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Black_prince [1.1K]
3 years ago
15

Scores on a college entrance examination are normally distributed with a mean of 500 and a standard deviation of 100. What perce

nt of people who write this exam obtain scores between 350 and 650?
Mathematics
1 answer:
Alla [95]3 years ago
3 0

Answer:

The percentage is P(350 <  X  650 ) = 86.6\%

Step-by-step explanation:

From the question we are told that

   The population mean is  \mu  =  500

     The standard deviation is  \sigma  =  100

The  percent of people who write this exam obtain scores between 350 and 650    

    P(350 <  X  650 ) =  P(\frac{ 350 -  500}{ 100}

Generally  

               \frac{X -  \mu }{\sigma }  =  Z (The \  standardized \  value \ of  \  X )

   P(350 <  X  650 ) =  P(\frac{ 350 -  500}{ 100}

   P(350 <  X  650 ) =  P(-1.5

   P(350 <  X  650 ) =  P(Z < 1.5) -  P(Z <  -1.5)

From the z-table  P(Z <  -1.5 )  =  0.066807

   and P(Z < 1.5  ) =  0.93319

=>    P(350 <  X  650 ) =  0.93319 -  0.066807

=>  P(350 <  X  650 ) = 0.866

Therefore the percentage is  P(350 <  X  650 ) = 86.6\%

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secA=1/cosA


tgA=sinA/cosA


sin^2A/1/cos^2A-sin^2A/cos^2A


sin^2A.cos^2A/cos2A

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6 0
2 years ago
Which expression is equivalent??? help!
dexar [7]

Answer:

The equivalent expression for the given expression \sqrt[3]{256x^{10}y^{7} } is

4x^{3} y^{2}(\sqrt[3]{4xy} )

Step-by-step explanation:

Given:

\sqrt[3]{256x^{10}y^{7} }

Solution:

We will see first what is Cube rooting.

\sqrt[3]{x^{3}} = x

Law of Indices

(x^{a})^{b}=x^{a\times b}\\and\\x^{a}x^{b} = x^{a+b}

Now, applying above property we get

\sqrt[3]{256x^{10}y^{7} }=\sqrt[3]{(4^{3}\times 4\times (x^{3})^{3}\times x\times (y^{2})^{3}\times y   )} \\\\\textrm{Cube Rooting we get}\\\sqrt[3]{256x^{10}y^{7} }= 4\times x^{3}\times y^{2}(\sqrt[3]{4xy}) \\\\\sqrt[3]{256x^{10}y^{7} }= 4x^{3}y^{2}(\sqrt[3]{4xy})

∴ The equivalent expression for the given expression \sqrt[3]{256x^{10}y^{7} } is

4x^{3} y^{2}(\sqrt[3]{4xy} )

5 0
3 years ago
A well mixed tank with a capacity of 1500 gals originally contains 1000 gallons of freshwater. One pipe containing 1/2 lb of sal
Illusion [34]

Answer:

400 lb of salt

Step-by-step explanation:

Let us assume the water flows into the rank for x minutes.

There is an initial of 1000 gallons of water in the tank and water flows in through one pipe at 4 gal/min and through another pipe at 6 gal/min. In x minute, the amount of water in the tank = 1000 + 4x + 6x = 1000 + 10x

Water flows out at 5 gal/min, therefore in x minute the amount of water in the tank = 1000 + 10x - 5x = 1000 + 5x

The tank begins to overflow when it is full (has reached 1500 gallons). Therefore:

1500 = 1000 + 5x

5x = 1500 - 1000

5x = 500

x = 100 minutes.

1/2 lb salt per gallon flows into the tank at 4 gal/min and 1/3 lb of salt is flowing in at 6 gal/min, in 100 min the amount of salt that entered the tank = 4 gal/min × 100 min × 1/2 lb/gal + 6 gal/min × 100 min × 1/3 lb/gal= 400 lb

Therefore the amount of salt is in the tank when it is about to overflow = 400 lb of salt

7 0
3 years ago
Draw as many different triangles as possible that have two sides of length 4 cm and a 45° angle. Clearly mark the side lengths a
Basile [38]

Student Verified! ✅

Answer:

The answer is 3!

5 0
2 years ago
A store holiday sale has an item marked down by $10 with an additional discount of 25% off the
ValentinkaMS [17]

Answer:

C : $58.36

Step-by-step explanation:

Given:

A store holiday sale has an item marked down by $10.

Discount on new price = 25%

Final price was $36.27.

Question asked:

what was the original ?

Solution:

Let original price = x

New price = x - 10

Original price - marked down amount - discount amount = $36.27

x - 10 -25\% of(x - 10 ) = 36.27\\

x - 10 - \frac{25}{100} (x - 10) = 36.27\\x - 10 - 0.25(x - 10) = 36.27\\x - 10 -0.25x +2.5 = 36.27\\0.75x - 7.5 = 36.27\\

Adding both side by 7.5

0.75x = 43.77\\

Dividing both side by 0.75

x = 58.36

Therefore, the original price was $58.36

5 0
2 years ago
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