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4 years ago

(easy) If ΔEFG ~ ΔLMN with a ratio of 3:1, which of the following is true?

Mathematics

2 answers:

alexandr402 [8]4 years ago

7 0

Answer:

segment EG over segment LN equals segment FG over segment MN

Step-by-step explanation:

we know that

If two figures are similar, then the ratio of its corresponding sides is proportional and its corresponding angles are congruent

In this problem

The corresponding sides are

EF and LM

EG and LN

FG and MN

The corresponding angles are

∠E≅∠L

∠F≅∠M

∠G≅∠N

therefore

EF/LM=EG/LN=FG/MN=3/1

Alla [95]4 years ago

5 0

Answer:

C: Segment EG over segment LN equals segment FG over MN.

Step-by-step explanation:

We are given that $\triangle EFG \sim\traingle LMN$ with ratio 3:1

We have to find the true statement about two similar triangles in given options

When two triangle are similar

Then ratios of all sides of one triangle to its corresponding all sides of another triangle are equal.

Therefore, Corresponding side of EF is LM

Corresponding side of FG is MN

Corresponding side of EG is LN

Ratio

$\frac{EF}{LM}=\frac{FG}{MN}=\frac{EG}{LN}=\frac{3}{1}$

Hence, segment FG over segment MN equals to segment EG over segment LN.

Therefore, option C is true.

Answer : C: Segment EG over segment LN equals segment FG over MN.

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3 years ago

Read 2 more answers

Convert the Cartesian equation x 2 + y 2 + 2y = 0 to a polar equation.

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3 0

3 years ago

The ages of a random sample of five university professors are 39, 54, 61, 72, and 59. Using this information, find a 99% confide

kondor19780726 [428]

Answer:

99% confidence interval for the population standard deviation = (74.97 , 635.20).

Step-by-step explanation:

We are given that the ages of a random sample of five university professors are 39, 54, 61, 72 and 59. Also, it is provided that the ages of university professors are normally distributed.

So, firstly the pivotal quantity for 99% confidence interval for the population standard deviation is given by;

P.Q. = $\frac{(n-1)s^{2} }{\sigma^{2} }$ ~ $\chi^{2} __n_-_1$

where, s = sample standard deviation

$\sigma$ = population standard deviation

n = sample of university professors = 5

Also, $s^{2}$ = $\frac{\sum (X-\bar X)^{2} }{n-1}$ = 144.5

So, 99% confidence interval for population standard deviation, $\sigma$ is;

P(0.2070 < $\chi^{2} __5_-_1$ < 14.86) = 0.99 {As the table of $\chi^{2}$ at 4 degree of freedom

gives critical values of 0.2070 & 14.86}

P(0.2070 < $\frac{(n-1)s^{2} }{\sigma^{2} }$ < 14.86) = 0.99

P( $\frac{ 0.2070}{(n-1)s^{2} }$ < $\frac{1 }{\sigma^{2} }$ < $\frac{ 14.86}{(n-1)s^{2} }$ ) = 0.99

P( $\frac{ (n-1)s^{2}}{14.86 }$ < $\sigma^{2}$ < $\frac{ (n-1)s^{2}}{0.2070 }$ ) = 0.99

99% confidence interval for $\sigma^{2}$ = ( $\frac{ (n-1)s^{2}}{14.86 }$ , $\frac{ (n-1)s^{2}}{0.2070 }$ )

= ( $\frac{ (5-1) \times 144.5^{2}}{14.86 }$ , $\frac{ (5-1) \times 144.5^{2}}{0.2070 }$ )

= (5620.525 , 403483.092)

99% confidence interval for $\sigma$ = ( $\sqrt{5620.525}$ , $\sqrt{403483.092}$ )

= (74.97 , 635.20)

Therefore, 99% confidence interval for the population standard deviation of the ages of all professors at the university is (74.97 , 635.20).

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pantera1 [17]

Answer:

Step-by-step explanation:

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