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Gekata [30.6K]
3 years ago
5

Type SSS, SAS, ASA, SAA, or HL to justify why the two larger triangles are congruent.

Mathematics
1 answer:
Otrada [13]3 years ago
5 0

Answer:

SSS

Step-by-step explanation:

You might be interested in
On one day a pet store sells 2 birds 6 gerbils 3 fish and 3 hamsters whats the data and relative frequency as a percent?
bonufazy [111]

Answer:

Birds F_b=14.3\%

Gerbils F_g=42.9\%

Fish F_f=21.4\%

Hamsters F_h=21.4\%

Step-by-step explanation:

From the question we are told that:

Sales

Number of birds  n_b=2

Number of gerbils  n_g=6

Number of fish  n_f=3

Number of hamsters  n_h=3  

Generally the frequency for birds F_b is mathematically given b

F_b=\frac{n_b}{n}*100

F_b=\frac{2}{14}*100

F_b=14.3\%

Generally the frequency for Gerbils F_g is mathematically given b

F_g=\frac{n_g}{n}*100

F_g=\frac{6}{14}*100

F_g=42.9\%

Generally the frequency for fish F_f is mathematically given b

F_f=\frac{n_f}{n}*100

F_f=\frac{3}{14}*100

F_f=21.4\%

Generally the frequency for fish F_h is mathematically given b

F_h=\frac{n_h}{n}*100

F_h=\frac{3}{14}*100

F_h=21.4\%

5 0
3 years ago
Have another question! will give brainliest again
stepan [7]

Answer:

-64

Step-by-step explanation:

If I represent the worker's position relative to the surface of the earth then

if we let Δy = the change of position.

Δy = 0 -64 \\ Δy = -64

5 0
3 years ago
PQ=RQ a=? <br> Please need help!!
Svetllana [295]
A=100°
as 180-140= 40 , 40 x 2 = 80 & 180-80 = 100° = a
5 0
3 years ago
A bathtub is the shape of a rectangular prism. The dimensions are 5 1/2 ft times 2 1/2 ft times 2 3/4 ft. What is the volume of
Sav [38]

Answer:

37 \frac{13}{16}  ft^3

Step-by-step explanation:

A rectangular prism is also know as a cuboid.

The dimensions of the rectangular prism shaped bath tub are 5½ ft * 2½ ft * 2¾ ft.

The volume of a cuboid (rectangular prism) is given as:

V = L * B * H

Hence, the volume of the bath tub is:

V = 5½  * 2½ * 2¾

V = 11/2 * 5/2 * 11/4

V = 605 / 16 = 37 \frac{13}{16}  ft^3

The volume of the bath tub is 37 \frac{13}{16}  ft^3

5 0
3 years ago
A representative from the National Football League's Marketing Division randomly selects people on a random street in Kansas Cit
Orlov [11]

Using the binomial distribution, we have that:

a) 0.1024 = 10.24% probability that the marketing representative must select 4 people to find one who attended the last home football game.

b) 0.2621 = 26.21% probability that the marketing representative must select more than 6 people to find one who attended the last home football game.

c) The expected number of people is 4, with a variance of 20.

For each person, there are only two possible outcomes. Either they attended a game, or they did not. The probability of a person attending a game is independent of any other person, which means that the binomial distribution is used.

Binomial probability distribution  

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}  

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • p is the probability of a success on a single trial.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

The expected number of <u>trials before q successes</u> is given by:

E = \frac{q(1-p)}{p}

The variance is:

V = \frac{q(1-p)}{p^2}

In this problem, 0.2 probability of a finding a person who attended the last football game, thus p = 0.2.

Item a:

  • None of the first three attended, which is P(X = 0) when n = 3.
  • Fourth attended, with 0.2 probability.

Thus:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{3,0}.(0.2)^{0}.(0.8)^{3} = 0.512

0.2(0.512) = 0.1024

0.1024 = 10.24% probability that the marketing representative must select 4 people to find one who attended the last home football game.

Item b:

This is the probability that none of the first six went, which is P(X = 0) when n = 6.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{6,0}.(0.2)^{0}.(0.8)^{6} = 0.2621

0.2621 = 26.21% probability that the marketing representative must select more than 6 people to find one who attended the last home football game.

Item c:

  • One person, thus q = 1.

The expected value is:

E = \frac{q(1-p)}{p} = \frac{0.8}{0.2} = 4

The variance is:

V = \frac{0.8}{0.04} = 20

The expected number of people is 4, with a variance of 20.

A similar problem is given at brainly.com/question/24756209

3 0
2 years ago
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