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mixer [17]
3 years ago
7

What is 20% of 45? 15 points

Mathematics
2 answers:
vladimir2022 [97]3 years ago
5 0
The correct answer is 9. If you want it explained go here -->


https://www.mathpapa.com/algcalc2/?utm_expid=69051716-33.zMO_92bXSGuMCLhKsWRggg.3&utm_referrer=https...

hope this helps!
seropon [69]3 years ago
3 0
I believe it is 9 :)
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BE is the midsegment of triangle ACD. The value of x is:
Anna35 [415]
The midsegment theorem says that the midsegment is half the third side and parallel as well. 20*2 is 40 so x = 35
6 0
3 years ago
The graph shows a car's value as a function of its age.
otez555 [7]
The car was worth $10,000.

Hope this helps!
3 0
3 years ago
Read 2 more answers
Need help <br> please solve<br><br> 2 log39
Alexeev081 [22]
You just need a calculator.
The answer is 3.1821

Source: A caluculator
4 0
4 years ago
Why does the equation (x-4)^2-28=8 have two solutions?
Aliun [14]
<h3>Answer:</h3>

For your first answer it is because they are two different ways to solve the equation

<h3>Step-by-step explanation:</h3>

For Example!

<h2>Solution 1:</h2>

(x-4)² – 28 = 8The 8 is positive making it a different equation (this is like absolute value). (I am assuming you know the answer to the probelm)

<h2>Solution 2:</h2>

(x-4)² – 28 = -8

The 8 could be negative meaning that when you add the 28 to the right side it is -8+28 which will be a negative!

<h3>I hope this helps</h3><h2>IF IT DOES HELP PLEASE GIVE IT A BRAINLIEST!</h2>
4 0
2 years ago
Read 2 more answers
7 in 10 auto accidents involve a single vehicle. Suppose 14 accidents are randomly selected. (Round your answers to five decimal
Tpy6a [65]

Answer:

Step-by-step explanation:

Assuming a binomial distribution for the number of auto accidents. If 7 in 10 auto accidents involve a single vehicle, the probability, p = 7/10 = 0.7

Then the probability that the accident involved multiple vehicles is

q = 1 - p = 1 - 7/10 = 3/10 = 0.3

Since 14 accidents are randomly selected, n = 14

The formula for binomial distribution is expressed as

P(x = r) = nCr × q^(n - r) × p^r

a) we want to determine P(x = 4) =

P(x = 4) = 14C4 × 0.3^(14 - 4) × 0.7^4

P(x = 4) = 0.00142

b) we want to find P(x lesser than or equal to 4) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P( x = 4)

P(x = 0) = 14C0 × 0.3^(14 - 0) × 0.7^0 = 0.00000004783

P(x = 1) = 14C1 × 0.3^(14 - 1) × 0.7^1 = 0.00011

P(x = 2) = 14C2 × 0.3^(14 - 2) × 0.7^2 = 0.00002

P(x = 3) = 14C3 × 0.3^(14 - 3) × 0.7^3 = 0.00022

P(x = 4) = 0.00142

P(x lesser than or equal to 4) = 0.00000004783 + 0.00011 + 0.00002 + 0.00022 + 0.00142 = 0.00177

c) p = 0.7, q = 0.3

P(x = 5)= 14C5 × 0.7^(14 - 5) × 0.3^5 = 0.19631

6 0
3 years ago
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