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Andre45 [30]
2 years ago
6

We can use algebraic operations to rewrite expressions, including polynomial expressions. When we rewrite polynomial expressions

, the representations look different but still represent the same expression.
Where else have you seen this type of transformation—when something appears different, but its meaning remains the same?
Mathematics
1 answer:
Salsk061 [2.6K]2 years ago
6 0

Similar transformations to the application of algebraic operations to rewrite expressions that form representations that appear different from their parent expressions, can be seen in the following fields;

Geometry; Rigid transformations such as rotation, can transform a rectangle to appear to look like a kite

Physics; The heating of a liquid, can transform its physical state from liquid to gas, however, the gas formed by heating the liquid represent the same material chemically

Learn more about transformations here;

brainly.com/question/7161333

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Triangles H J K and L M N are congruent. Triangle H J K is rotated about point H to form triangle L N M. Triangle L M N is highe
Sauron [17]

Answer:

Translate K to N and rotate about K until HK lies on the line containing LN.

Step-by-step explanation:

Hope this helps bb's!!

Lots of love and peace your way!!  <3

7 0
2 years ago
Read 2 more answers
The service time distribution describes the probability P that the service time of the customer will be no more than T hours. If
Molodets [167]

Answer: option d.

Step-by-step explanation:

You have the following formul given in the problem:

p=1-e^{-mt}

You know that:

The number of  customers serviced in an hour by the technical support representative is 6 costumbers, therefore:

m=6

As the problem asked for the probability that  a costumber will be on hold less than 30 minutes, we know that:

t=0.5

Substitute the values above into the formula.

Then, you obtain:

p=1-e^{-(6)(0.5)}=0.95 or 95%

7 0
2 years ago
To increase sales, a local donut shop began putting an extra donut in some of the boxes. Customers are unaware of which boxes ha
Bingel [31]

Answer:

Step-by-step explanation:

Hello!

Part A

First, determine your study variable:

X: Number of boxes with an extra donut in a sample of eight.

To see if the variable has a Binomial distribution you have to check if the binomial criteria are met:

1. The number of observation of the trial is fixed (In this case n = 8, the boxes each customer bought make the sample)

2. Each observation in the trial is independent, this means that none of the trials will affect the probability of the next trial (In this case, the amount of donuts in one box does not affect on the probability of the next box having an extra donut)

3. The probability of success in the same from one trial to another (In this case our "success" will be that the box has an extra donut, according to the owners claim that is 1/7; ρ=0,14)

So X≈ Bi (n;ρ)

Where n represents the sample (n=8) and ρ is the probability of success (ρ=0.14)

Part B

The mean of the binomial distribution is E(X)= nρ

E(X)= 8 * 0.14= 1.12

The mean of the distribution is also called the expected value. You'd expect that 1.12 boxes have an extra donut.

The variance of the binomial distribution is V(X)= nρ(1 - ρ)

V(X)= 8*0.14*(1 - 0.14)= 0.9632

Its square root is the standard deviation

√V(X)= 0.98

The standard deviation is a measure of dispersion, it indicates how much the values ​​of the distribution of the central value are separated. In this case, 0.98 indicates that the distribution of the number of boxes with an extra donut is far from the expected value.

Part C

Two of the eight customers buy a box with an extra donut, symbolically:

P(X=2) = P(X≤2) - P(X≤1)= 0.91 - 0.68 = 0.22

There is a 22% chance that two customers bought a box with an extra donut.

Compute:

P(X≥2)= 1 - P(X<2)= 1 - P(X≤1)= 1 - 0.68= 0.32

There is a 32% chance that two or more customers bought a box with an extra donnut.

I hope it helps!

3 0
2 years ago
Shawn has 4 nickles. How many walnuts can he buy if he spends all 4 nickels?
Roman55 [17]

Answer:

This really depends.

Step-by-step explanation:

If a walnut costs 10 cents, he can buy 2 walnuts.

If a walnut costs 5 cents, he can buy 4.

If a walnut costs 20 cents, he can buy 3.

3 0
2 years ago
Let Y denote a geometric random variable with probability of success p. a Show that for a positive integer a, P(Y &gt; a) = qa .
lakkis [162]

Answer:

a) For this case we can find the cumulative distribution function first:

F(k) = P(Y \leq k) = \sum_{k'=1}^k P(Y =k')= \sum_{k'=1}^k p(1-p)^{k'-1}= 1-(1-p)^k

So then by the complement rule we have this:

P(Y>a) = 1-F(a)= 1- [1-(1-p)^a]= 1-1 +(1-p)^a = (1-p)^a = q^a

b) P(Y>a)= q^a

P(Y>b) = q^b

So then we have this using independence:

P(Y> a+b) = q^{a+b}

We want to find the following probability:

P(Y> a+b |Y>a)

Using the definition of conditional probability we got:

P(Y> a+b |Y>a)= \frac{P(Y> a+b \cap Y>a)}{P(Y>a)} = \frac{P(Y>a+b)}{P(Y>a)} = \frac{q^{a+b}}{q^a} = q^b = P(Y>b)

And we see that if a = 2 and b=5 we have:

P(Y> 2+5 | Y>2) = P(Y>5)

c) For this case we use independent identical and with the same distribution experiments.

And the result for part b makes sense since we are interest in find the probability that the random variable of interest would be higher than an specified value given another condition with a value lower or equal.

Step-by-step explanation:

Previous concepts

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

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

If we define the random of variable Y we know that:

Y\sim Geo (1-p)

Part a

For this case we can find the cumulative distribution function first:

F(k) = P(Y \leq k) = \sum_{k'=1}^k P(Y =k')= \sum_{k'=1}^k p(1-p)^{k'-1}= 1-(1-p)^k

So then by the complement rule we have this:

P(Y>a) = 1-F(a)= 1- [1-(1-p)^a]= 1-1 +(1-p)^a = (1-p)^a = q^a

Part b

For this case we can use the result from part a to conclude that:

P(Y>a)= q^a

P(Y>b) = q^b

So then we have this assuming independence:

P(Y> a+b) = q^{a+b}

We want to find the following probability:

P(Y> a+b |Y>a)

Using the definition of conditional probability we got:

P(Y> a+b |Y>a)= \frac{P(Y> a+b \cap Y>a)}{P(Y>a)} = \frac{P(Y>a+b)}{P(Y>a)} = \frac{q^{a+b}}{q^a} = q^b = P(Y>b)

And we see that if a = 2 and b=5 we have:

P(Y> 2+5 | Y>2) = P(Y>5)

Part c

For this case we use independent identical and with the same distribution experiments.

And the result for part b makes sense since we are interest in find the probability that the random variable of interest would be higher than an specified value given another condition with a value lower or equal.

8 0
2 years ago
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