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faltersainse [42]
2 years ago
13

At an art auction, Giannina's painting sold for over $300. Write an inequality that describes P, the price of Geannina's paintin

g in dollars.
Mathematics
2 answers:
Korvikt [17]2 years ago
7 0

We are given : Giannina's painting sold for over $300.

Let the price of Geannina's painting in dollars = P.

<em>Note: We are given a word " over " in above statement.</em>

Over word represents more than. So, we need to use more than inequality symbol to write an inequality for the given statement.

Let us rephrase the given statement.

" Price P is greater than $300".

Now, we can write an inequality :

P> 300.

<h3>Therefore, the required inequality for given statement is P > 300.</h3>
SVEN [57.7K]2 years ago
7 0
If P is the price of the painting, and the painting is over $300, the inequality is P>300
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mariarad [96]

Answer:

The answer is \frac{7\pi }{12}^{c}=105^{0}

Step-by-step explanation:

The expression is

\frac{7\pi }{12}^{c}\\= \frac{7\pi }{12}*\frac{180}{\pi }^{0} \,;[1^{c}=\frac{180}{\pi}^{0}]\\=7*15\\=105^{0}

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MAVERICK [17]

Answer:

Pepe weighs 4600 grams.

Step-by-step explanation:

5.4 kilograms converted into grams is 5400 grams. Since Pepe is 800 grams lighter it would be 5400 - 800 which equals 4600 grams.

5.4 kilograms = 5400 grams

5400 - 800 = 4600

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2 years ago
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Gekata [30.6K]

Answer:

Expand the brackets, and simplify.

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Aleksandr-060686 [28]

Answer:

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Step-by-step explanation:

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3 years ago
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(3 points) Blades of grass Suppose that the heights of blades of grass are Normally distributed and independent, with each heigh
NemiM [27]

The final answer is:

a) P( Y < 42.5 )  = 0.8541

b) P( 39.5 < Y < 40.5 ) = 0.1670.

What is the normal distribution?

A continuous probability distribution for a real-valued random variable in statistics is known as a normal distribution or Gaussian distribution.

If x follows a normal distribution with mean μ and standard deviation σ then the distribution of

\sum_{i =1}^{n}x_{i}  follows an approximately normal distribution with a mean n\mu and standard deviation \sqrt{n }\sigma

let x be the height of blades of grass

x follows normal distribution with mean = μ = 4 and standard deviation = σ = 0.75.

Y = x1 + x2 +...........+x10

Y = \sum_{i =1}^{10}x_{i}

Distribution of Y is normal with,

Mean = \mu _{y}=10*4 = 40 and standard deviation = \sigma _{y}=\sqrt{10}*0.75 = 2.3717

a)

P( Y < 42.5 )

Using normal distribution formmula,

f(x)= {\frac{1}{\sigma\sqrt{2\pi}}}e^{- {\frac {1}{2}} (\frac {x-\mu}{\sigma})^2}

=NORMDIST( x, mean, SD , 1 )      

=NORMDIST(42.5, 40, 2.3717, 1 )

=0.8541

P( Y < 42.5 )  = 0.8541

b)

P( 39.5 < Y < 40.5 ) = P( Y < 40.5 ) - P( Y < 39.5 )

Using normal distribution formmula,

f(x)= {\frac{1}{\sigma\sqrt{2\pi}}}e^{- {\frac {1}{2}} (\frac {x-\mu}{\sigma})^2}

P( Y < 40.5 )  =NORMDIST(40.5, 40, 2.3717, 1 ) = 0.5835

P( Y < 39.5 ) = NORMDIST(39.5, 40, 2.3717, 1 ) = 0.4165

P( 39.5 < Y < 40.5 ) = 0.5835 - 0.4165  = 0.1670

P( 39.5 < Y < 40.5 ) = 0.1670

Hence, the final answer is:

a) P( Y < 42.5 )  = 0.8541

b) P( 39.5 < Y < 40.5 ) = 0.1670.

To learn more about the normal distribution visit,

brainly.com/question/4079902

#SPJ4

5 0
1 year ago
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