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3 years ago

Inequalities always have just one solution. True or False

Mathematics

1 answer:

amm18123 years ago

8 0

Answer:true

Step-by-step explanation:

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A piece of cardboard is 15 inches by 30 inches. A square is to be cut from each corner and the sides folded up to make an open-t

solmaris [256]

Answer:

Maximum volume = 649.519 cubic inches

Step-by-step explanation:

A rectangular piece of cardboard of side 15 inches by 30 inches is cut in such that a square is cut from each corner. Let x be the side of this square cut. When it was folded to make the box the height of box becomes x, length becomes (30-2x) and the width becomes (15-2x).

Volume is given by

V = $V = Length\times Width\times Height\\V = (30 - 2x)(15-2x)x= 4x^3-90x^2+450x\\So,\\V(x) = 4x^3-90x^2+450x$

First, we differentiate V(x) with respect to x, to get,

$\frac{d(V(x))}{dx} = \frac{d(4x^3-12x^2+9x)}{dx} = 12x^2 - 180x +450$

Equating the first derivative to zero, we get,

$\frac{d(V(x))}{dx} = 0\\\\12x^2 - 180x +450 = 0$

Solving, with the help of quadratic formula, we get,

$x = \displaystyle\frac{5(3+\sqrt{3})}{2}, \frac{5(3-\sqrt{3})}{2}$ ,

Again differentiation V(x), with respect to x, we get,

$\frac{d^2(V(x))}{dx^2} = 24x - 180$

At x =

$\displaystyle\frac{5(3-\sqrt{3})}{2}$ ,

$\frac{d^2(V(x))}{dx^2} < 0$

Thus, by double derivative test, the maxima occurs at

x = $\displaystyle\frac{5(3-\sqrt{3})}{2}$ for V(x).

Thus, largest volume the box can have occurs when $x = \displaystyle\frac{5(3-\sqrt{3})}{2}}$ .

Maximum volume =

$V(\displaystyle\frac{5(3-\sqrt{3})}{2}) = (30 - 2x)(15-2x)x = 649.5191\text{ cubic inches}$

8 0

3 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:

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

80 miles:

This means that $n = 80$ . So

$\mu = 0.01(80) = 0.8$

The probability of no damaging accidents is P(X = 0). So

$P(X = 0) = \frac{e^{-0.8}*(0.8)^{0}}{(0)!} = 0.4493$

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

6 0

2 years ago

5. principal = $450 rate = 5% time = 4 years simple interest = _ compound interest =

tatiyna

Answer:

simple interest :90

Step-by-step explanation:

Step 1: Multiply 450 with the interest rate which is 0.05 so that equal 22.5

Step 2: Multiply the 22.5 with the 4 years

So Multiply principal with the interest rate with the time

MAYBE

4 0

3 years ago

What is x in 7x-17=18

Evgen [1.6K]

Answer:

x=5

Step-by-step explanation:

7x-17=18

7x=18+17

7x=35

x=35/7

x=5

5 0

3 years ago

What is the missing term in the factorization?

jasenka [17]

Answer:

the missing term is 5

Step-by-step explanation:

Hi there !

12x² - 75 =

= 3(4x² - 25)

use formula a² - b² = (a - b)(a + b)

= 3(2x - 5)(2x + 5)

= 3(2x + 5)(2x - 5)

the missing term is 5

Good luck !

4 0

3 years ago