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3 years ago

Please give me the correct answer

Mathematics

2 answers:

11Alexandr11 [23.1K]3 years ago

6 0

Answer:

Step-by-step explanation:

i think the 4th is the answer

Lyrx [107]3 years ago

5 0

Bottom right is correct answer

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Ariyonne claims (3,6) that is the point of intersection of the lines y=4x-2 and y=1/2x+5 Is she correct? How do you know?

yan [13]

Answer:

The point of intersection of the lines is $(2,6)$ .

Thus, Ariyonne's claim is incorrect.

Step-by-step explanation:

Given equations:

1) $y=4x-2$

2) $y=\frac{1}{2}x+5$

To find the point of intersection, we need to solve the given system of equations.

In order to solve the system we will use substitution method.

We will substitute value of $y$ from first equation in the second equation and solve for $x$

So, we have

$4x-2=\frac{1}{2}x+5$

Multiplying both sides by 2 to remove fraction.

$2\times(4x-2)=2\times(\frac{1}{2}x+5)$

Using distribution.

$8x-4=x+10$

Subtracting both sides by $x$

$8x-4-x=x+10-x$

$7x-4=10$

Adding 4 to both sides.

$7x-4+4=10+4$

$7x=14$

Dividing both sides by 7.

$\frac{7x}{7}=\frac{14}{7}$

$x=2$

We can find value of $y$ by plugging in $x=2$ in the first equation.

$y=4(2)-2$

$y=8-2$

$y=6$

So, the point of intersection of the lines is $(2,6)$ .

Thus, Ariyonne's claim is incorrect.

3 0

3 years ago

6. Which of the following function pairs are inverses?

Papessa [141]

Answer: f(x) = 2x - 3 and f(x) = -4x

3 0

4 years ago

Read 2 more answers

You select a sample of 50 scores from a population of 2,000 scores. You compute the range and standard deviation on the sample o

nalin [4]

Answer:

The measure of dispersion which is likely to vary most between your first and second samples is the range.

Step-by-step explanation:

The range and standard deviation of a data are measures of dispersion, i.e. they measure the degree to which the data is dispersed.

The formula to compute the range is:

$Range=X_{max}-X_{min}$

The formula to compute the sample standard deviation is:

$s=\sqrt{\frac{1}{n-1}\sum (X-\bar X)^{2} }$

The sample size is: <em>n</em> = 50.

As the sample size is large (n = 50 > 30) the sample standard deviation (s) can be used to approximate the population standard deviation (σ). Thus, whatever the sample values be both the standard deviations can be used to approximate the population standard deviation. Hence, it can be said that both the sample standard deviations are approximately equal.

Whereas the range of the two samples are very likely to vary since it is based on the minimum and maximum value of the data. For both the samples the minimum and maximum value may be differ. Thus providing different range values.

Thus, the measure of dispersion which is likely to vary most between your first and second samples is the range.

5 0

4 years ago

"You are dating Moon rocks based on their proportions of uranium-238 (half-life of about 4.5 billion years) and its ultimate dec

Olenka [21]

Step-by-step explanation:

Initial mass of the isotope = x

Time taken by the sample to decay its mass by 41% = t

Formula used :

$N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}$