What is the measure of the missing length?

3. Consider the functions f(x) = x2 + 10x – 5, g(x) = 8x + 1, and h(x) = 3x – 4. List the functions from least to greatest value

weeeeeb [17]

<u>x = 3</u>
f(x) = x²+ 10x - 5
f(3) = (3)² + 10(3) - 5
f(3) = 9 + 30 - 5
f(3) = 39 - 5
f(3) = 34

g(x) = 8x + 1
g(3) = 8(3) + 1
g(3) = 25

h(x) = 3x - 4
h(3) = 3(3) - 4
h(3) = 9 - 4
h(3) = 5

<u>x = 6
</u>f(x) = x² + 10x - 5
f(6) = (6)² + 10(6) - 5
f(6) = 36 + 60 - 5
f(6) = 96 - 5
f(6) = 91

g(x) = 8x + 1
g(6) = 8(6) + 1
g(6) = 48 + 1
g(6) = 49

h(x) = 3x - 4
h(6) = 3(6) - 4
h(6) = 18 - 4
h(6) = 14

I would explain to someone that you don't need to do any calculations to know the order of the functions when x is equal to 15 by knowing that f(x) is equal to 370, g(x) is equal to 121, and h(x) is equal to 41 to know that it is easy finding the function of x without calculating the answer.

4 0

3 years ago

Graph - 9x + 6y = 18 using linear standard form

postnew [5]

Answer:

The graph is attached Below and the plotting is given below.

Step-by-step explanation:

Given:

-9x + 6y = 18

Solution:

To draw a line on a graph the required minimum two points but here we will have three points as point A, point B, and point C.

For point A

Put x = -4 in the given equation we get

-9×-4 + 6y = 18

6y = 18-36

∴ $y=\frac{-18}{3}\\ \\y=-3$

∴ Point A ≡ ( -4, -3 ).

For point B

Put x = -2 in the given equation we get

-9×-2 + 6y = 18

6y = 18 - 18

6y = 0

∴ $y=\frac{0}{3}\\ \\y=0$

∴ Point B ≡ ( -2, 0 ).

For point C

Put x = 0 in the given equation we get

-9×0 + 6y = 18

6y = 18

∴ $y=\frac{18}{3}\\ \\y=3$

∴ Point C ≡ ( 0, 3 ).

Now we have Point A ,B and C join it and you will have Line.

3 0

3 years ago

Carol went on a holiday to Bristol.

Leviafan [203]

Answer:

658.91$

Step-by-step explanation:

Find the total cost

flight cost 397$

staying at the hotel for 6 night cost 6 x 28.43£

= 170.58£ = 233.69$

breakfast for 4 morning cost 4 x 5.15£= 20.6£

= 28.22$

Total cost is 397$+233.69$+28.22=658.91$

5 0

2 years ago

A highway traffic condition during a blizzard is hazardous. Suppose one traffic accident is expected to occur in each 60 miles o

o-na [289]

Answer:

a) 0.3408 = 34.08% probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway.

b) 0.0879 = 8.79% probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway.

c) 0.4493 = 44.93% probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway.

Step-by-step explanation:

We have the mean for a distance, which means that the Poisson distribution is used to solve this question. For item b, the binomial distribution is used, as for each blizzard day, the probability of an accident will be the same.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

Suppose one traffic accident is expected to occur in each 60 miles of highway blizzard day.

This means that $\mu = \frac{n}{60}$ , in which 60 is the number of miles.

(a) What is the probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway?

$n = 25$ , and thus, $\mu = \frac{25}{60} = 0.4167$

This probability is:

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.4167}*(0.4167)^{0}}{(0)!} = 0.6592$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.6592 = 0.3408$

0.3408 = 34.08% probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway.

(b) Suppose there are six blizzard days this winter. What is the probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway?

Binomial distribution.

6 blizzard days means that $n = 6$

Each of these days, 0.6592 probability of no accident on this stretch, which means that $p = 0.6592$ .

This probability is P(X = 2). So

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

$P(X = 2) = C_{6,2}.(0.6592)^{2}.(0.3408)^{4} = 0.0879$

0.0879 = 8.79% probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway.

(c) If the probability of damage requiring an insurance claim per accident is 60%, what is the probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway?

Probability of damage requiring insurance claim per accident is of 60%, which means that the mean is now:

$\mu = 0.6\frac{n}{60} = 0.01n$