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baherus [9]
4 years ago
13

Lorea is designing the top of a quilt which measures 2,160 square inches. Triangles will cover 432 square inches of the quilt, a

nd squares will cover 180 square inches. The rest of the quilt will be parallelograms with a base of 2 inches and a height of 1.5 inches. How many parallelograms should she cut to complete the quilt?
Mathematics
2 answers:
timurjin [86]4 years ago
7 0
2160 - 180 - 432 = 612
Each parallelogram is 3 square inches
divide by 3
612/3 = 264 parallelograms to complete the quilt.
kari74 [83]4 years ago
7 0

Answer:it’s 516

Step-by-step explanation:

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Let​ T: set of real numbers R Superscript nℝnright arrow→set of real numbers R Superscript mℝm be a linear​ transformation, and
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Answer:

\{T(v_1), T(v_2), T(v_3)\} is linearly dependent set.

Step-by-step explanation:

Given:  \{v_1,v_2,v_3\} is a linearly dependent set in set of real numbers R

To show: the set \{T(v_1), T(v_2), T(v_3)\} is linearly dependent.

Solution:

If \{v_1,v_2,v_3,...,v_n\} is a set of linearly dependent vectors then there exists atleast one k_i:i=1,2,3,...,n such that k_1v_1+k_2v_2+k_3v_3+...+k_nv_n=0

Consider k_1T(v_1)+k_2T(v_2)+k_3T(v_3)=0

A linear transformation T: U→V satisfies the following properties:

1. T(u_1+u_2)=T(u_1)+T(u_2)

2. T(au)=aT(u)

Here, u,u_1,u_2∈ U

As T is a linear transformation,

k_1T(v_1)+k_2T(v_2)+k_3T(v_3)=0\\T(k_1v_1)+T(k_2v_2)+T(k_3v_3)=0\\T(k_1v_1+k_2v_2+k_3v_3)=0\\

As \{v_1,v_2,v_3\} is a linearly dependent set,

k_1v_1+k_2v_2+k_3v_3=0 for some k_i\neq 0:i=1,2,3

So, for some k_i\neq 0:i=1,2,3

k_1T(v_1)+k_2T(v_2)+k_3T(v_3)=0

Therefore, set \{T(v_1), T(v_2), T(v_3)\} is linearly dependent.

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Step-by-step explanation:

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Step-by-step explanation:

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Construct the confidence interval for the population mean mu. c = 0.90​, x = 16.9​, s = 9.0​, and n = 45. A 90​% confidence inte
Georgia [21]

Answer:

The  90%  confidence interval for population mean is   14.7  <  \mu  <  19.1

Step-by-step explanation:

From the question we are told that

   The sample mean is  \= x  =  16.9

    The confidence level is  C  =  0.90

     The sample size is  n  =  45

     The standard deviation

Now given that the confidence level is  0.90 the  level of significance is mathematically evaluated as

       \alpha =  1-0.90

       \alpha  =  0.10

Next we obtain the critical value of  \frac{\alpha }{2}  from the standardized normal distribution table. The values is  Z_{\frac{\alpha }{2} } =  1.645

The  reason we are obtaining critical values for \frac{\alpha }{2}  instead of  that of  \alpha  is because \alpha  represents the area under the normal curve where the confidence level 1 - \alpha (90%)  did not cover which include both the left and right tail while \frac{\alpha }{2}  is just considering the area of one tail which is what we required calculate the margin of error

  Generally the margin of error is mathematically evaluated as

        MOE  =  Z_{\frac{\alpha }{2} } *  \frac{\sigma }{\sqrt{n} }

substituting values

         MOE  = 1.645*  \frac{ 9 }{\sqrt{45} }

         MOE  = 2.207

The  90%  confidence level interval is mathematically represented as

      \= x  -  MOE  <  \mu  <  \= x  +  MOE

substituting values

     16.9 -  2.207  <  \mu  <  16.9 +  2.207

    16.9 -  2.207  <  \mu  <  16.9 +  2.207

     14.7  <  \mu  <  19.1

         

3 0
3 years ago
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