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Arada [10]
3 years ago
15

I need an answer asap xoxo, anastasia

Mathematics
1 answer:
Lorico [155]3 years ago
6 0
The answer is 0.1 repeating (that is the one with the line over it) because you divide 5 by 45 and that is what u get
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1. When the point is reflected over x-axis, what will<br> the new coordinates of the new point be?
stich3 [128]

Answer:

(x , y) ----> (-x , y)

Step-by-step explanation:

(x , y) ----> (-x , y)

Example for 1. :

(-4,2) ---> (4,2)

4 0
2 years ago
Factor the following quadratic function (put the lesser value first).<br> x2 + 5x + 6
Anestetic [448]

Answer: Evaluate

5x+x_{2} +6

3 0
2 years ago
Is (3,4) a solution to the system of<br> equations. <br> X + y = 7<br> x - 2y = - 5
Anuta_ua [19.1K]

Answer: Yes, the point (3,4) is a solution to the system.

===================================================

Proof of this:

Replace x with 3 and y with 4 in the first equation

x+y = 7

3+4 = 7

7 = 7

This confirms the first equation. Repeat for the second equation

x-2y = -5

3-2(4) = -5

3 - 8 = -5

-5 = -5

We get true equations for both when we plug in (x,y) = (3,4). This confirms it is a valid solution to the system of equations. It turns out it's the only solution to this system of equations. Visually, the two lines cross at the single location (3,4).

8 0
3 years ago
When a number x is multiplied by 6, the result is 38 .
serg [7]
X = 19/3

________
________
6 0
3 years ago
The U.S. government has devoted considerable funding to missile defense research over the past 20 years. The latest development
Bad White [126]

Answer:

a) Let the random variable X= "number of these tracks where SBIRS detects the object." in order to use the binomial probability distribution we need to satisfy some conditions:

1) Independence between the trials (satisfied)

2) A value of n fixed , for this case is 20 (satisfied)

3) Probability of success p =0.2 fixed (Satisfied)

So then we have all the conditions and we can assume that:

X \sim Bin(n =20, p=0.8)

b) X \sim Bin(n =20, p=0.8)

c) P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456

d) P(X \geq 15) = P(X=15)+ .....+P(X=20)

P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456

P(X=16)=(20C16)(0.8)^{16} (1-0.8)^{20-16}=0.218

P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205

P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137

P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058

P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012

P(X\geq 15)=0.804208

e) E(X) = np = 20*0.8 = 16

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

Part a

Let the random variable X= "number of these tracks where SBIRS detects the object." in order to use the binomial probability distribution we need to satisfy some conditions:

1) Independence between the trials (satisfied)

2) A value of n fixed , for this case is 20 (satisfied)

3) Probability of success p =0.2 fixed (Satisfied)

So then we have all the conditions and we can assume that:

X \sim Bin(n =20, p=0.8)

Part b

X \sim Bin(n =20, p=0.8)

Part c

For this case we just need to replace into the mass function and we got:

P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456

Part d

For this case we want this probability: P(X\geq 15)

And we can solve this using the complement rule:

P(X \geq 15) = P(X=15)+ .....+P(X=20)

P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456

P(X=16)=(20C16)(0.8)^{16} (1-0.8)^{20-16}=0.218

P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205

P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137

P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058

P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012

P(X\geq 15)=0.804208

Part e

The expected value is given by:

E(X) = np = 20*0.8 = 16

5 0
3 years ago
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