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likoan [24]
3 years ago
6

What is .006 in word form

Mathematics
2 answers:
givi [52]3 years ago
6 0
The answer is "six-thousandths."
julia-pushkina [17]3 years ago
3 0
It is six thousandths
You might be interested in
Petra jogs 2 miles in 24 mins. at this rate how long would it take her to jog 7 miles?
Wewaii [24]
The answer would be 84 because 2miles is equal to 24 minutes so you would break that in half down to one mile. One mile would be 12 minutes. So you would multiply 12 times 7 which equals 84 which would be your answer.
7 0
2 years ago
Are there any vertical, complimentary, or any supplementary angles here ? And label for ex. 1 supplementary
kumpel [21]

<u>Supplementary:</u> 1 and 2 & 3 and 4

Supplementary angles: A pair of angles whose sum is equal to 180°.

<u />

<u>Vertical:</u> 1 and 3 & 2 and 4

Vertical angles: A pair of opposite angles that are made by two intersecting lines.

<u>Complimentary:</u> None

Complimentary angles: A pair of angles whose sum is equal to 90°.

Hope this helps! :D

7 0
3 years ago
Suppose 52% of the population has a college degree. If a random sample of size 563563 is selected, what is the probability that
amm1812

Answer:

The value is  P(| \^ p -  p| < 0.05 ) = 0.9822

Step-by-step explanation:

From the question we are told that

    The population proportion is  p =  0.52

     The sample size is  n  =  563      

Generally the population mean of the sampling distribution is mathematically  represented as

           \mu_{x} =  p =  0.52

Generally the standard deviation of the sampling distribution is mathematically  evaluated as

       \sigma  =  \sqrt{\frac{ p(1- p)}{n} }

=>      \sigma  =  \sqrt{\frac{ 0.52 (1- 0.52 )}{563} }

=>      \sigma  =   0.02106

Generally the  probability that the proportion of persons with a college degree will differ from the population proportion by less than 5% is mathematically represented as

            P(| \^ p -  p| < 0.05 ) =  P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 ))

  Here  \^ p is the sample proportion  of persons with a college degree.

So

 P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = P(\frac{[[0.05 -0.52]]- 0.52}{0.02106} < \frac{[\^p - p] - p}{\sigma }  < \frac{[[0.05 -0.52]] + 0.52}{0.02106} )

Here  

    \frac{[\^p - p] - p}{\sigma }  = Z (The\ standardized \  value \  of\  (\^ p - p))

=> P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = P[\frac{-0.47 - 0.52}{0.02106 }  <  Z  < \frac{-0.47 + 0.52}{0.02106 }]

=> P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = P[ -2.37 <  Z  < 2.37 ]

=>  P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = P(Z <  2.37 ) - P(Z < -2.37 )

From the z-table  the probability of  (Z <  2.37 ) and  (Z < -2.37 ) is

  P(Z <  2.37 ) = 0.9911

and

  P(Z <  - 2.37 ) = 0.0089

So

=>P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) =0.9911-0.0089

=>P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = 0.9822

=> P(| \^ p -  p| < 0.05 ) = 0.9822

3 0
3 years ago
Solve the quadratic equation below for the exact values of x:<br> 4x^2-5=75
grandymaker [24]

Step-by-step explanation:

4 {x}^{2}  - 5 = 75 \\ 4 {x}^{2}  = 75 + 5 \\ 4 {x}^{2}  = 80 \\  {x}^{2}  =  \frac{80}{4}  \\  {x}^{2}  = 20 \\ x = \pm \sqrt{20}  \\ x = \pm 2 \sqrt{5}

6 0
3 years ago
If triangle ABC is similar to triangle DEC, find the values of x and y.
Mumz [18]

The values are x=8 and y=35, if the given ΔABC and ΔDEC are equal, it is obtained by Pythagoras theorem.

Step-by-step explanation:

The given are,

           From ΔABC,

                    AB= 6

                    BC= 10

                    AC = x

           From ΔDEC,

                    CD= 28

                    DE= 21

                    CE = y

Step:1

        Pythagoras theorem from  ΔABC,

                    BC^{2}=AB^{2} + AC^{2}...............(1)

       Substitute the values,

                   10^{2} = 6^{2} + AC^{2}

                  100 = 36 + AC^{2}

                   AC^{2} = 100 - 36

                           = 64

                    AC = \sqrt{64}

                    AC = 8

                  AC = x = 8

Step:2

        Pythagoras theorem for ΔDEC,

               CE^{2} = CD^{2}  + DE^{2}................(2)

       From the values,

              CE^{2} = 28^{2} + 21^{2}

             CE^{2} = 784 + 441

                     = 1225

              CE = \sqrt{1225}

              CE = 35

             CE = y = 35

Result:

         The values are x=8 and y=35, if the given ΔABC and ΔDEC are equal.

5 0
3 years ago
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