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mash [69]
2 years ago
10

Pls help with this math problem

Mathematics
1 answer:
valentinak56 [21]2 years ago
7 0
-1.5. -1 1/2. -3/2






yeah so its this i’m 99.98% sure
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The Han dynasty lasted from 206 B.C. to 9 A.D. How many years did the dynasty last?
cricket20 [7]

Answer:

The Han dynasty last was, 215 years

Step-by-step explanation:

Given statement: The Han dynasty lasted from 206 B.C. to 9 A.D.

To find how many years did the dynasty last.

B.C. represents Before Christ

A.D. represents After Death

To find the difference between two dates,

  • When the endings are the same you subtract
  • When they are different you add

You can see that the endings in the given 206 B.C. and 9 A.D. are B.C. and A.D. are different.

Just add these 206 and 9 we get;

206 + 9 = 215 years

Therefore, 215 years did the Han dynasty last.


4 0
3 years ago
The probability that a single radar station will detect an enemy plane is 0.65.
taurus [48]

Answer:

a) We need 4 stations to be 98% certain that an enemy plane flying over will be detected by at least one station.

b) If seven stations are in use, the expected number of stations that will detect an enemy plane is 4.55.

Step-by-step explanation:

For each station, there are two two possible outcomes. Either they detected the enemy plane, or they do not. This means that we can solve this problem using concepts of the binomial probability distribution.

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 combinatios 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.

In this problem, we have that:

The probability that a single radar station will detect an enemy plane is 0.65. This means that n = 0.65.

(a) How many such stations are required to be 98% certain that an enemy plane flying over will be detected by at least one station?

This is the value of n for which P(X = 0) \leq 0.02.

n = 1.

P(X = 0) = C_{1,0}.(0.65)^{0}.(0.35)^{1} = 0.35

n = 2

P(X = 0) = C_{2,0}.(0.65)^{0}.(0.35)^{2} = 0.1225

n = 3

P(X = 0) = C_{3,0}.(0.65)^{0}.(0.35)^{3} = 0.0429

n = 4

P(X = 0) = C_{4,0}.(0.65)^{0}.(0.35)^{4} = 0.015

We need 4 stations to be 98% certain that an enemy plane flying over will be detected by at least one station.

(b) If seven stations are in use, what is the expected number of stations that will detect an enemy plane?

The expected number of sucesses of a binomial variable is given by:

E(x) = np

So when n = 7

E(x) = 7*(0.65) = 4.55

If seven stations are in use, the expected number of stations that will detect an enemy plane is 4.55.

6 0
3 years ago
Write the prime factorization of 30. 
Ostrovityanka [42]
30|2
15|3
5|5
1

30=2\cdot3\cdot5
7 0
3 years ago
Write an equation of the translated function of the parent function f(x)=|x| with the following transformations:
MrRa [10]

Answer:

Option b

Step-by-step explanation:

To write the searched equation we must modify the function f (x) = | x | in the following way:

1. Do y = f(x + 4)

This operation horizontally shifts the function f(x) = | x | by a factor of 4 units to the left on the x axis.

y = | x +4 |

2. Do y = f (\frac{1}{4}x)

This operation horizontally expands the function f (x) = | x | in a factor of 4 units. y = |\frac{1}{4}x + 1|

3. Do y = f(x)-4

This operation vertically shifts the function f (x) = | x | by a factor of 4 units down on the y-axis.

y = |\frac{1}{4}x +1| -4

4. After these transformations the function f(x) = | x | it looks like:

f(x) = |\frac{1}{4}x +1| -4

Therefore the correct option is option b. You can verify that your vertex is at point (-4, -4) by making f (-4)

f(-4) = |\frac{1}{4}(-4) +1| -4 = -4

8 0
4 years ago
*you get 16 points*PLEASE ANSWER i’m kinda failing so help
Alecsey [184]
The answer is four because there's one line that is placed on the origin which makes it no solution because there's not supposed to be a line on the origin and number 1 is a solution because there are neither lines plotted on the origin and has parallel lines
4 0
3 years ago
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