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Sonja [21]
4 years ago
6

Can someone help me with this question? It's a simple question compared to all the other questions on this app. Many appreciatio

ns!!

Mathematics
1 answer:
IgorLugansk [536]4 years ago
5 0
Do 25×6
then divide that by 160
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Plz help ill mark brainlist
spin [16.1K]

Answer:

ITS (A)

Step-by-step explanation:

PLEASE MARK ME AS BRAINLIST

8 0
3 years ago
For what values of fxis * x ^ 2 - 36 = 5x true?
jeka57 [31]

Answer:

Step-by-step explanation:

x ² - 36 = 5x

x²-5x-36=0

a=1  b= -5  c=-36

calculate the discriminent b²-4ac ......continu

5 0
4 years ago
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Find The Value x ~<br><img src="https://tex.z-dn.net/?f=%20%5Clarge%20%5Csf%20%5Cunderline%20%5Cbold%7B4x%20%2B%2026%20%3D%2016%
Nikitich [7]

The value of x in the algebraic equation is: -5/2.

<h3>How do you Find the Value of a Variable in an Algebraic Equation?</h3>

Given an algebraic equation, to find the unknown value of x, solve by isolating x in the equation.

Given:

4x + 26 = 16

Subtract 26 from both sides

4x = 16 - 26

4x = -10

Divide both sides by 4

x = -10/4

x = -5/2

Therefore, the value of x in the algebraic equation is: -5/2.

Learn more about algebraic equation on:

brainly.com/question/2164351

6 0
3 years ago
Read 2 more answers
Plz help answer if you know!
Musya8 [376]

Answer:

20,500

Step-by-step explanation:

2,500 * 5 = 12,500

33,000 - 12,500 = 20,500

8 0
3 years ago
Read 2 more answers
The weights of the chocolate in Hershey Kisses are normally distributed with a mean of 4.5338 g and a standard deviation of 0.10
dusya [7]

The distribution is not a standard normal distribution.

We know that the variable (X) usually obeys and follows a standard normal distribution provided that the summation of (X) is zero and the variance of the random variable V(X) = 1.

Mathematically:

E(X) = 0 and V(X) = 1

Given that;

  • the mean which follows a normal distribution (X) = 4.5338 g, and
  • the standard deviation SD(X) = 0.1039 g

As such, the distribution is not a standard normal distribution.

Therefore, we can conclude that the distribution is not a standard normal distribution.

Learn more about standard normal distribution here:

brainly.com/question/11876263?referrer=searchResults

7 0
3 years ago
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