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3 years ago

In the figure, △ABC is similar to △DEF. what is m∠1

Mathematics

1 answer:

Nady [450]3 years ago

7 0

Answer:

C. 139°

Step-by-step explanation:

Given:

m<A = 62°

m<B = 77°

Required:

Find m<1

Solution:

Since ∆ABC is similar to ∆DEF, therefore:

<A ≅ <D, which means m<D = 62°

<B ≅ <E, which means m<E = 77°

<C ≅ <F

Therefore, based on exterior angle theorem:

m<1 = m<D + m<E

m<1 = 62° + 77° (substitution)

m<1 = 139°

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Find the greatest common factor of these three expressions. 5v2 , 14v4 , and 35

a_sh-v [17]

Question

Find the greatest common factor of these three expressions. 5v2 , 14v4 , and 35

Answer:

Step-by-step explanation:

assuming that 5v2 is 5², and 14v4 is 14^4

5² = 25

14^4 = 38416

35 = 35

The factors of 25 are: 1, 5, 25

The factors of 35 are: 1, 5, 7, 35

The factors of 38416 are: 1, 2, 4, 7, 8, 14, 16, 28, 49, 56, 98, 112, 196, 343, 392, 686, 784, 1372, 2401, 2744, 4802, 5488, 9604, 19208, 38416

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the greatest common factor is 1

6 0

3 years ago

Suppose we have a population whose proportion of items with the desired attribute is p = 0:5. (a) If a sample of size 200 is tak

crimeas [40]

Answer:

a. If a sample of size 200 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 41.26%.

b. If a sample of size 100 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 30.5%.

Step-by-step explanation:

This problem should be solved with a binomial distribution sample, but as the size of the sample is large, it can be approximated to a normal distribution.

The parameters for the normal distribution will be

$\mu=p=0.5\\\\\sigma=\sqrt{p(1-p)/n} =\sqrt{0.5*0.5/200}= 0.0353$

We can calculate the z values for x1=0.47 and x2=0.51:

$z_1=\frac{x_1-\mu}{\sigma}=\frac{0.47-0.5}{0.0353}=-0.85\\\\z_2=\frac{x_2-\mu}{\sigma}=\frac{0.51-0.5}{0.0353}=0.28$

We can now calculate the probabilities:

$P(0.47$

If a sample of size 200 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 41.26%.

b) If the sample size change, the standard deviation of the normal distribution changes:

$\mu=p=0.5\\\\\sigma=\sqrt{p(1-p)/n} =\sqrt{0.5*0.5/100}= 0.05$

We can calculate the z values for x1=0.47 and x2=0.51:

$z_1=\frac{x_1-\mu}{\sigma}=\frac{0.47-0.5}{0.05}=-0.6\\\\z_2=\frac{x_2-\mu}{\sigma}=\frac{0.51-0.5}{0.05}=0.2$

We can now calculate the probabilities:

$P(0.47$

If a sample of size 100 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 30.5%.

8 0

3 years ago

For the telephone bill, Etisalat charges you a fixed amount of 50 AED each month, plus 0.25 AED per minute of usage. My Bill las

IgorLugansk [536]

Answer:

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3 0

3 years ago

Assume each figure shown has the same orientation. Which figure is the image of square LMNP after a translation of

umka21 [38]

Figure 4 is the image of the square LMNP after the translation.

<u>Step-by-step explanation:</u>

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L (-3,1)

M(-1,1)

N(-1,-1)

P(-3,-1)

after translation of (x,y) → (x+5, y -3) the coordinates of the image obtained as,

L'(2,-2)

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N'(4,-4)

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3 years ago

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Bumek [7]

Answer:

Step-by-step explanation:

4 0

3 years ago