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

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=? 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 trials before q successes 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