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alexira
3 days ago
13

Laurence, the stadium manager, spends $22,000 on a broadcast TV ad, $30,000 on a cable TV ad, and $6,000 on a radio ad. What per

centage of broadcast TV advertising did he do? (Round up)
Mathematics
1 answer:
Zina [9.1K]3 days ago
7 0

38 % of the total budget goes to advertising on broadcast TV

Solution:

Based on the details,

Expenditure for broadcast TV advertising = $ 22, 000

Expenditure for cable TV advertising = $ 30, 000

Expenditure for radio advertising = $ 6000

Need to determine: the proportion of broadcast TV advertising

Total expenditure = $ 22, 000 + $ 30, 000 + $ 6000

Total expenditure = $ 58000

proportion of broadcast TV advertising = \frac{\text{ amount spend on broadcast tv}}{\text{ total amount}} \times 100

\rightarrow \frac{22000}{58000} \times 100\\\\\rightarrow 0.3793 \times 100 = 37.93 \approx 38

Hence 38 % of the total budget is allocated to broadcast TV advertising

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In a certain town, 60 families own tv sets, 85 own scooters, 70th own refrigerators, and 95 own radio sets. One hundred and thir
tester [8842]

Answer:

310

Step-by-step explanation:

The highest possible number is achieved when each household has just one appliance. When calculating the total, we find ...

60 + 85 + 70 + 95 = 310

The maximum possible number of families is 310.

5 0
1 month ago
A box contains 288 sugar cubes. You open the top of the box and count 12 cubes across the front and 6 cubes along a side. How ma
Zina [9171]
Hello!

To determine how many cubes are present in one layer,

multiply the amount across by the side length.

12 * 6 = 72

Then divide by the total number of cubes

288 / 72 = 4

There are 4 layers

The result is 4

I hope this is helpful!
6 0
19 hours ago
. Donny had a 10-foot board. He used the board to make two shelves. One shelf measured 4 feet 7 inches wide and the other shelf
tester [8842]
The length of the remaining piece measures 2 feet and 3 inches.
3 0
10 days ago
The diagram shows the cross section of a cylindrical pipe with water lying in the bottom.
PIT_PIT [9117]
Hello

So my brother shared this on Yahoo
To start, draw a line connecting the circle's center to one endpoint of the chord (water surface) and again to the point with maximum depth. This creates a right triangle. The length of the side up to the water-surface measures 5 cm, and the hypotenuse is 7 cm.

<span>Now, perform the following steps: </span>

<span>Calculate angle θ at the right triangle's corner using: cos θ = 5/7 ⇒ θ = cos ˉ¹ (5/7) </span>

<span>Thus, θ is approximately 44.4°, leading to the angle subtended at the circle center by the water surface being roughly 88.8°. </span>

<span>To find the shaded area, subtract the triangular area above the water in your diagram from the area of the sector. </span>

<span>Shaded area ≃ 88.8/360*area of circle - ½*7*7*sin88.8° </span>
<span>= 88.8/360*π*7² - 24.5*sin 88.8° </span>
<span>which is approximately 13.5 cm² </span>
<span>(using area of ∆ = ½.a.b.sin C for the triangle) </span>



<span>b) </span>

<span>The water's volume is equal to the cross-sectional area multiplied by the length </span>
<span>≈ 13.5 * 30 cm³ </span>
<span>≈ 404 cm³</span>

I hope this was helpful
5 0
14 days ago
Machines A and B always operate independently and at their respective constant rates. When working alone, Machine A can fill a p
Inessa [9000]

Answer:

The value of x is \frac{10}{3} hours.

Step-by-step explanation:

Machine A = 5 hours

Machine B = x hours

When both machines work together, it takes 2 hours.

Using the formula: \frac{T}{A} + \frac{T}{B} = 1

where:

T represents the total time worked by both machines

A represents the time taken by Machine A

B represents the time taken by Machine B

\frac{2}{5} + \frac{2}{x} = 1

Multiplying all terms by the common denominator (5B),

5x(\frac{2}{5} + \frac{2}{x} = 1)

2x + 10 = 5x.

Combining like terms yields:

10 = 5x - 2x

10 = 3x;

To isolate x, divide both sides by 3.

\frac{3x}{3} = \frac{10}{3}

x = \frac{10}{3} hours.

Thus, Machine B will require \frac{10}{3} hours to fill the lot.

7 0
23 days ago
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