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3 years ago

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Mathematics

1 answer:

USPshnik [31]3 years ago

4 0

Answer:

$17,500 at 8%
$17,500 at 14%

Step-by-step explanation:

The fraction that needs to be invested at the higher rate is ...

(11% - 8%)/(14% -8%) = 3%/6% = 1/2

Half the money should be invested at each of the rates:

$17,500 for 8% return

$17,500 for 14% return.

_____

If you let x represent the amount at the higher rate, then the total return is ...

14%x + 8%(35000 -x) = 11%(35000)

(14% -8%)x = (11% -8%)(35000) . . . . subtract 8%·35000

x = (11% -8%)/(14% -8%)×35000 . . . . divide by the coefficient of x

Note that this formula is exactly the one we started with, above.

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Determine the infinite limit. lim x→−9− x + 8 x + 9

Serga [27]

If given
$\lim_{x \to \ -9} (-x+8x+9)$
then
$\lim_{x \to \ -9} (-x+8x+9)=-54.$

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3 years ago

Calling All Mathematicians Please Please Please Help On This Question!!!!!!!!!!!!!!

Fiesta28 [93]

Use the distance formula, AB=√[(1-(-9)]²+[(-1-(-3)]=√[100+4]=√104
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3 years ago

Table show step by step how do you simplify the algebraic expression 3(x-5)+4x. <br>Justify step 4

ale4655 [162]

<h3>Answer: Commutative Property (choice C)</h3>

===================================================

Explanation:

Here are all of the justifications for each step

Step 1. None needed. This is the original expression
Step 2. Distributive property
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6 0

3 years ago

Please help me if you don't mind work out the problem I'll give you brainly

svp [43]

Answer:

C) Izzy

Step-by-step explanation:

Justin typed 90 words in 2 minute:

work

405/9=45 words a minute

45*2=90

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420/10=42 words a minute

42*2=84

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588/12= 49 words a minute

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I hope this helps :)

6 0

3 years ago

Read 2 more answers

The probability that two people have the same birthday in a room of 20 people is about 41.1%. It turns out that

salantis [7]

Answer:

a) Let X the random variable of interest, on this case we know that:

$X \sim Binom(n=20, p=0.411)$

This random variable represent that two people have the same birthday in just one classroom

b) We can find first the probability that one or more pairs of people share a birthday in ONE class. And we can do this:

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

And we can find the individual probability:

$P(X=0) = (20C0) (0.411)^0 (1-0.411)^{20-0}=0.0000253$

And then:

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

And since we want the probability in the 3 classes we can assume independence and we got:

$P= 0.99997^3 = 0.9992$

So then the probability that one or more pairs of people share a birthday in your three classes is approximately 0.9992

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

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".

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:

$nCx=\frac{n!}{(n-x)! x!}$

Solution to the problem

Part a

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

$X \sim Binom(n=20, p=0.411)$

This random variable represent that two people have the same birthday in just one classroom

Part b

We can find first the probability that one or more pairs of people share a birthday in ONE class. And we can do this:

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

And we can find the individual probability:

$P(X=0) = (20C0) (0.411)^0 (1-0.411)^{20-0}=0.0000253$

And then:

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

And since we want the probability in the 3 classes we can assume independence and we got:

$P= 0.99997^3 = 0.9992$

So then the probability that one or more pairs of people share a birthday in your three classes is approximately 0.9992

4 0

4 years ago