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3 years ago

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Answer two questions about Equations AAA and BBB:\begin{aligned} A.&&5x&=3x \\\\ B.&&5&=3 \end{aligned}A

.B.5x5=3x=31) How can we get Equation BBB from Equation AAA?

1 answer:

max2010maxim [7]3 years ago

6 0

Answer:

mulitply and divide by both sides

and yes

Step-by-step explanation:

You might be interested in

Over the past decade, the mean number of hacking attacks experienced by members of the Information Systems Security Association

fredd [130]

Answer:

$P(X>600)=P(\frac{X-\mu}{\sigma}>\frac{600-\mu}{\sigma})=P(Z>\frac{600-510}{14.28})=P(z>6.302)$

And we can find this probability using the complement rule and the normal standard distribution and we got:

$P(z>6.302)=1-P(z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the number of attacks of a population, and for this case we know the distribution for X is given by:

$X \sim N(510,14.28)$

Where $\mu=510$ and $\sigma=14.28$

We are interested on this probability

$P(X>600)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>600)=P(\frac{X-\mu}{\sigma}>\frac{600-\mu}{\sigma})=P(Z>\frac{600-510}{14.28})=P(z>6.302)$

And we can find this probability using the complement rule and the normal standard distribution and we got:

$P(z>6.302)=1-P(z$

7 0

3 years ago

Cher wraps 40 presents in 2 hours. At that rate, how many presents would she wrap in 15 minutes

gavmur [86]

Answer:

5 presents

Step-by-step explanation:

40 presents in 2 hours

20 presents in 1 hour (divide by 2)

20 presents in 60 minutes ( 1 hour is 60 minutes )

5 presents in 15 minutes ( divide by 4)

6 0

3 years ago

The fence around a circular garden is 18 feet long. What is the area of the garden to the nearest square foot? Use 3.14 for pi.

Sidana [21]

Answer:

A = 26 ft²

Step-by-step explanation:

To find the area of a circle, use the formula A = πr²

We know π = 3.14 and C = 18.

Use the circumference to find "r". C = 2πr

C = 2πr

18 = 2(3.14)r

18 = 6.28r

r = 18/6.28 Exact answer

r ≈ 2.866 Rounded to three decimal places

You can use either of the two bolded "r" values. I will use the rounded answer in the formula for area.

Substitute π = 3.14 and r = 2.866.

A = πr²

A = (3.14)(2.866)² Solve exponents first

A = 3.14(8.213956) Multiply

A = 25.7918218... Unrounded answer

A ≈ 26 Rounded to nearest square foot

Add the units back in.

<h3>A = 26 ft²</h3>

7 0

3 years ago

Find the probability of the following events , when a dice is thrown once:

Fudgin [204]

Answer:

Step-by-step explanation:

s a die is rolled once, therefore there are six possible outcomes, i.e., 1,2,3,4,5,6.

(a) Let A be an event ''getting a prime number''.

Favourable cases for a prime number are 2,3,5,

i.e., n(A)=3

Hence P(A)=n(A)n(S)=36=12

(b) Let A be an event ''getting a number between 3 and 6''.

Favourable cases for events A are 4 or 5.

i.e., n(A)=2

P(A)=n(A)n(S)=26=13

(c) Let A be an event ''a number greater than 4''.

Favourable cases of events A are 5, 6.

i.e., n(A)=2

P(A)=n(A)n(S)=26=13

(d) Let A be the event of getting a number at most 4.

∴ A={1,2,3} ⇒ n(A)=4,n(S)=6

∴ Required probability =n(A)n(S)=42=23

(e) Let A be the event of getting a factor of 6.

∴ A={1,2,36} ⇒ n(A)=4,n(A)=6

∴ Required probability =46=23

(ii) Since, a pair of dice is thrown once, so there are 36 possible outcomes. i.e.,

(a) Let A be an event ''a total 6''. Favourable cases for a total of 6 are (2,4), (4,2), (3,3), (5,1), (1,5).

i.e., n(A)=5

Hence P(A)=n(A)n(S)=536

(b) Let A be an event ''a total of 10n. Favourable cases for total of 10 are (6,4), (4,6), (5,5).

i.e., n(A)=5

P(A)=n(A)n(S)=336=112

(c) Let A be an event ''the same number of the both the dice''. Favourable cases for same number on both dice are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).

i.e., n(A)=6

P(A)=n(A)n(S)=636=16

(d) Let A be an event ''of getting a total of 9''. Favourable cases for a total of 9 are (3,6), (6,3), (4,5), (5,4).

i.e., n(A)=4

P(A)=n(A)n(S)=436=19

(iii) We have, n(S) = 36

(a) Let A be an event ''a sum less than 7'' i.e., 2,3,4,5,6.

Favourable cases for a sum less than 7 ar

7 0

3 years ago

Read 2 more answers

What polynomial must be added to x^2-2x+6 so that the sum is 3x^2+7x?

slavikrds [6]

2x^2+9x-6 i found this by subtracting 3x^2+7x from the other equation

5 0

3 years ago