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2 years ago

I NEED HELP PLEASEEEEEEEEEEE

Mathematics

2 answers:

RoseWind [281]2 years ago

6 0

Answer:

2x + 4y = 12 and y = -1/2x + 3

Ratling [72]2 years ago

5 0

2x +4y=12 and y=-1/2+3

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Stella [2.4K]

2x-1 < x+3

5x-1 > 6-2x

x-5 < 0

solve the inequality for x.

x < 4

x > 1

X < 5

find the intersection

X ∑ {1,4}

Alternative form: {x I 1 < x < 4}

∑ indicates summation and is used as a shorthand notation for the sum of terms that follow a pattern. For example, the sum of the first 4 squared integers, 12+22+32+42, follows a simple pattern: each term is of the form i2, and we add up values from i=1 to i=4.

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9 months ago

0.99 0.89 7/8 least to greatest

katovenus [111]

From least to greatest 7/8, 0.89, 0.99

6 0

3 years ago

Consider writing onto a computer disk and then sending it through a certifier that counts the number of missing pulses. Suppose

Furkat [3]

Answer:

a) 0.164 = 16.4% probability that a disk has exactly one missing pulse

b) 0.017 = 1.7% probability that a disk has at least two missing pulses

c) 0.671 = 67.1% probability that neither contains a missing pulse

Step-by-step explanation:

To solve this question, we need to understand the Poisson distribution and the binomial distribution(for item c).

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 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.

Poisson mean:

$\mu = 0.2$

a. What is the probability that a disk has exactly one missing pulse?

One disk, so Poisson.

This is P(X = 1).

$P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164
$

0.164 = 16.4% probability that a disk has exactly one missing pulse

b. What is the probability that a disk has at least two missing pulses?

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

In which

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

$P(X = 0) = \frac{e^{-0.2}*0.2^{0}}{(0)!} = 0.819$

$P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164
$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.819 + 0.164 = 0.983$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.983 = 0.017$

0.017 = 1.7% probability that a disk has at least two missing pulses

c. If two disks are independently selected, what is the probability that neither contains a missing pulse?

Two disks, so binomial with n = 2.

A disk has a 0.819 probability of containing no missing pulse, and a 1 - 0.819 = 0.181 probability of containing a missing pulse, so $p = 0.181$

We want to find P(X = 0).

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

$P(X = 0) = C_{2,0}.(0.181)^{0}.(0.819)^{2} = 0.671$

0.671 = 67.1% probability that neither contains a missing pulse

8 0

3 years ago

Can someone help me please

Anastaziya [24]

The first step to solve this problem is to compute first for the area of the bigger rectangle:

A = LW

A = 12m (10m)

A = 120 m^2

Next step is to find the area of the smaller rectangle:

A = LW

A = 7m (2m)

A = 14 m^2

The last step is to deduct the area of the smaller rectangle to the area of the larger rectangle:

Area of larger rectangle – Area of the smaller rectangle = Area of the shaded region

120 m^2 – 14 m^2 = 106 m^2

Therefore, the area of the shaded region is 106 square meters.

7 0

3 years ago

What is the solution 4x+6<18

mash [69]

Answer:

x < 3

Step-by-step explanation:

4x +6 < 18

add and subtract 6 from both sides

4x + 6 - 6 < 18 -6

4x < 12

x< 12/4

X < 3

4 0

3 years ago