4.3-4.4 section of pre-calculus problem in link please also tell how you got your answer if can .

Pentagon VWXYZ is shown on the coordinate grid. A student reflected pentagon VWXYZ across the y-axis to create pentagon V'W'X'Y'

sukhopar [10]

Answer: (x, y) —> (-x, y)

Step-by-step explanation:

This is because the translation rule for reflection across the y-axis is (x, y) —> (-x, y)

4 0

3 years ago

Read 2 more answers

+4 What is the solution of 2x-15 - <0? • -45x51 0 -4 -45 -4

soldier1979 [14.2K]

Answer:

alto 5.340040 to costco

Step-by-step explanation:

7 0

3 years ago

WHAT IS 3+3=ANSWER NEEDED

VikaD [51]

Answer:

i think its a uhh a carrot?? no the answer is 6

Step-by-step explanation:

beetlejuice

5 0

2 years ago

Consider a nuclear power plant that produces 1200 MW of power and has a conversion efficiency of 34 percent (that is, for each u

cestrela7 [59]

Answer:

The amount of nuclear fuel required is 1.24 kg.

Step-by-step explanation:

From the principle of mass energy equivalence we know that energy generated by mass 'm' in an nuclear plant is

$E=m\cdot c^2$

where

'c' is the speed of light in free space

Since the power plant operates at 1200 MW thus the total energy produced in 1 year equals

$E=1200\times 10^6\times 3600\times 24\times 365=3.8\times 10^{16}Joules$

Thus using the energy produced in the energy equivalence we get

$3.8\times 10^{16}=mass\times (3\times 10^{8})^2\\\\\therefore mass=\frac{3.8\times 10^{16}}{9\times 10^{16}}=0.422kg$

Now since the efficiency of conversion is 34% thus the fuel required equals

$mass_{required}=\frac{0.422}{0.34}=1.24kg$

8 0

3 years ago

2. Suppose that you are looking for a student at your college who lives within five miles of you. You know that 55% of the 25,00

svlad2 [7]

Using the binomial distribution, it is found that:

a) There is a 0.0501 = 5.01% probability that you need to contact four people.

b) You expect to contact 1.82 students until you find one who lives within five miles of you.

c) The standard deviation is of 1.22 students.

d) There is a 0.3369 = 33.69% probability that 3 of them live within five miles of you.

e) It is expected that 2.75 students live within five miles of you.

For each student, there are only two possible outcomes. Either they live within 5 miles of you, or they do not. The probability of a student living within 5 miles of you is independent of any other student, which means that the binomial distribution is used to solve this question.

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:

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

In this problem:

55% of the students live within five miles of you, thus $p = 0.55$ .

Item a:

This probability is P(X = 0) when n = 3(none of the first three living within five miles of you) multiplied by 0.55(the fourth does live within five miles), hence:

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

$P(X = 0) = C_{3,0}.(0.55)^{0}.(0.45)^{3} = 0.091125$