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Neko [114]
3 years ago
5

I need assistant with these parallelogram problems​

Mathematics
1 answer:
trasher [3.6K]3 years ago
8 0

Problem 6a

  1. Angle DCB is supplementary to angle ADC and angle ABC. Reason: Adjacent angles in a parallelogram are always supplementary.
  2. AE = EC. Reason: The diagonals of any parallelogram always bisect each other. In other words, the diagonals cut each other in half.
  3. DC = AB. Reason: Opposite sides of a parallelogram are congruent.
  4. angle DCB = angle DAB. Reason: Opposite angles of a parallelogram are congruent.
  5. CD = AB. Reason: segment DC is the same as segment CD. The order of the endpoints does not matter when it comes to listing a segment. Opposite sides of a parallelogram are congruent.

=================================================

Problem 6b

<h3>Answers:</h3><h3>angle L = 129</h3><h3>angle S = 51</h3><h3>angle C = 129</h3>

----------

Work Shown:

angle U = 51 (given)

angle S = angle U (Opposite angles of a parallelogram are congruent)

L+U = 180 (adjacent angles in parallelogram are supplementary)

L = 180-U

angle L = 180-51

angle L = 129

angle C = angle L (Opposite angles of a parallelogram are congruent)

angle C = 129

=================================================

Problem 6c

<h3>Answers:</h3><h3>x = 33/4 = 8.25</h3><h3>y = 50</h3>

------------------

Work Shown:

TI across the top is 4x+9

NA across the bottom is 42

TI = NA since opposite sides of a parallelogram are congruent

4x+9 = 42

4x = 42-9

4x = 33

x = 33/4

x = 8.25

------

angle TAN = 3y

angle TIN = 150

angle TAN = angle TIN (opposite angles of parallelogram are congruent)

3y = 150

y = 150/3

y = 50

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What is the value of z in the equation 2(4z − 9 − 7) = 166 − 46?
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2(4z - 9 - 7) = 166 - 46 ← simplify both sides

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3 years ago
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According to an NRF survey conducted by BIGresearch, the average family spends about $237 on electronics (computers, cell phones
Usimov [2.4K]

Answer:

(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is 0.0537.

(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is 0.0023.

(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is 0.1101.

Step-by-step explanation:

We are given that according to an NRF survey conducted by BIG research, the average family spends about $237 on electronics in back-to-college spending per student.

Suppose back-to-college family spending on electronics is normally distributed with a standard deviation of $54.

Let X = <u><em>back-to-college family spending on electronics</em></u>

SO, X ~ Normal(\mu=237,\sigma^{2} =54^{2})

The z score probability distribution for normal distribution is given by;

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

where, \mu = population mean family spending = $237

           \sigma = standard deviation = $54

(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is = P(X < $150)

        P(X < $150) = P( \frac{X-\mu}{\sigma} < \frac{150-237}{54} ) = P(Z < -1.61) = 1 - P(Z \leq 1.61)

                                                             = 1 - 0.9463 = <u>0.0537</u>

The above probability is calculated by looking at the value of x = 1.61 in the z table which has an area of 0.9463.

(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is = P(X > $390)

        P(X > $390) = P( \frac{X-\mu}{\sigma} > \frac{390-237}{54} ) = P(Z > 2.83) = 1 - P(Z \leq 2.83)

                                                             = 1 - 0.9977 = <u>0.0023</u>

The above probability is calculated by looking at the value of x = 2.83 in the z table which has an area of 0.9977.

(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is given by = P($120 < X < $175)

     P($120 < X < $175) = P(X < $175) - P(X \leq $120)

     P(X < $175) = P( \frac{X-\mu}{\sigma} < \frac{175-237}{54} ) = P(Z < -1.15) = 1 - P(Z \leq 1.15)

                                                         = 1 - 0.8749 = 0.1251

     P(X < $120) = P( \frac{X-\mu}{\sigma} < \frac{120-237}{54} ) = P(Z < -2.17) = 1 - P(Z \leq 2.17)

                                                         = 1 - 0.9850 = 0.015

The above probability is calculated by looking at the value of x = 1.15 and x = 2.17 in the z table which has an area of 0.8749 and 0.9850 respectively.

Therefore, P($120 < X < $175) = 0.1251 - 0.015 = <u>0.1101</u>

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4 years ago
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Answer:

Step-by-step explanation:

I'm assuming that strange looking thing is a 9, so begin by multiplying both sides by that denominator to get:

x-9=\sqrt{x}+3 then subtract 3 from both sides to get

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x^2-24x+144=x and get everything on one side to solve for x by factoring:

x^2-25x+144=0 and factor that however it is you have learned to factor second-degree polynomials to get that

x = 16 and x = 9. We have to check for extraneous solutions because anytime you manipulate an equation, as we did by squaring both sides, you run the risk of these solutions that actually don't work when you plug them back into the original equation. Let's try 16 first:

\frac{16-9}{\sqrt{16}+3 }=1  If 16 is a solution, then this statement will be mathematically true.

\frac{7}{4+3}=1  and   \frac{7}{7}=1 so 16 works. Let's try 9:

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x = 16 is the only solution.

5 0
3 years ago
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