Can someone help me please

Give the equation for each of the following statements.

emmasim [6.3K]

Answer:

Step-by-step explanation:

1. Y varies jointly as R and T

Y = k*R*T

Y = kRT

Where,

k = constant of proportionality

2. B is directly proportional to the square of A and inversely proportional to the cube of C.

B = k(A²) / C³

B = kA²/C³

Where,

k = constant of proportionality

4 0

3 years ago

TUI

anygoal [31]

Answer:

B and C

When you divide 64000 by 2 and 480000 by 15, they all equal 32000.

4 0

4 years ago

The table shows the hourly cookie sales by students in grades 7 and 8 at the school's annual bake sale. Grade 7 Grade 8 20 21 15

Naddika [18.5K]

Answer:

1. The interquartile range for the grade 7 data is 6.

2. The interquartile range for the grade 8 data is 6.

3. The difference of the medians of the two data sets is 2.

4. The difference is about 1/3 times the interquartile range of either data set.

Step-by-step explanation:

The hourly cookie sales by students in grades 7 and 8 at the school's annual bake sale is given by the following table.

Grade 7 Grade 8

20 21

15 29

30 14

24 19

18 24

21 25

The data set for grade 7 is

20, 15, 30, 24, 18, 21

Arrange the data in ascending order.

15, 18, 20, 21, 24, 30

Divide the data in four equal parts.

(15), 18, (20), (21), 24,( 30)

$Q_1=18, Median=\frac{20+21}{2}=20.5, Q_3=24$

The interquartile range for the grade 7 data is

$IQR=Q_3-Q_1=24-18=6$

Therefore the interquartile range for the grade 7 data is 6.

The data set for grade 8 is

21, 29, 14, 19, 24, 25

Arrange the data in ascending order.

14, 19, 21, 24, 25, 29

Divide the data in four equal parts.

(14), 19, (21), (24), 25, (29)

$Q_1=19, Median=\frac{21+24}{2}=22.5, Q_3=25$

The interquartile range for the grade 8 data is

$IQR=Q_3-Q_1=25-19=6$

Therefore the interquartile range for the grade 8 data is 6.

The difference of the medians of the two data sets is

$D=22.5-20.5=2$

Therefore the difference of the medians of the two data sets is 2.

Let the difference is about x times the interquartile range of either data set.

The IQR of each data is 6.

$D=x(IQR)$

$2=x(6)$

$\frac{2}{6}=x$

$\frac{1}{3}=x$

Therefore the difference is about 1/3 times the interquartile range of either data set.

8 0

3 years ago

Please help, fast‍♀️‍♀️

kramer

Inequalities are used to express unequal expressions.

The inequalities from the word problems are:

$\mathbf{m - 3.5 \le -2}$ .
$\mathbf{0 \ge 2x + 1}$ .
$\mathbf{-\frac 12 \ge 2k - 4}$

The statements from the inequalities are:

-4 is not a solution to $\mathbf{x + 8 < -3}$
-6 is not a solution to $\mathbf{10 \le 3 - m}$
-1 is not a solution to $\mathbf{-3x \le -12.5}$

Graph b represents $\mathbf{x > -7}$

<h3>The word problems</h3>

<u>1. A number minus 3.5 is less than or equal to -2</u>

The statement can be broken down into the following expressions

$\mathbf{A\ number\ minus\ 3.5 \to m - 3.5}$

$\mathbf{less\ than\ or\ equal\ to\ -2 \to \le -2}$

So, when the expressions are brought together, we have:

$\mathbf{m - 3.5 \le -2}$

<u>2. Zero is greater than or equal to twice a number x plus 1</u>

The statement can be broken down into the following expressions

$\mathbf{Zero\ is\ greater\ than\ or\ equal\ to \to 0 \ge }$

$\mathbf{twice\ a\ number\ x\ plus\ 1\ \to 2x + 1}$

So, when the expressions are brought together, we have:

$\mathbf{0 \ge 2x + 1}$

<u />

<u>3. -1/2 is at least twice a number k minus 4</u>

The statement can be broken down into the following expressions

$\mathbf{-\frac 12\ is\ at\ least \to -\frac 12 \ge }$

$\mathbf{twice\ a\ number\ k\ minus\ 4\ \to 2k - 4}$

So, when the expressions are brought together, we have:

$\mathbf{-\frac12 \ge 2k - 4}$

None of the options is correct

<h3>The solutions</h3>

<u>4. Tell whether -4 is a solution to x + 8 < -3</u>

We have:

$\mathbf{x + 8$

Subtract 8 from both sides

$\mathbf{x + 8 - 8$

$\mathbf{x$

The above inequality means that:

-4 is not a solution, because -4 is greater than -11

<u>5. Tell whether -6 is a solution to 10 <= 3 - m</u>

We have:

$\mathbf{10 \le 3 - m}$

Subtract 3 from both sides

$\mathbf{10 -3\le 3 - 3 - m}$

$\mathbf{7 \le - m}$

Multiply both sides by -1 (the inequality sign changes)

$\mathbf{-7 \ge m}$

Make m the subject

$\mathbf{m \le -7}$

The above inequality means that:

-6 is not a solution, because -6 is greater than -7

<u>6. Tell whether -1 is a solution to -3x <= -12.5</u>

We have:

$\mathbf{-3x \le -12.5}$

Divide both sides by -3 (the inequality sign changes)

$\mathbf{x \ge 4\frac16}$

The above inequality means that:

<em>x is greater than or equal to </em> $\mathbf{4\frac16}$ <em />

-1 is not a solution, because -1 is less than $\mathbf{4\frac16}$ <em />

<h3>The graph</h3>

The inequality is given as: $\mathbf{x > -7}$

The less than sign (>) means that:

The graph would use an open circle
The arrow must point to the right

Only graph b satisfies this condition

Hence, the graph of $\mathbf{x > -7}$ is graph b