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Travka [436]
3 years ago
14

Simplify: (–7x – 5x4 + 5) – (–9x4 – 5 – 9x). 4x4 + 2x + 2 + 10 –16x4 + 8x + 10 –16x4 – 8x 4x4 + 2x + 10

Mathematics
1 answer:
fenix001 [56]3 years ago
5 0

Answer:

=−4.4x^8+3.6x^5−35x^4+5x+37

Step-by-step explanation:

please mark me brainliest!

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Given the function g(x) = –x^2+ 4x + 9, determine the average rate of change of
Ghella [55]

Answer:

9

Step-by-step explanation:

4 0
3 years ago
Yuson used her calculator to solve the equation 4/5x - 8 = 3. She entered the following on her screen and got an incorrect answe
Firlakuza [10]

Answer:

55/4

Explanation:

Let's solve your equation step-by-step.

4 /5x−8=3

Step 1: Add 8 to both sides.

4 /5x−8+8=3+8

4 /5 x = 11

Step 2: Multiply both sides by 5/4.

( 5 /4 )*( 4 /5 x)=( 5 /4 )*(11)

x=  55/4

Answer:

x=

55 /4

Hope that helps :)

 

4 0
3 years ago
Read 2 more answers
Write to explain how you can break apart 16*34 into four simpler multiplication problems
____ [38]
16 is 10 +6, and 34 is 30+4
10*30=300
10*4=40
6*30=180
4*6=24
add them up: 300+40+180+24=544

Using an array to break
<span>https://www.lake.k12.fl.us/cms/lib05/FL01000799/Centricity/Domain/7908/Topic%204%20Centers%204th%20g...</span>


8 0
3 years ago
A person invest $10,000 into a bank the bank pays 4.75% interest compounded semi annually. To the nearest 10th of a year, how lo
Mama L [17]

Answer:

T is 13.9 years to the nearest 10th of a year

Step-by-step explanation:

In this question, we are to calculate the number of years at which someone who invests a particular amount will have a particular amount based on compound interest.

To calculate the number of years, what we do is to use the compound interest formula.

Mathematically,

A = P(1+ r/n) ^nt

Where A is the final amount after compounding all interests which is $19,200 according to the question

P is the initial amount invested which is $10,000 according to the question

r is the rate which is 4.75% according to the question = 4.75/100 = 0.0475

n is the number of times per year in which interest is compounded. This is 2 as interest is compounded semi-annually

t= ?

we plug these values;

19200 = 10,000(1+0.0475/2)^2t

divide through by 10,000

1.92 = (1+0.02375)^2t

1.92 = (1.02375)^2t

We find the log of both sides

log 1.92 = log [(1.02375)^2t)

log 1.92 = 2tlog 1.02375

2t = log 1.92/log 1.02375

2t = 27.79

t = 27.79/2

t = 13.89 years

The question asks to give answer to the nearest tenth of a year and thus t = 13.9 years

7 0
3 years ago
Let U1, ..., Un be i.i.d. Unif(0, 1), and X = max(U1, ..., Un). What is the PDF of X? What is EX? Hint: Find the CDF of X first,
Kryger [21]

Answer:

E(X)= n \int_{0}^1 x^n dx = n [\frac{1}{n+1}- \frac{0}{n+1}]=\frac{n}{n+1}

Step-by-step explanation:

A uniform distribution, "sometimes also known as a rectangular distribution, is a distribution that has constant probability".

We need to take in count that our random variable just take values between 0 and 1 since is uniform distribution (0,1). The maximum of the finite set of elements in (0,1) needs to be present in (0,1).

If we select a value x \in (0,1) we want this:

max(U_1, ....,U_n) \leq x

And we can express this like that:

u_i \leq x for each possible i

We assume that the random variable u_i are independent and P)U_i \leq x) =x from the definition of an uniform random variable between 0 and 1. So we can find the cumulative distribution like this:

P(X \leq x) = P(U_1 \leq 1, ...., U_n \leq x) \prod P(U_i \leq x) =\prod x = x^n

And then cumulative distribution would be expressed like this:

0, x \leq 0

x^n, x \in (0,1)

1, x \geq 1

For each value x\in (0,1) we can find the dendity function like this:

f_X (x) = \frac{d}{dx} F_X (x) = nx^{n-1}

So then we have the pdf defined, and given by:

f_X (x) = n x^{n-1} , x \in (0,1)  and 0 for other case

And now we can find the expected value for the random variable X like this:

E(X) =\int_{0}^1 s f_X (x) dx = \int_{0}^1 x n x^{n-1}

E(X)= n \int_{0}^1 x^n dx = n [\frac{1}{n+1}- \frac{0}{n+1}]=\frac{n}{n+1}

6 0
3 years ago
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