone [24]3 years ago
4 0

Answer:

A random movie is between 1.8 and 2.0 hours.

z=1.8-1.9/0.3= -0.33= 0.3707

z=2.0-1.9/0.3= 0.33= 0.6293

0.6293-0.3707= 0.2586

Therefore, the chance a random movie is between 1.8 and 2.0 hours long is 0.2586.

A movie is longer than 2.3 hours.

z=2.3-1.9/0.3 =1.33 =0.9082

1-0.9082= 0.0918

Therefore, the chance a movie is longer than 2.3 hours is 0.9082.

The length of movie that is shorter than 94% of the movies

z=x-1.9/0.3= 0.94

z=x-1.9/0.3= 1-0.94= 0.06

=x-1.9/0.3= 0.06= -1.56

x= 1.432

Therefore, the length of the movie that is shorter than 94% of the movies about 1.4 hours.  

Step-by-step explanation:

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Step-by-step explanation:

We are given the following information in the question:

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4 0
3 years ago
Kay has 28 coins which includes nickels and dimes if the total is 2.35$, how many of each coin does Kay have?
m_a_m_a [10]

Answer:9 nickels and 19 dimes.

Step-by-step explanation:

Let n be the number of nickels Kay has.

Let d be the number of dimes Kay has.

Value of 1 nickel is $0.05

Value of n nickels is $0.05n

Value of 1 dime is $0.1

Value of d nickels is $0.1dn

Given that Kay has a total of 28 coins.

So,n+d=28                                                ...(i)

Given that Kay has a total value of $2.35

So,0.05n+0.1d=2.35                                 ...(ii)

Using (i) and (ii),

0.05n+0.1(28-n)=2.35\\0.05n+2.8-0.1n=2.35\\2.8-2.35=0.05n\\0.45=0.05n\\n=9

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Rom4ik [11]
<h2>Steps:</h2>

So firstly, I will be factoring by grouping. For this, factor x⁶ - 9x⁴ and -x² + 9 separately. Make sure that they have the same quantity on the inside of the parentheses:

x^4(x^2-9)-1(x^2-9)=0

Now, you can rewrite the equation as:

(x^4-1)(x^2-9)=0

However, it's not completely factored. Next, we will apply the formula for the difference of squares, which is x^2-y^2=(x+y)(x-y) . In this case:

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Next, we will apply the difference of squares once more with the second factor as such:

x^2-1=(x+1)(x-1)\\\\(x^2+1)(x+1)(x-1)(x+3)(x-3)=0

<h2>Answer:</h2>

<u>The factored form of this equation is: (x^2+1)(x+1)(x-1)(x+3)(x-3)=0</u>

8 0
4 years ago
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