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Gnom [1K]
3 years ago
8

To which graph does the point (−1, 4) belong?

Mathematics
2 answers:
Verdich [7]3 years ago
6 0
Insert x = -1 and y = 4 into the first inequality:-
4 <= -(-1) + 4 = 5  

so this fits


Its A
marta [7]3 years ago
4 0

Answer: First one

Step-by-step explanation:

Due to slope form which is y=mx + b.

-1 = x and y = 4.

It does not show the one because it would be equal to X, but it does show the negative

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A jar has 16 marbles, 10 green, 2 blue, 4 red. what is the probability of randomly choosing a red marble then without putting it
marta [7]

The probability of drawing a red marble and then a green marble is:

P = 1/6.

<h3>How to find the probability?</h3>

In the jar we have a total of 16 marbles, such that:

  • 10 are green.
  • 2 are blue
  • 4 are red.

First, the probability of getting a red marble is give by the quotient between the number of red marbles and the total number of marbles:

p = 4/16 = 1/4

Now we need to draw a green one, the probability is computed in the same way, but this time there are 15 marbles, because we already took one.

q = 10/15

The joint probability (first drawing red, then green) is given by the product between the individual probabilities:

P = p*q = (1/4)*(10/15) = 10/60 = 1/6

If you want to learn more about probability:

brainly.com/question/25870256

#SPJ1

8 0
1 year ago
Select the two values of x that are roots of this equation.<br> x2 - 5x + 2 = 0
Margaret [11]

Answer:

C and D

{x}^{2}  - 5x + 2 = 0 \\ x =  \frac{5± \sqrt{17} }{2}

8 0
3 years ago
Read 2 more answers
I NEED HELP PLEASEE
xenn [34]

\bf \cfrac{1+cot^2(\theta )}{1+csc(\theta )}=\cfrac{1}{sin(\theta )} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{1+cot^2(\theta )}{1+csc(\theta )}\implies \cfrac{1+\frac{cos^2(\theta )}{sin^2(\theta )}}{1+\frac{1}{sin(\theta )}}\implies \cfrac{~~\frac{sin^2(\theta )+cos^2(\theta )}{sin^2(\theta )}~~}{\frac{sin(\theta )+1}{sin(\theta )}}\implies \cfrac{~~\frac{1}{sin^2(\theta )}~~}{\frac{sin(\theta )+1}{sin(\theta )}}

\bf \cfrac{1}{\underset{sin(\theta )}{~~\begin{matrix} sin^2(\theta ) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }}\cdot \cfrac{~~\begin{matrix} sin(\theta ) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{sin(\theta )+1}\implies \cfrac{1}{sin^2(\theta )+sin(\theta )} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \cfrac{1+cot^2(\theta )}{1+csc(\theta )}\ne \cfrac{1}{sin(\theta )}~\hfill

5 0
3 years ago
If ORBIT is coded BEOVG, then how is PLANET coded? /
arlik [135]

Answer:

1. CYNARG

Step-by-step explanation:

This a example of Caesar's Cipher, in which each letter in the original word leads to a ciphered letter according to the following equation:

C = (P + o) \text{mod} 26

In which C is the index of the Ciphered letter in the alphabet, P is the index of the original letter and o is the offset.

Finding the offset:

O is coded B

O is the 15th letter in the alphabet, so P = 15.

B is the 2nd letter in the alphabet, so C = 2

C = (P + o) \text{mod} 26

2 = (15 + o) \text{mod} 26

|o = |2-15| \text{mod} 26

o = 13

So

C = (P + 13) \text{mod} 26

PLANET:

P

P is the 16th letter in the alphabet.

C = (16 + 13) \text{mod} 26 = 3

So P is coded C.

L

L is the 12th letter in the alphabet:

C = (12 + 13) \text{mod} 26 = 25

L is coded Y(25th letter in the alphabet)

A

A is the 1st letter in the alphabet

C = (1 + 13) \text{mod} 26 = 14

A is coded N

N

N is the 14th letter in the alphabet

C = (14 + 13) \text{mod} 26 = 1

N is coded A

E

E is the 5th letter in the alphabet

C = (5 + 13) \text{mod} 26 = 18

E is coded R

So the correct answer is:

1. CYNARG

4 0
3 years ago
Suppose we choose7 objects without replacement from 9 objects
stiks02 [169]

Answer:

<h2>181 440</h2>

Step-by-step explanation:

We have 9 choices for Choosing the first object

              8 choices for Choosing the 2nd object

              .

              .

              .

             3 choices for Choosing the seventh object

Therefore if we want to choose7 objects without replacement from 9 objects we have : 9×8×7×6×5×4×3 = 181 440

Also ,we can calculate it this way :

9P7 = 181 440

7 0
2 years ago
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