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d1i1m1o1n [39]
3 years ago
12

5a - 2b -3 + 2b - 6a simplify the expression and then write an equivalent expression. Please anyone help me it's my first questi

on! :3
Mathematics
2 answers:
Anastasy [175]3 years ago
6 0

Answer:

-(a + 3)

Step-by-step explanation:

5a - 2b -3 + 2b - 6a

5a - 6a - 2b + 2b - 3

-a - 3

-(a + 3)

svet-max [94.6K]3 years ago
3 0

Step-by-step explanation:

simplify the expression into like terms

so 5a-2b-3+2b-6a (add all a and b together)

this will make

-a-3

I know how to do the equivalent sorry :(

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emmainna [20.7K]
I'm assuming that subtraction sign is mean to be an equals sign. If that is true than the answer is 403.75. 
4 0
3 years ago
Solve the problem. Use the Central Limit Theorem.The annual precipitation amounts in a certain mountain range are normally distr
bazaltina [42]

Answer:

0.8944 = 89.44% probability that the mean annual precipitation during 25 randomly picked years will be less than 112 inches.

Step-by-step explanation:

To solve this question, we use the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 109.0 inches, and a standard deviation of 12 inches.

This means that \mu = 109, \sigma = 12

Sample of 25.

This means that n = 25, s = \frac{12}{\sqrt{25}} = 2.4

What is the probability that the mean annual precipitation during 25 randomly picked years will be less than 112 inches?

This is the p-value of Z when X = 112. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{112 - 109}{2.4}

Z = 1.25

Z = 1.25 has a p-value of 0.8944.

0.8944 = 89.44% probability that the mean annual precipitation during 25 randomly picked years will be less than 112 inches.

7 0
3 years ago
What is 27 percent of 190 bushels?
Neko [114]
Multiply 190 by 0.27 and you should get your answer
3 0
3 years ago
Factor the expression.
nalin [4]
6x^2+14x+4
First factor out all numerical factors (=2 in this case)
2(3x^2+7x+2)
look for m,n such that m*n=3*2, m+n=7 => m=6, n=1
2(3x^2+6x  + 1x+2)
Factor 3x^2+6x into 3x(x+2)
2( 3x(x+2)+1(x+2) )
factor out common factor (x+2)
2(x+2)(3x+1)
=>
6x^2+14x+4=2(x+2)(3x+1)
8 0
3 years ago
Find the values of p and q in the arithmetic progression-12,p,q,18​
Mashutka [201]

Answer:

p=14

q=16

Step-by-step explanation:

if,  

           a,b,c,d..........  (Arithmetic progression)

           b-a=c-b=d-c

      The differences of the adjacent elements are equal.

6 0
2 years ago
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