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larisa86 [58]
2 years ago
14

Given: mTRV = 60° mTRS = (4x)° Prove: x = 30 What is the missing reason in step 3? substitution property of equality angle addit

ion postulate subtraction property of equality addition property of equality
Mathematics
2 answers:
mihalych1998 [28]2 years ago
8 0

Answer: Angle addition postulate.

Explanation: If \angle TRV= 60^{\circ} and \angle TRS=4x^{\circ}

here, if we have to prove x=30

If there is a condition that TR is a line which meets with the line segment VS at point R then by the Angle addition postulate,  we can say that \angle TRV+\angle TRS=180^{\circ}⇒x=30^{\circ}

But,

In option (1) substitution property of equality

If there is condition that  \angle TRV=\angle TRS

then we can use  substitution property of equality,

And, in this case 4x^{\circ}=60^{\circ}⇒x=15^{\circ}

which is wrong. So, we can not use this property here.

In option (3) subtraction property of equality

There is no use of this property to find the value x.

In option (4) addition property of equality

There is no use of this property to find the value x.

Lina20 [59]2 years ago
8 0

Answer: The answer is the second option

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2 years ago
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In USA Today (Sept. 5, 1996), the results of a survey involving the use of sleepwear while traveling were listed as follows: Mal
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Answers:

a) 0.018.

b) 0.614.

c) 0.166.

d) 0.479

Explanations:

1) Table

----------------------Male ----------- Female ---------- Total

Underwear ----- 0.220 ---------- 0.024 ---------- 0.244

Nightgown ----- 0.002 ---------- 0.180 ------------ 0.182

Nothing --------- 0.160 ----------- 0.018 ------------ 0.178

Pajamas --------- 0.102 ---------- 0.073 ----------- 0.175

T-shirt ------------ 0.046 --------- 0.088 ------------ 0.134

Other --------------0.084 --------- 0.003 ------------ 0.087

2) Check that it is a table of relative frequencies:

total = 0.244 + 0.182 + 0.178 + 0.175 + 0.134 + 0.087 = 1 ⇒ indeed this is a table of relative frequencies.

That means that each data is a probabiilty.

3) Soluitions:

(a) What is the probability that a traveler is a female who sleeps in the nothing?

relative frequency of females that use nothing = 0.018

(b) What is the probability that a traveler is male?

Probability that a traveler is a male = relative frequency of males = 0.220 + 0.002 + 0.160 + 0.102 + 0.046 + 0.084 = 0.614

(c) Assuming the traveler is male, what is the probability that he sleeps in pajamas?

It is the relative frequency of the males that sleep in pajamas divided by the relative frequency of males = 0.102 / 0.614 = 0.166

(d) What is the probability that a traveler is male if the traveler sleeps in pajamas or a T-shirt?

It is the relative frequency of males that sleep in pajamas plus that of the males that sleep in T-shirt, divided the total relative frequency of travelers that sleep in pajamas or T-shirt

i) relative frequency of males that sleeps in pajamas plus that of the males that sleeps in T-shirt = 0.102 + 0.046 = 0.148

ii) relative frequency of travelers that sleep in pajamas or T-shirt = 0.102 + 0.046 + 0.073 + 0.088 = 0.309

iii) quotient = 0.148 / 0.309 = 0.479

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2 years ago
The average annual amount American households spend for daily transportation is $6312 (Money, August 2001). Assume that the amou
lions [1.4K]

Answer:

(a) The standard deviation of the amount spent is $3229.18.

(b) The probability that a household spends between $4000 and $6000 is 0.2283.

(c) The range of spending for 3% of households with the highest daily transportation cost is $12382.86 or more.

Step-by-step explanation:

We are given that the average annual amount American households spend on daily transportation is $6312 (Money, August 2001). Assume that the amount spent is normally distributed.

(a) It is stated that 5% of American households spend less than $1000 for daily transportation.

Let X = <u><em>the amount spent on daily transportation</em></u>

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

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

where, \mu = average annual amount American households spend on daily transportation = $6,312

           \sigma = standard deviation

Now, 5% of American households spend less than $1000 on daily transportation means that;

                      P(X < $1,000) = 0.05

                      P( \frac{X-\mu}{\sigma} < \frac{\$1000-\$6312}{\sigma} ) = 0.05

                      P(Z < \frac{\$1000-\$6312}{\sigma} ) = 0.05

In the z-table, the critical value of z which represents the area of below 5% is given as -1.645, this means;

                           \frac{\$1000-\$6312}{\sigma}=-1.645                

                            \sigma=\frac{-\$5312}{-1.645}  = 3229.18

So, the standard deviation of the amount spent is $3229.18.

(b) The probability that a household spends between $4000 and $6000 is given by = P($4000 < X < $6000)

      P($4000 < X < $6000) = P(X < $6000) - P(X \leq $4000)

 P(X < $6000) = P( \frac{X-\mu}{\sigma} < \frac{\$6000-\$6312}{\$3229.18} ) = P(Z < -0.09) = 1 - P(Z \leq 0.09)

                                                            = 1 - 0.5359 = 0.4641

 P(X \leq $4000) = P( \frac{X-\mu}{\sigma} \leq \frac{\$4000-\$6312}{\$3229.18} ) = P(Z \leq -0.72) = 1 - P(Z < 0.72)

                                                            = 1 - 0.7642 = 0.2358  

Therefore, P($4000 < X < $6000) = 0.4641 - 0.2358 = 0.2283.

(c) The range of spending for 3% of households with the highest daily transportation cost is given by;

                    P(X > x) = 0.03   {where x is the required range}

                    P( \frac{X-\mu}{\sigma} > \frac{x-\$6312}{3229.18} ) = 0.03

                    P(Z > \frac{x-\$6312}{3229.18} ) = 0.03

In the z-table, the critical value of z which represents the area of top 3% is given as 1.88, this means;

                           \frac{x-\$6312}{3229.18}=1.88                

                         {x-\$6312}=1.88\times 3229.18  

                          x = $6312 + 6070.86 = $12382.86

So, the range of spending for 3% of households with the highest daily transportation cost is $12382.86 or more.

8 0
2 years ago
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