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3 years ago

There are 58 girls in a junior class of 146 students. find the ratio of girls

1 answer:

7 0

For ever 146 Students, there are 58 females.

73:28 is the simplified ratio, 28 being the female students!

Hope this helps!

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The stopping distance d of an automobile is directly proportional to the square of its speed s. On one road, a car requires 75 f

Anton [14]

Answer:

The car requires 192 feet to stop from a speed of 48 miles per hour on the same road

Step-by-step explanation:

Direct proportion means that two quantities increase or decrease in the same ratio

If y is directly proportional to x (y ∝ x) , then $\frac{y_{1}}{y_{2}}=\frac{x_{1}}{x_{2}}$ <em>OR</em> y = k x, where k is the constant of proportionality

∵ The stopping distance d of an automobile is directly

proportional to the square of its speed s

- That means d ∝ s²

∴ $\frac{d_{1}}{d_{2}}=\frac{(s_{1})^{2}}{(s_{2})^{2}}$

∵ A car requires 75 feet to stop from a speed of 30 miles per hour

∴ d = 75 feet

∴ s = 30 miles/hour

- Change the mile to feet

∵ 1 mile = 5280 feet

∴ 30 miles/hour = 30 × 5280 = 158400 feet/hour

∵ The car require to stop from a speed of 48 miles per hour

on the same road

- Change the mile to feet

∴ 48 miles/hour = 48 × 5280 = 253440 feet/hour

∵ $\frac{d_{1}}{d_{2}}=\frac{(s_{1})^{2}}{(s_{2})^{2}}$

- Substitute the values of $d_{1}$ by 75 feet, $s_{1}$ by 158400 feet/hour

and $s_{2}$ by 253440 feet/hour

∴ $\frac{75}{d_{2}}=\frac{(158400)^{2}}{(253440)^{2}}$

∴ $\frac{75}{d_{2}}=\frac{25}{64}$

- By using cross multiplication

∴ 25 × $d_{2}$ = 75 × 64

- Divide both sides by 25

∴ $d_{2}$ = 192 feet

The car requires 192 feet to stop from a speed of 48 miles per hour on the same road

4 0

2 years ago

Find the missing side length, indicate whether the side lengths form a Pythagorean triple, and identify the correct explanation.

saw5 [17]

Step-by-step explanation:

Pythagorean Formula = a2 + b2 = C

which is a = Side of triangle

b = Side of Triangle

c = Hypotenuse

4 0

2 years ago

Read 2 more answers

Use matrices to solve the system of equations if possible. Use Gaussian elimination with back substitution or gauss Jordan elimi

CaHeK987 [17]

In matrix form, the system is given by

$\begin{bmatrix} -1 & 1 & -1 \\ 2 & -1 & 1 \\ 3 & 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -20 \\ 29 \\ 29 \end{bmatrix}$

I'll use G-J elimination. Consider the augmented matrix

$\left[ \begin{array}{ccc|c} -1 & 1 & -1 & -20 \\ 2 & -1 & 1 & 29 \\ 3 & 2 & 1 & 29 \end{array} \right]$

• Multiply through row 1 by -1.

$\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 2 & -1 & 1 & 29 \\ 3 & 2 & 1 & 29 \end{array} \right]$

• Eliminate the entries in the first column of the second and third rows. Combine -2 (row 1) with row 2, and -3 (row 1) with row 3.

$\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 5 & -2 & -31 \end{array} \right]$

• Eliminate the entry in the second column of the third row. Combine -5 (row 2) with row 3.

$\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 3 & 24 \end{array} \right]$

• Multiply row 3 by 1/3.

$\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 1 & 8 \end{array} \right]$

• Eliminate the entry in the third column of the second row. Combine row 2 with row 3.

$\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 8 \end{array} \right]$

• Eliminate the entries in the second and third columns of the first row. Combine row 1 with row 2 and -1 (row 3).

$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 9 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 8 \end{array} \right]$

Then the solution to the system is

$\boxed{x=9, y=-3, z=8}$

If you want to use G elimination and substitution, you'd stop at the step with the augmented matrix

$\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 1 & 8 \end{array} \right]$

The third row tells us that $z=8$ . Then in the second row,

$y-z = -11 \implies y=-11 + 8 = -3$

and in the first row,

$x-y+z=20 \implies x=20 + (-3) - 8 = 9$

5 0

1 year ago

Which is an exponential decay function f (x)=3/4 (7/4)x f (x)=2/3 (4/5)-x. F (x)=3/2 (8/7)-x f (x)=1/3 (9/2)x

dsp73

The given functions are
$1.\, f(x)= \frac{3}{4}( \frac{7}{4})^{x} \\ \\2.\,f(x)= \frac{2}{3}( \frac{4}{5} ) ^{-x}\\\\3.\, f(x)= \frac{3}{2} ( \frac{8}{7} )^{-x}\\\\4.\,f(x)= \frac{1}{3} ( \frac{9}{2} )^{x}$

Evaluate the functions.
1. Because 7/4 > 1 and the exponent is positive,
the function does not decay.
2.Because 4/5 < 1 and the exponent is negative,
the function does not decay.
3. Because 8/7 > 1 and the exponent is negative,
the function decays.
4. Because 9/2 > 1 and the exponent is positive,
the function does not decay.

A composite plot the functions verifies the answer.

Answer: $f(x)= \frac{3}{2} ( \frac{8}{7} )^{-x}$

7 0

3 years ago

Read 2 more answers

Video game system and several games are sold for $664. The cost of the games is 3 times as much is a cost in system. Find the co

Ede4ka [16]

Answer:

games=$483

system=$161

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers