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VARVARA [1.3K]
2 years ago
9

URGENT Enter the values for the highlighted variables that show how to subtract the rational expressions correctly: StartFractio

n 2 Over x squared minus 36 EndFraction minus StartFraction 1 Over x squared + 6 x EndFraction = StartFraction 2 Over (x + 6) (x minus 6) EndFraction minus StartFraction 1 Over x (x + a) EndFraction. = StartFraction b x Over (x + 6) (x minus 6) x EndFraction minus StartFraction x minus c Over (x + 6) (x minus 6) x EndFraction. = StartFraction d x minus x + e Over (x + 6) (x minus 6) x EndFraction. = StartFraction x + f Over (x + 6) (x minus 6) x EndFraction. = StartFraction g Over x (x minus 6) EndFraction a = b = c = d = e = f = g =
Mathematics
1 answer:
Marrrta [24]2 years ago
3 0

The value of the highlighted variables ave been determined a =6, b= 2 ,c= 6 , d = 2  , e = 6 , f = 6, g =1

<h3>What is an Expression ?</h3>

An expression are mathematical statement consisting of variables , constants and mathematical operators .

The given expression is

\rm \dfrac{2}{x^2-36} - \dfrac{1}{x^2 +6x} = \dfrac{2}{(x+6)(x-6)}-\dfrac{1}{x(x+a)}\\\\= \dfrac{bx}{x(x+6)(x-6)}-\dfrac{x-c}{(x+6)(x+6)x}\\\\= \dfrac{dx-x+e}{x(x+6)(x-6)}\\\\= \dfrac{x+f}{x(x+6)(x-6)}\\\\= \dfrac{g}{x(x-6)}

a = 6

b= 2

c= 6

d = 2

e = 6

f = 6

g =1

Therefore the value of the highlighted variables ave been determined.

To know more about Expression

brainly.com/question/14083225

#SPJ1

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Suppose that in one metropolitan area, 25% of all home- owners are insured against earthquake damage. Four home-owners are to be
emmasim [6.3K]

Answer:

Step-by-step explanation:

Given that,

25% home owners are insecure of earthquake problem

If we select 4 home owners at random

let X denote the number among the four who have earthquake insurance

Let find the probability distribution of X

Let S- denotes a home owner who has insurance

Let F denotes a home owner who does not have insurance.

Then, P(S) =25% = 0.25

P(F) = 1 — P(S) = 1 —0.25

P(F) = 0.75

The possible outcomes of X is

X(0) = If no person has insurance

X(1) = If only 1 person has insurance

X(2) = If only 2 persons has insurance

X(3) = If only 3 persons has insurance

X(4) = If only 4 persons has insurance.

The cardinality of the sample space is n(C) = 2ⁿ = 2⁴ => 16

So, the sample space is given as

{FFFF, FFFS, FFSF, FFSS, FSFF, FSFS, FSSF, FSSS, SFFF, SFFS, SFSF, SFSS, SSFF, SSFS, SSSF, SSSS}

For X=0, the possible is {FFFF} i.e. no insurance, the one without insurance.

P(X) = 0.25×0.25×0.25×0.25

P(X) = 0.25⁴

P(X) = 0.00390625

For X=1, the possible outcome are

FFFS, FFSF, FSFF, SFFF

P(X=1) = 0.25³•0.75 + 0.25³•0.75 + 0.25³•0.75 + 0.25³•0.75

P(X=1) = 0.046875

For X=2 the possible outcomes are

FFSS, FSFS, FSSF, SFFS, SFSF, SSFF,

P(X=2)=0.25²•0.75²+0.25²•0.75²+ 0.25²•0.75²+ 0.25²•0.75²+ 0.25²•0.75²+ 0.25²•0.75²

P(X=2) = 0.2109375

For X=3 the possible outcomes are

FSSS, SFSS, SSFS, SSSF

P(X=3) = 0.25•0.75³+0.25•0.75³+ 0.25•0.75³+0.25•0.75³

P(X=3) = 0.421875

X=4 the possible outcomes are

SSSS

P(X=4) = 0.75×0.75×0.75×0.75

P(X=4) = 0.31640625.

Or

Using normal distribution

P(X=k) = ⁿCk • 0.25^k • 0.75^(4-k)

So,

P(X=0) = 4C0 • 0.25^0 • 0.75^4

P(X=0) = 0.31640625

P(X=1) = 4C1 • 0.25^1 • 0.75^3

P(X=1) = 0.421875

P(X=2) = 4C2 • 0.25^2 • 0.75^2

P(X=2) = 0.2109375

P(X=3) = 4C3 • 0.25^3 • 0.75^1

P(X=3) = 0.046875

P(X=4) = 4C4 • 0.25^4 • 0.75^0

P(X=4) = 0.00390625.

6 0
3 years ago
Solve the radical equation.<br> sqrt 25 – 40x = X – 10
Citrus2011 [14]

-40x=-10-20

-40x=-35

x= 7/8

5 0
3 years ago
3.2g of sugar is needed to make 8 cakes. How much sugar is needed for 12 cakes?
4vir4ik [10]

3.2s:8c

4.8s:12c

Therefore, you would need 4.8g of sugar to make 12 cakes. This was solved by using ratios.

8 0
3 years ago
Read 2 more answers
The weights of newborn baby boys born at a local hospital are believed to have a normal distribution with an average weight of 5
Lisa [10]

Answer: 0.4987

Step-by-step explanation:

Given : The weights of newborn baby boys born at a local hospital are believed to have a normal distribution with

Mean : \mu=5\text { lbs}

Standard deviation : \sigma= 2\text{ lbs}

Let X be the random variable that represents the weight of randomly selected student .

Z score : z=\dfrac{x-\mu}{\sigma}

For x = 5 lbs

z=\dfrac{5-5}{2}=0

For x = 11 lbs

z=\dfrac{11-5}{2}=3

By using the standard normal distribution table ,  the probability that the weight of the newborn baby boy will be between 5 lbs and 11 lbs :-

P(5

Hence, the probability that the weight of the newborn baby boy will be between 5 lbs and 11 lbs =0.4987

5 0
3 years ago
Can I get the answer
timama [110]
Yes because 8/10 can be simplified to 4/5, which has same denominator as 3/5.
6 0
3 years ago
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