All categories

3 years ago

Evaluate the algebraic expressions for the given value of each variable

Mathematics

1 answer:

baherus [9]3 years ago

4 0

Plug in 4 for r and 1 for s

r+s^2
4+1^2

Evaluate exponents first (order of operations)
4+1

Add
5

Final answer: 5

You might be interested in

A particular concentration of a chemical found in polluted water has been found to be lethal to 20% of the fish that are exposed

torisob [31]

Answer:

a) $P(X=14)=(20C14)(0.8)^{14} (1-0.8)^{20-14}=0.109$

b) $P(X\geq 10) = 1-P(X \leq 9) = 0.9994$

c) $P(X \leq 16)= 1-P(X>16) =1-P(X \geq 17)= 1- [P(X=17) +...+P(X=20)]=0.589$

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

$Var(X) = np(1-p) = 20*0.8*(1-0.8) = 3.2$

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".

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

$X \sim Binom(n=20, p=1-0.2=0.8)$

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

For this case we want this probability:

$P(X=14)$

And using the mass function we have this:

$P(X=14)=(20C14)(0.8)^{14} (1-0.8)^{20-14}=0.109$

Part b

For this case we want this probability:

$P(X \geq 10)$

And we can find this using the complement rule:

$P(X\geq 10) = 1-P(X$

$P(X=0)=(20C0)(0.8)^{0} (1-0.8)^{20-0}=1.05x10^{-14}$

$P(X=1)=(20C1)(0.8)^{1} (1-0.8)^{20-1}=8.39x10^{-13}$

$P(X=2)=(20C2)(0.8)^{2} (1-0.8)^{20-2}=3.19x10^{-11}$

$P(X=3)=(20C3)(0.8)^{3} (1-0.8)^{20-3}=7.65x10^{-10}$

$P(X=4)=(20C4)(0.8)^{4} (1-0.8)^{20-4}=1.30x10^{-8}$

$P(X=5)=(20C5)(0.8)^{5} (1-0.8)^{20-5}=1.66x10^{-7}$

$P(X=6)=(20C6)(0.8)^{6} (1-0.8)^{20-6}=1.66x10^{-6}$

$P(X=7)=(20C7)(0.8)^{7} (1-0.8)^{20-7}=1.33x10^{-5}$

$P(X=8)=(20C8)(0.8)^{8} (1-0.8)^{20-8}=8.65x10^{-5}$

$P(X=9)=(20C9)(0.8)^{9} (1-0.8)^{20-9}=0.00046$

And if we replace we got:

$P(X\geq 10) = 1-P(X \leq 9) = 0.9994$

Part c

For this case we want this probability:

$P(X \leq 16)$

And we can use the complement rule like this:

$P(X \leq 16)= 1-P(X>16) =1-P(X \geq 17)= 1- [P(X=17) +...+P(X=20)]$

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

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

$P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.0576$

$P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.0115$

And if we replace we got:

$P(X \leq 16)= 1-P(X>16) =1-P(X \geq 17)= 1- [P(X=17) +...+P(X=20)]=0.589$

Part d

The expected value is given by:

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

$Var(X) = np(1-p) = 20*0.8*(1-0.8) = 3.2$

3 0

3 years ago

5x-2=3x+40 answer plz :')

noname [10]

X=21

5x(-3x) -2= 3x(-3x) +40

2x-2(+2)=40(+2)

2x=42

X=21

7 0

3 years ago

Help me out please!!! 50 points!!!

german

Step-by-step explanation:

I assume they intersect at a right angle (90°).

then the intersection is a rectangle with one side being 12 m, and the diagonal being 13 m long.

the diagonal is the Hypotenuse (baseline) of a right-angled triangle with the length and the width of the rectangle as legs.

so, by using Pythagoras we have

diagonal² = length² + width²

13² = 12² + width²

169 = 144 + width²

25 = width²

width = 5 m

so, Oakdale Avenue is 5 m wide.

3 0

3 years ago

Read 2 more answers

What procedure can be used to solve the equation StartFraction x Over 19.3 EndFraction = 38.6, and what is the solution?

kozerog [31]

Answer: it’s number D I think let me know if I’m wrong

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

4. A line with a slope of -2 passes

Shalnov [3]

Answer:

y=-2x+5..…...........

6 0

3 years ago