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Orlov [11]
2 years ago
5

The stadium owner is considering an offer from Reliable Tools to change the name to Reliable Stadium. The company is offering to

pay $750,000 per year. If the stadium already receives $160,000 for music promotions and accepts this offer, how much will it receive in sponsorship rights each year in total?
Mathematics
2 answers:
defon2 years ago
6 0
The answer is probably 590,000 because isn't it subtraction
Dmitry [639]2 years ago
4 0

Amount stadium receives if it changes the name from Reliable Tools to Reliable Stadium = $750,000 per year

Amount stadium receives from music promotions = $160,000

Total amount received each year = Amount stadium receives if it changes the name from Reliable Tools to Reliable Stadium + Amount stadium receives from music promotions

Total amount = $750,000 + $160,000 = $910000

So it will receive $910000 in sponsorship rights each year in total.

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LekaFEV [45]

Answer:

plz explain correctly please

8 0
2 years ago
In equilateral ∆ABC with side a, the perpendicular to side AB at point B intersects extension of median AM in point P. What is t
Sergeeva-Olga [200]

Answer:

Perimeter  = (2 + √3)·a

Step-by-step explanation:

Given: ΔABC is equilateral and AB = a

The diagram is given below :

AM is a median , PB ⊥ AB , PM = b

Now, by using properties of equilateral triangle, median is perpendicular bisector and each angle is of 60°.

We get, ∠AMB = 90°. So, by linear pair ∠AMB + ∠PMB = 180° ⇒ ∠PMB = 90°. Also, ∠ABC = 60° and ∠ABP = 90° (given) So, ∠PBM = 30°

Since, AM is perpendicular bisector of BC. So,

MB = \frac{a}{2}

Now in ΔAMB , By using Pythagoras theorem

AB^{2}=AM^{2}+MB^{2}\\AM^{2}=AB^{2}-MB^{2}\\AM^{2}=a^{2}-(\frac{a}{2})^{2}\\AM=\frac{\sqrt{3}\cdot a}{2}

Now, in ΔBMP :

sin\thinspace 30^{o}=\frac{\text{Perpendicular}}{\text{Hypotenuse}}\\\\sin\thinspace 30^{o}=\frac{\text{MB}}{\text{PB}}\\\\PB=\frac{\text{MB}}{\text{sin 30}}\\\\PB=\frac{\frac{a}{2}}{\frac{1}{2}}\implies PB = a\\\\tan\thinspace 30^{o}=\frac{\text{Perpendicular}}{\text{Base}}\\\\tan\thinspace 30^{o}=\frac{\text{MB}}{\text{PM}}\\\\PM=\frac{\text{MB}}{\text{tan 30}}\\\\PM=\frac{\frac{a}{2}}{\frac{1}{\sqrt3}}\implies PM=b= \frac{\sqrt{3}\cdot a}{2}

Perimeter of ABM = AB + PB + PM + AM

\text{Perimeter = }a+a+b+ \frac{\sqrt{3}\cdot a}{2}\\\\=2\cdot a + \frac{\sqrt{3}\cdot a}{2} +\frac{\sqrt{3}\cdot a}{2}\\\\=2\cdot a +\sqrt{3}\cdot a\\\\=(2+\sqrt3})\cdot a

Hence, Perimeter of ΔABP = (2 + √3)·a units

3 0
2 years ago
At Central High School, 28% of the students are Freshmen, 26% are Sophomores, 24% are Juniors, and 22% are Seniors. The student
Jet001 [13]

Answer:

(D) This is a stratified random sample because a separate random sample is selected from each class

Step-by-step explanation:

A sample of size n is defined to be a stratified random sample if it is selected from a population which has been divided into a number of non overlapping groups or sub populations called strata, such that part of the sample is drawn at random from each stratum.

It is to be emphasized that good stratification requires that each of these strata should be internally homogeneous but externally should differ from one another.

The advantage of stratified sampling is low cost , greater accuracy and better coverage.

If we analyze the given scenario a sample is selected from each strata depicting its corresponding  percentage in the population.  It is internally homogeneous but externally different.

4 0
2 years ago
A) The ratio 20 minutes to 1 hour can be written in the form 1:n.
Fudgin [204]

Answer:

a) 1:3

b) 9:225,000

Step-by-step explanation:

A)

1 hour is equal to 60 minutes. Written as a fraction it would look like

\frac{20}{60} which would be equal to \frac{1}{3}, or 1:3

B)

Just multiply both sides of the ratio by 9, you will get 9:225,000

Hope I helped, please mark as brainliest! Have a good day!

7 0
2 years ago
If e = 9 cm and f = 40 cm, what is the length of g?
rewona [7]

(9,40,41) is a Pythagorean Triple, farther down the list than teachers usually venture.

Answer: D. 41 cm

There's a subset of Pythagorean Triples where the long leg is one less than the hypotenuse,

a^2+b^2 = (b+1)^2

a^2 + b^2 = b^2 + 2b +1

a^2=2b+1

So we get one for every odd number, since the square of an odd number is odd and the square of an even number is even.

b = (a^2 - 1)/2

a=3, b=(3^2-1)/2=4, c=b+1=5

a=5, b=(5^2-1)/2 =12, c = 13

a=7, b=24, c=25

a=9, b=40, c=41

a=11, b=60, c=61

a=13, b=84, c=85

It's good to be able to recognize Pythagorean Triples when we see them.  

Otherwise we'd have to work the calculator:

√(9² + 40²) = √1681 = 41

8 0
2 years ago
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