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3 years ago

Are the figures below similar? Why or why not? Determine whether the triangles shown in the image are similar.

Mathematics

2 answers:

Komok [63]3 years ago

5 0

Answer:

the Answer is <u>B)no because the corresponding angles are not congruent</u>

Step-by-step explanation:

Vika [28.1K]3 years ago

3 0

Answer:

B) no, because the corresponding angles are not congruent

Step-by-step explanation:

Similar triangles must have proportional sides, as well as congruent corresponding angles. In this instance, we can see that the angles are not <em>congruent</em>, and so there is no need to solve for proportion.

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26×6 using distributive property

Elina [12.6K]

Answer:

156

Step-by-step explanation:

26=20+6

(20×6) + (6×6)

=120+36

=156

7 0

4 years ago

Please Help me it would mean a lot! TYSM!!!

MA_775_DIABLO [31]

Step-by-step explanation:

A and B works because both are 8x+4y

hope it helped

5 0

3 years ago

Mark and Damien were in a hot dog contest. Mark ate 12 hot dogs. Damien ate 4/6 as many hot dogs as Mark. How many hot dogs did

mario62 [17]

Answer:

Step-by-step explanation:

4/6 of 12 is 8

7 0

3 years ago

Items are inspected for flaws by two quality inspectors. Both inspectors inspect every item and the probability that an item has

Genrish500 [490]

Answer:

a)0.976

b)0.00926

c)0.2402

d)0.35

Step-by-step explanation:

Let $X_i$ be an item passed by inspector i

Let Y be the event that there is a fault in an item

The probability that an item has a flaw is 0.1 i.e. P(Y)=0.1

If a flaw is present ,it will be detected by the first inspector with probability 0.92 i.e. $P(\bar{X_1}|Y)=0.92$

So, $P(X_1|Y)=1-0.92=0.08$

If a flaw is present ,it will be detected by the second inspector with probability 0.7 i.e. $P(\bar{X_2}|Y)=0.7$

So, $P(X_2|Y)=1-0.7=0.3$

If an item does not have a flaw, it will be passed by the first inspector with probability 0.95 i.e. $P(X_1|\bar{Y}) = 0.95$

So, $P(\bar{X_1}|\bar{Y}) = 1-0.95=0.05$

If an item does not have a flaw, it will be passed by the second inspector with probability 0.8 i.e. $P(X_2|\bar{Y}) = 0.8$

So, $P(\bar{X_2}|\bar{Y}) = 1-0.8=0.2$

a)P(found by atleast one inspector | It has flaw )=1-P(found by none inspector | It has flow )

P(found by atleast one inspector | It has flaw )= $1-P(X_1|Y) P(X_2|Y)$

P(found by atleast one inspector | It has flaw )= $1-0.08 \times 0.3$

P(found by atleast one inspector | It has flaw )=0.976

Hence the probability that it will be found by at least one of the two inspectors if it has flaw is 0.976

b) $P(Y|X_1)=\frac{P(X_1|Y) P(Y)}{P(X_1|Y) P(Y)+P(X_1|\bar{Y}) P(\bar{Y})}$

$P(Y|X_1)=\frac{0.08 \times 0.1}{0.08 \times 0.1+0.95 \times 0.9}=0.00926$

C)P( two inspectors draw different conclusions on the same item)= $P(X_1 \cap \bar{X_2} \cap Y)+P(\bar{X_1} \cap X_2 \cap Y)+P(X_1 \cap \bar{X_2} \cap \bar{Y})+P(\bar{X_1} \cap X_2 \cap \bar{Y})$

P( two inspectors draw different conclusions on the same item)=0.2402

$P(Y|(X_1 \cap X_2))=\frac{P(Y \cap X_1 \cap X_2)}{P(X_1 \cap X_2)}\\P(Y|(X_1 \cap X_2))=\frac{P(Y \cap X_1 \cap X_2)}{P(X_1 \cap X_2 \cap Y)+P(X_1 \cap X_2 \cap \bar{Y})}\\P(Y|(X_1 \cap X_2))=0.35$

3 0

3 years ago

What is the solution of the system? use a graph. check your answer. y=2x+4, y=x+2

vladimir1956 [14]

Answer:

see picture with work shown

4 0

3 years ago