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3 years ago

HELP PLEASE !!

Mathematics

2 answers:

Crazy boy [7]3 years ago

7 0

Answer:

D i hope this is right and helps

Step-by-step explanation:

Fudgin [204]3 years ago

3 0

Answer:

im pretty sure its d

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In a continuous series, mean(x) = 80, assumed mean(A) = 60 and<br>EF=40 then find the value of EFd.

Aneli [31]

Using the given information, the value of Σfd is 800

<h3>Calculating mean using Assumed mean </h3>

From the question, we are to determine the value of Σfd

The formula for mean, using the assumed mean method is given by

$\bar x = A + \frac{\sum fd}{\sum f}$

Where $\bar x$ is the mean

A is the assumed mean

From the given information,

$\bar x = 80$

$A = 60$

$\sum f = 40$

Putting the parameters into the equation, we get

$80 = 60 + \frac{\sum fd}{40}$

$80-60 = \frac{\sum fd}{40}$

$20 = \frac{\sum fd}{40}$

$\sum fd = 20 \times 40$

$\sum fd = 800$

Hence, the value of Σfd is 800

Learn more on Mean here: brainly.com/question/20118982

#SPJ1

8 0

2 years ago

4, 9, 13, 18,<br> The common difference 'd' is

Ivanshal [37]

Answer:

There is no common difference

Step-by-step explanation:

4 9 13 18

diff 5 4 5

Because the difference between 9 and 13 is 4

and the difference in the other terms is 5

there is no common differeence

7 0

2 years ago

PLEASE HELP ME, ILL BRAINLIEST YOU ill give u 30 points if you help

lisov135 [29]

Answer:

$x = 58$

$y = 90$

Step-by-step explanation:

This is a rhombus becuase it a quadralterial and it has four right triangles, and it diagonals bisect each others so it cant be a trapezoid, and paralleolgram. and it doesnt have four right corner angles so it cant be a square or rectangle or kite.

A rhombus inner angle measure is 90 degrees.

$y = 90$

This means the upper left angle of the top triangle measure is 58 becuase triangle interior theorem.

Since a rhombus bisects angles, angle x measure 58.

7 0

3 years ago

Read 2 more answers

What is the solution to this equation? 2x+3=x-4

g100num [7]

Answer:

x = -7

Step-by-step explanation:

2x+3 = x-4

2x-x = -4-3

x = -7

hope it helps ☺️

4 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