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2 years ago

Thank you so much, really.

Mathematics

1 answer:

Nataly_w [17]2 years ago

4 0

If a point is a solution to the system, then the point (x, y) coordinate will satisfy both the linear equation. Then the correct option is D.

<h3>What is the solution to the equation?</h3>

The distribution of weights to the variables involved that establishes the equilibrium in the calculation is referred to as a result.

A relationship between two or more parameters that, when shown on a graph, produces a linear model. The degree of the variable will be one.

If a point is a solution to the system, then the point (x, y) coordinate will satisfy both the linear equation.

Then the correct option is D.

More about the solution of the equation link is given below.

brainly.com/question/545403

#SPJ1

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Any help would be nice

mrs_skeptik [129]

Answer:

I think it's C.

Step-by-step explanation:

A. and D. don't make any sense so I assume they're incorrect. B. just seems wrong also by a small margin.

7 0

4 years ago

The ages of the Oscar-winners for best actor and actress for the years 1970 through 1990 are given below: Women: 34, 34,26,37,42

Artemon [7]

Answer:

See explanation below.

Step-by-step explanation:

Numerical way

We can calculate the mean and the standard deviation from a sample with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s= \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

We have the following dataset for women: 34,34,26,37,42,41,35,31,41,33,30,74,33,49,38,61,21,41,26,80,72

And the mean and deviation are:

$\bar X_F = 41.857$

$s_F = 16.399$

We can calculate the median ordering the data like this:

21,26,26,30,31,33,33,34,34,35,37,38,41,41,41,42,49,61,72,74,80

And the median would be on the 11 position of the data ordered, for this case is:

$median_F = 37$

$Minimum_F = 21 , Maximum_F = 80$

And the range is given by: $Range_F = 80-21=59$

We can also calculate the coefficient of variation given by:

$\hat{CV} =\frac{s}{\bar X} = \frac{16.399}{41.857}=0.392$

We have the following dataset for men:

43, 40,48,48,56,38,60,32,40,42,37,76,39,55,45,35,61,33,51,32,42

And the mean and deviation are:

$\bar X_M = 45.381$

$s_M = 11.218$

We can calculate the median ordering the data like this:

32, 32,33,35,37,38,39,40,40,42,42,43,45,48,48,51,55,56,60,61,76

And the median would be on the 11 position of the data ordered, for this case is:

$median_M = 42$

$Minimum_M = 32 , Maximum_F = 76$

And the range is given by: $Range_M = 76-32=44$

We can also calculate the coefficient of variation given by:

$\hat{CV} =\frac{s}{\bar X} = \frac{11.218}{45.381}=0.247$

We se that the mean for the male groups is higher than the mean for female group. The deviation is higher for the female group comapred to the male group. We have more variation for the female group as we can see on the range and the coeffcient of variation.

Graph

We can see the histograms for each group on the figures attached.

As we can see the histogram for males is skewed to the right with the median lower than the mean.

By the other hand the histogram for the female group seems to be skewed to the right also but with missing values on the interval 50-60, for this reason we have more variation compared to the male group.

7 0

4 years ago

write the equation of a line that is perpendicular to y=3z-2 and that passes through the point (-9,5)

MariettaO [177]

Y=-1/3+2 should be what you are looking for.

7 0

3 years ago

What types of quadrilaterals have congruent diagonals?

labwork [276]

Rectangle and Square

8 0

3 years ago

3. In an auditorium, 1/6 of the students are fifth graders, 1/3 are fourth graders, and 1/4 of the remaining students are second

Katen [24]

<h3>Further explanation</h3>

<u>Given:</u>

In an auditorium,

$\frac{1}{6}$ of the students are fifth graders,
$\frac{1}{3}$ are fourth graders, and
$\frac{1}{4}$ of the remaining students are second graders.

There are 96 students in the auditorium.

<u>Question:</u>

How many second graders are there?

<u>The Process:</u>

The least common multiple (LCM) of 3 and 6 is 6.

Let us draw the diagram of all students.

$96 \div 6 \ units = 16 \rightarrow \boxed{16}\boxed{16}\boxed{16}\boxed{16}\boxed{16}\boxed{16} = 96 \ students$

$\frac{1}{6}$ of the students are fifth graders, or 1 of 6 units above, that is $\boxed{16} = 16 \ students$
$\frac{1}{3} = \frac{2}{6}$ are fourth graders, or 2 of 6 units above, that is $\boxed{16}\boxed{16} = 32 \ students$

The remainder is 6 units - (1 unit + 2 units) = 3 units, that is $\boxed{16}\boxed{16}\boxed{16} = 48 \ students$ . Or, 96 - 16 - 32 = 48 students.

$\frac{1}{4}$ of the remaining students are second graders.

Let us count how many second graders are there.

$\boxed{ \ \frac{1}{4} \times 48 \ students = ? \ }$

Thus, there are 12 second-graders in the auditorium.

- - - - - - - - - -

Quick Steps

$\boxed{ \ \frac{1}{4} \times \bigg(1 - \frac{1}{6} - \frac{1}{3} \bigg) \times 96 = ? \ }$

$\boxed{ \ = \frac{1}{4} \times \bigg(\frac{6}{6} - \frac{1}{6} - \frac{2}{6} \bigg) \times 96 \ }$

$\boxed{ \ = \frac{1}{4} \times \frac{3}{6} \times 96 \ }$

$\boxed{ \ = \frac{3}{24} \times 96 \ }$

We crossed out 24 and 96.

$\boxed{ \ = 3 \times 4 \ }$

$\boxed{\boxed{ \ = 12 \ students \ }}$

<h3>Learn more</h3>

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Keywords: In an auditorium, 1/6 of the students, fifth graders, 1/3, fourth, 1/4, the remaining students, second, 96, how many, second graders

3 0

3 years ago

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