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Leto [7]
3 years ago
8

Please help me I need help for this

Mathematics
2 answers:
Likurg_2 [28]3 years ago
7 0
4a is the greatest common factor
amid [387]3 years ago
5 0

Answer:

the greatest common factor is 4a

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Solve the simultaneous Equation using the two <br> methods<br>4m=n +7<br>3m+ 4n+9=0​
GrogVix [38]

Elimination method:

4m = n + 7

3m + 4n + 9 = 0

<em>First, let's get the equations in the same form.</em>

4m - n - 7 = 0

3m + 4n + 9 = 0

<em>Now let's make multiply the first equation by 4 so we can eliminate n.</em>

16m - 4n - 28 = 0

+3m + 4n + 9 = 0

<em>Now we can add the equations.</em>

16m + 3m - 4n + 4n - 28 + 9 = 0

19m + 0n - 19 = 0

19m - 19 = 0

19m = 19

<em>m = 1</em>

<em>Now we put m back into one (or both) of the original equations.</em>

4(1) = n + 7

4 = n + 7

<em>n = -3</em>

<em>If you plug m into the other equation, you get the same result.</em>

<em />

Substitution method:

4m = n + 7

3m + 4n + 9 = 0

<em>With this method, we plug one of the equations into the other one. I'm going to use m in the second equation as a substitute for m in the second equation.</em>

3m + 4n + 9 = 0

3m = -4n - 9

m = (-4/3)n - 3

<em>Now I can substitute the right side into the first equation like so:</em>

4[(-4/3)n - 3] = n + 7

(-16n)/3 - 12 = n + 7

(-16n)/3 = n + 19

-16n = 3(n + 19)

-16n = 3n + 57

0 = 16n + 3n + 57

0 = 19n + 57

0 = 19n/19 + 57/19

0 = n + 3

<em>-3 = n</em>

<em>And then we put that back into one of the original equations.</em>

4m = n + 7

4m = -3 + 7

4m = 4

<em>m = 1</em>

Hopefully you learned something from this.

5 0
3 years ago
The distribution of actual weights of 8-ounce wedges of cheddar cheese produced at a dairy is Normal with mean 8.1 ounces and st
Sonja [21]

Answer:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

\bar X \sim N(\mu=8.1, \frac{0.2}{\sqrt{10}}=0.063)

approximately Normal, mean 8.1, standard deviation 0.063.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variable weights of 8-ounce wedges of cheddar cheese produced at a dairy. We know from the problem that the distribution for the random variable X is given by:

X\sim N(\mu =8.1,\sigma =0.2)

We take a sample of n=10 . That represent the sample size.

What can we say about the shape of the distribution of the sample mean?

From the central limit theorem we know that the distribution for the sample mean \bar X is also normal and is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

\bar X \sim N(\mu=8.1, \frac{0.2}{\sqrt{10}}=0.063)

approximately Normal, mean 8.1, standard deviation 0.063.

3 0
4 years ago
Find f(-3) if f (x) =|x+5|
vesna_86 [32]

Answer:

the answer is two because the absolute value delete the negative in result not in question

7 0
2 years ago
2/3 k is greater than or equal to -12
yKpoI14uk [10]
There is a possibility that k could be negative 18 as -18 times 2/3 equals -12
8 0
3 years ago
Plot the image of point P under a reflection across the y-axis.
tigry1 [53]

Answer:

Why are you sending questions at 9:15 in the morning.

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
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