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Talja [164]
3 years ago
8

How many solutions does this equation have ? 2(2x-1)=(x+1)+3(x-1)​

Mathematics
1 answer:
Lunna [17]3 years ago
6 0
Yep his right and it is shown to me good
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Help me plz on the question thx
GrogVix [38]
1/2 is the answer. I hope this helps
4 0
3 years ago
How do you simplify 3(j-2k)+4 ?​
masha68 [24]

Answer: 3j-6k+4

Step-by-step explanation:

First you have to distribute the 3 to the j and the -2k since they are both in the parentheses. So it would look like (3*j) +(-2k*3)+4 and if you simplify that it is 3j-2k+4. After that there are no like terms so that is your answer.

5 0
2 years ago
Life Expectancies In a study of the life expectancy of people in a certain geographic region, the mean age at death was years an
Sphinxa [80]

Answer:

The probability that the mean life expectancy of the sample is less than X years is the p-value of Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}, in which \mu is the mean life expectancy, \sigma is the standard deviation and n is the size of the sample.

Step-by-step explanation:

To solve this question, we need to understand 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.

We have:

Mean \mu, standard deviation \sigma.

Sample of size n:

This means that the z-score is now, by the Central Limit Theorem:

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

Find the probability that the mean life expectancy will be less than years.

The probability that the mean life expectancy of the sample is less than X years is the p-value of Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}, in which \mu is the mean life expectancy, \sigma is the standard deviation and n is the size of the sample.

8 0
2 years ago
The difference of two numbers is 90 and their quotient is 10
bija089 [108]
To solve this, set up two equations using the information you're given. Let's call our two numbers a and b:
1) D<span>ifference of two numbers is 90
a - b (difference of two numbers) = 90

2) The quotient of these two numbers is 10
a/b (quotient of the two numbers) = 10


Now you can solve for the two numbers.
1) Solve the second equation for one of the variables. Let's solve for a:
a/b = 10
a = 10b

2) Plug a =10b into the first equation and solve for the value of b:
a - b = 90
10b - b = 90
9b = 90
b = 10

3) Using b = 10, plug it back into one of the equations to find the value of a. I'll plug it back into the first equation:
a - b = 90
a - 10 = 90
a = 100

-------

Answer: The numbers are 100 and 10</span>
3 0
3 years ago
What is x/5+7=16 gonna be?
Margarita [4]

x=45

Step-by-step explanation:

x/5+7-7=16-7

5/1×x/5=9×5/1

x=45

6 0
3 years ago
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