Ask question

Ask question

All categories

2 years ago

8

The front row in a movie theater has 23 seats. If you were asked to sit in the seat the occupied the median position, in which s

eat would you have to sit?

2 answers:

Natalka [10]2 years ago

8 0

You would have to sit in seat 12

Irina18 [472]2 years ago

6 0

Not 100% sure but I thing the 12th cuz median means middle so if you occupied the middle seat out of 23 then 11 +11 = 22 so the middle seat whould be the 12th

You might be interested in

Complete the table by writing the fraction as a decimal and as a percent

aniked [119]

Answer:

Decimal = 0.6

Percentage = 60%

Step-by-step explanation:

Let's simplify the fraction.

$\frac{120}{200} = \frac{60}{100} = 0.6$

Hence the Decimal is 0.6

To express 120/200 in percentage , we have to multiply the fraction with 100.

$\frac{120}{200} \times 100 = \frac{120}{2} = 60\%$

Hence the percentage value is 60%

7 0

3 years ago

The school production of Our Town' was a big success. For opening night. 434 tickets were sold. Students paid $4.00 each, while

Alex73 [517]

Answer:

286 students attended

148 non students attended

Step-by-step explanation:

Given

$Tickets = 434$

$Students = \$4.00$

$Non-Students = \$6.00$

$Total = \$2032.00$

Solving (a): Number of students

Represent students with S and non students with N

So:

$S + N = 434$ --- (1)

$4S + 6N = 2032$ --- (2)

Make N the subject of formula in (1)

$N = 434 - S$

Substitute 434 - S for N in (2)

$4S +6(434 - S) = 2032$

Open Bracket

$4S + 2604 - 6S = 2032$

Collect Like Terms

$4S - 6S = 2032 - 2604$

$-2S = -572$

Divide through by -2

$S = 286$

<em>Hence; 286 students attended</em>

Solving (b):

Recall that

$N = 434 - S$

$N = 434 - 286$

$N = 148$

<em>148 non students attended</em>

4 0

2 years ago

City A had a population of 10000 in the year 1990. City A’s population grows at a constant rate of 3% per year. City B has a pop

Georgia [21]

Answer:

City A and city B will have equal population 25years after 1990

Step-by-step explanation:

Given

Let

$t \to$ years after 1990

$A_t \to$ population function of city A

$B_t \to$ population function of city B

<u>City A</u>

$A_0 = 10000$ ---- initial population (1990)

$r_A =3\%$ --- rate

<u>City B</u>

$B_{10} = \frac{1}{2} * A_{10}$ ----- t = 10 in 2000

$A_{20} = B_{20} * (1 + 20\%)$ ---- t = 20 in 2010

Required

When they will have the same population

Both functions follow exponential function.

So, we have:

$A_t = A_0 * (1 + r_A)^t$

$B_t = B_0 * (1 + r_B)^t$

Calculate the population of city A in 2000 (t = 10)

$A_t = A_0 * (1 + r_A)^t$

$A_{10} = 10000 * (1 + 3\%)^{10}$

$A_{10} = 10000 * (1 + 0.03)^{10}$

$A_{10} = 10000 * (1.03)^{10}$

$A_{10} = 13439.16$

Calculate the population of city A in 2010 (t = 20)

$A_t = A_0 * (1 + r_A)^t$

$A_{20} = 10000 * (1 + 3\%)^{20}$

$A_{20} = 10000 * (1 + 0.03)^{20}$

$A_{20} = 10000 * (1.03)^{20}$

$A_{20} = 18061.11$

From the question, we have:

$B_{10} = \frac{1}{2} * A_{10}$ and $A_{20} = B_{20} * (1 + 20\%)$

$B_{10} = \frac{1}{2} * A_{10}$

$B_{10} = \frac{1}{2} * 13439.16$

$B_{10} = 6719.58$

$A_{20} = B_{20} * (1 + 20\%)$

$18061.11 = B_{20} * (1 + 20\%)$

$18061.11 = B_{20} * (1 + 0.20)$

$18061.11 = B_{20} * (1.20)$

Solve for B20

$B_{20} = \frac{18061.11}{1.20}$

$B_{20} = 15050.93$

$B_{10} = 6719.58$ and $B_{20} = 15050.93$ can be used to determine the function of city B

$B_t = B_0 * (1 + r_B)^t$

For: $B_{10} = 6719.58$

We have:

$B_{10} = B_0 * (1 + r_B)^{10}$

$B_0 * (1 + r_B)^{10} = 6719.58$

For: $B_{20} = 15050.93$

We have:

$B_{20} = B_0 * (1 + r_B)^{20}$

$B_0 * (1 + r_B)^{20} = 15050.93$

Divide $B_0 * (1 + r_B)^{20} = 15050.93$ by $B_0 * (1 + r_B)^{10} = 6719.58$

$\frac{B_0 * (1 + r_B)^{20}}{B_0 * (1 + r_B)^{10}} = \frac{15050.93}{6719.58}$

$\frac{(1 + r_B)^{20}}{(1 + r_B)^{10}} = 2.2399$

Apply law of indices

$(1 + r_B)^{20-10} = 2.2399$

$(1 + r_B)^{10} = 2.2399$ --- (1)

Take 10th root of both sides

$1 + r_B = \sqrt[10]{2.2399}$

$1 + r_B = 1.08$

Subtract 1 from both sides

$r_B = 0.08$

To calculate $B_0$ , we have:

$B_0 * (1 + r_B)^{10} = 6719.58$

Recall that: $(1 + r_B)^{10} = 2.2399$

So:

$B_0 * 2.2399 = 6719.58$

$B_0 = \frac{6719.58}{2.2399}$

$B_0 = 3000$

Hence:

$B_t = B_0 * (1 + r_B)^t$

$B_t = 3000 * (1 + 0.08)^t$

$B_t = 3000 * (1.08)^t$

The question requires that we solve for t when:

$A_t = B_t$

Where:

$A_t = A_0 * (1 + r_A)^t$

$A_t = 10000 * (1 + 3\%)^t$

$A_t = 10000 * (1 + 0.03)^t$

$A_t = 10000 * (1.03)^t$

and

$B_t = 3000 * (1.08)^t$

$A_t = B_t$ becomes

$10000 * (1.03)^t = 3000 * (1.08)^t$

Divide both sides by 10000

$(1.03)^t = 0.3 * (1.08)^t$

Divide both sides by $(1.08)^t$

$(\frac{1.03}{1.08})^t = 0.3$

$(0.9537)^t = 0.3$

Take natural logarithm of both sides

$\ln(0.9537)^t = \ln(0.3)$

Rewrite as:

$t\cdot\ln(0.9537) = \ln(0.3)$

Solve for t

$t = \frac{\ln(0.3)}{ln(0.9537)}$

$t = 25.397$

Approximate

$t = 25$

7 0

2 years ago

marin [14]

Answer:

Step-by-step explanation:

From the table attached,

x-intercept of the linear function is, the value of 'x' when f(x) = 0

x = 3 [x-intercept]

Function 'g' is the sum of 2 and the cube root of the sum of three time x and 1.

g(x) = $2+\sqrt[3]{3x+1}$

For x-intercept,

g(x) = 0

$2+\sqrt[3]{3x+1}=0$

3x + 1 = (-2)³

3x + 1 = -8

3x = -8 - 1

3x = -9

x = -3

Therefore, the x-intercept of function 'f' is different or greater than the x-intercept of function g.

4 0

2 years ago

What is the height of this relation: <br> h= -0.3d² + 1.2d + 1.5

marshall27 [118]

Answer:

h = -0.3d^2 + 1.2d + 1.5

Step-by-step explanation:

3 0

3 years ago