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2 years ago

Which would prove that ΔABC ~ ΔXYZ? Select two options.

Mathematics

2 answers:

xxTIMURxx [149]2 years ago

7 0

Answer:

StartFraction B A Over Y X = StartFraction B C Over Y Z EndFraction = StartFraction A C Over X Z EndFraction
StartFraction A C Over X Z EndFraction = StartFraction B A Over Y X EndFraction , Angle A is-congruent-to Angle X

Step-by-step explanation:

For similarity of triangles, it is sufficient to show corresponding sides are proportional (first choice). (SSS)

It is also sufficient to show two corresponding sides are proportional and the angle between them is congruent (third choice). (SAS)

_____

IMO, it would be much easier to read the answer choices if they were written using appropriate symbols:

BA/YX = BC/YZ = AC/XZ

AC/XZ = BA/YX, ∠A≅∠X

nataly862011 [7]2 years ago

4 0

Answer:

1 and 3

Step-by-step explanation:

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2 years ago

Which expression can be used to convert 100 USD to Japanese yen

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As of 12:04 EST U.S.

$1=<span>112.624847Yen

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2 years ago

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A company administers an "aptitude test for managers" to aid in selecting new management trainees. Prior experience suggests tha

RoseWind [281]

Answer:

0.94

Step-by-step explanation:

The question after this basically is:

<em>"If the applicant passes the "aptitude test for managers", what is the probability that the applicant will succeed in the management position?"</em>

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So,

P(successful if hired) = 60% = 0.6 [let it be P(x)]

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2 years ago

Exercise 6.12 presents the results of a poll where 48% of 331 Americans who decide to not go to college do so because they canno

tensa zangetsu [6.8K]

<h2>Answer with explanation:</h2>

As per given , we have

$\hat{p}=0.48$ , n=331

Critical value for 90% Confidence interval : $z_{\alpha/2}=1.645$

a) Confidence interval :

$\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$0.48\pm (1.645)\sqrt{\dfrac{0.48(1-0.48)}{331}}\\\\\approx 0.48\pm0.02746\\\\=(0.48-0.02746,\ 0.48+0.02746)\\\\=(0.45254,\ 0.50746)$

Hence, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot a ord it : $(0.45254,\ 0.50746)$

b) Margin of error : E=1.5%=0.015

Formula for sample size : $n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2$

For p =0.48 , we have

$n=0.48(1-0.48)(\dfrac{1.645}{0.015})^2=3001.88373333\approx3002$

Hence, the required sample size to survey = 3002

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2 years ago

Kirsty counts the number of peas in a sample of pea pods

RoseWind [281]

Mode is most occurring

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Median is middle number

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Median is 6

Mean is average

4+5+6+7+8 divided by five

Mean is 6

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2 years ago

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