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klemol [59]
3 years ago
7

Simplified form of this expression 5z-8-4z-3z+6

Mathematics
2 answers:
kvasek [131]3 years ago
7 0

Answer:

-2z-2

Step-by-step explanation:

5z-8-4z-3z+6

5z-4z-3z-8+6

-2z-2

Margarita [4]3 years ago
5 0

Answer:

-2z -2

Step-by-step explanation:

5z-8-4z-3z+6

5z -4z - 3z +6 - 8

-2z -2

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Use the method of completing the square to transform the quadratic equation into the equation form (x + p)2 = q. 3 + x - 3x2 = 9
Triss [41]
To <span>transform the quadratic equation into the equation form (x + p)2 = q we shall proceed as follows:
3+x-3x^2=9
putting like terms together we have:
-3x^2+x=6
dividing through by -3 we get:
x^2-x/3=-2
but
c=(b/2a)^2
c=(-1/6)^2=1/36
thus the expression will be:
x^2-x/3+1/36=-2+1/36
1/36(6x-1)</span>²=-71/36

the answer is:
1/36(6x-1)²=-71/36
7 0
3 years ago
Please help! posted picture of question
o-na [289]
C. 5x - 2y = 16 is your answer

plug in each point into the equation.

(2, -3)

x = 2, y = -3

5(2) - 2(-3) = 16
10 + 6 = 16
16 = 16 (True)

(4,2)

x = 4, y = 2

5(4) - 2(2) = 16
20 - 4 = 16
16 = 16 (True)


hope this helps
4 0
3 years ago
Read 2 more answers
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
Answers ASAP plsssssss
ASHA 777 [7]

Answer:

Step-by-step explanation:the answer is option A as 1 • 3 =3 and 3+4=7

6 0
2 years ago
(20 Points) !!!<br><br> 5/8 times 4/5<br> Fastest get Brainliest
OleMash [197]
5/8 * 4/5
5*4     20      1
----- = ----- =  ---
8*5    40       2

5/8 * 4/5 = 1/2 or .5
4 0
3 years ago
Read 2 more answers
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