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3 years ago

Y = -x -1 y = -2x + 7 find x and y

Mathematics

1 answer:

goldenfox [79]3 years ago

3 0

Answer:

Step-by-step explanation:

Y = -x -1..........(1)

y = -2x + 7....(2)

Equating the two equations

-x -1 = -2x + 7

-x + 2x = 7+1

x = 8

Putting x in (1)

y = -8 - 1

y = -9

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loris [4]

Answer:

6.19

Step-by-step explanation:

I hope this helps you

6 0

3 years ago

Please answer and show work.

andrew11 [14]

Simplifying
9 + -2x = 35

Solving
9 + -2x = 35

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-9' to each side of the equation.
9 + -9 + -2x = 35 + -9

Combine like terms: 9 + -9 = 0
0 + -2x = 35 + -9
-2x = 35 + -9

Combine like terms: 35 + -9 = 26
-2x = 26

Divide each side by '-2'.
x = -13

Simplifying
x = -13

4 0

4 years ago

A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 stude

EleoNora [17]

Answer:

a) $P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)$

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

b) $P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

c) $P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

d) $E(X) = 20*0.2= 4$

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".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=20, p=0.2)$

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

We want this probability:

$P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)$

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

Part b

We want this probability:

$P(X=4)$

And using the probability mass function we got:

$P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

Part c

We want this probability:

$P(X>3)$

We can use the complement rule and we got:

$P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

Part d

The expected value is given by:

$E(X) = np$

And replacing we got:

$E(X) = 20*0.2= 4$

3 0

3 years ago

10.25

Mrac [35]

C. 2g-19.5=40 I think i'm right

3 0

3 years ago

Read 2 more answers

I need the answers to 9-12! Tysm!

loris [4]

Answer:

9) 5y

10) 10 x² - 4

11) 2a ∧5 + 5b

12) 8m +4n + 2

Step-by-step explanation:

9) 8 y - 3 y = 5y

10) 6x² + 4 (x²-1) = 6x²+ 4x²-4 = 10x²-4

11) 4a∧5 - 2a∧5 +4b + b = 2a∧5 +5b

12) 8m +14-12+4n = 8m+4n+2

3 0

3 years ago

Y = -x -1 y = -2x + 7 find x and y​

Y = -x -1 y = -2x + 7 find x and y