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zubka84 [21]
4 years ago
10

What expression has the same value as -3/2-(2-3/8)+3/2

Mathematics
1 answer:
ella [17]4 years ago
5 0

Answer:

\dfrac{-3}{2}-(2-\dfrac{3}{8})+\dfrac{3}{2}=\dfrac{-13}{8}

Step-by-step explanation:

We need to find the value of expression \dfrac{-3}{2}-(2-\dfrac{3}{8})+\dfrac{3}{2}.

Firstly solving the second term as :

(2-\dfrac{3}{8})=\dfrac{16-3}{8}=\dfrac{13}{8}

Now the above expression becomes,

\dfrac{-3}{2}-(2-\dfrac{3}{8})+\dfrac{3}{2}\\=\dfrac{-3}{2}-\dfrac{13}{8}+\dfrac{3}{2}

-3/2 and +3/2 equals 0.

It means that, \dfrac{-3}{2}-(2-\dfrac{3}{8})+\dfrac{3}{2}=\dfrac{-13}{8}

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a game room has a floor that is 120 feet by 75 feet. A scale drawing of the floor on grid paper uses a scale of 1 unit:5 feet. W
Ghella [55]
For every 5 feet, the scale drawing is 1 unit.  So we divide the dimensions by 5 to get the dimensions of the scale drawing.

120/5 = 24

1:5 = x:120  There are 24 5's in 120 so the new ratio is 24:120

75/5 = 15

1:5 = x:75  There are 15 5's in 75 so the new ratio is 15:75

The dimensions of the scale drawing is 24 units by 15 units.
4 0
3 years ago
A manufacturing company produces two sizes of cylindrical containers that each have a height of 50 centimeters. The radius of Co
ira [324]

Answer:

Volume of B=62800cm^3

Step-by-step explanation:

We are given  

Height = 50 cm

Radius of A=16cm

Radius of Cylinder B is 25% more than that of A

Radius of B = 16 + \frac{25}{100}*16 cm =16+4= 20 cm

Volume of cylinder B = pi*r^2*h

                                   =3.14 * (20)^2*50

                                    =62800cm^3

3 0
3 years ago
Which value of x makes this equation true? 100-6x =160-10x
Verdich [7]

Answer:

x=15

Step-by-step explanation:

7 0
4 years ago
Read 2 more answers
The shape of the distribution of the time required to get an oil change at a 15-minute oil-change facility is unknown. However,
horsena [70]

Answer:

The mean oil-change time that would there be a 10% chance of being at or below is 15.46 minutes.

Step-by-step explanation:

We are given that records indicate that the mean time is 16.2 minutes, and the standard deviation is 3.4 minutes.

On a typical​ Saturday, the​ oil-change facility will perform 35 oil changes between 10 A.M. and 12 P.M. Assuming that data follows normal distribution.

Let \bar X = <u><em>sample mean time</em></u>

The z score probability distribution for sample mean is given by;

                          Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time = 16.2 minutes

            \sigma = standard deviation = 3.4 minutes

            n = sample size = 35 oil changes

<u>Now, the mean oil-change time that would there be a 10% chance of being at or below is given by;</u>

<u></u>

          P(X \leq x) = 0.10          {where x is required mean oil-change time}

          P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } \leq \frac{x-16.2}{\frac{3.4}{\sqrt{35} } } ) = 0.10

          P(Z \leq \frac{x-16.2}{\frac{3.4}{\sqrt{35} } } ) = 0.10

Now, in the z table the critical value of X which represents the below 10% of the probability area is given as -1.282, that means;

                  \frac{x-16.2}{\frac{3.4}{\sqrt{35} } }  =  -1.282

               x - 16.2  =  -1.282 \times {\frac{3.4}{\sqrt{35} } }

                         x  =  16.2 - 0.74 = <u>15.46 minutes</u>

Hence, the mean oil-change time that would there be a 10% chance of being at or below is 15.46 minutes.

8 0
3 years ago
Find the values of x for the following equations..<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2B4x-98%3D0" id="TexFormula1"
Travka [436]

Hello !

Look at the picture

7 0
3 years ago
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