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soldi70 [24.7K]
3 years ago
13

If the sales tax rate is 8.25% how much would pay for an item that costs $24.00?

Mathematics
1 answer:
Blizzard [7]3 years ago
8 0

Answer:

$25.98

Step-by-step explanation:

8.25% of 24 = 0.0825 × 24 = 1.98

1.98 + 24 = 25.98

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2x^(2) + xy + 2y^(2) = 5, (1, 1)<br> (ellipse)
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The original price of an audio jukebox was 280 it is on sale for 220 what percent discount does this represent
solong [7]

Answer:

The answer would be 12%

Step-by-step explanation:

250-220=30 and 30 is 12% of 250

4 0
3 years ago
Read 2 more answers
A rectangular area adjacent to a river is fenced​ in; no fence is needed on the river side. The enclosed area is 1500 square fee
ZanzabumX [31]

Answer:

a) C(x) = 15000/x + 6x +80

b) Domain of C(x)  { R x>0 }

Step-by-step explanation:

We have:  

Enclosed area = 1500 ft²   = x*y      from which     y  =  1500 / x    (a) where x is perpendicular to the river

Cost = cost of sides of fenced area perpendicular to the river  + cost of side parallel to river + cost of 4 post then

Cost = 10*y + 2*3*x + 4*20 and accoding to (a)  y = 1500/x

Then

C(x)  = 10* ( 1500/x ) + 6*x + 80

C(x) = 15000/x + 6x +80

Domain of C(x)  { R x>0 }

5 0
3 years ago
A city hosted a music festival that included three concerts. According to the sales database, 28% of the audience attended the f
Vitek1552 [10]

Answer:

P(X\cap Y\cap Z)=0.05

Step-by-step explanation:

From the question we are told that:

Percentage of audience in first concert P(X)=0.28

Percentage of audience in second concert P(Y)=0.42

Percentage of audience in third concert P(Z)=0.30

Audience Percentage at at-least one concert P(X \cup Y \cup Z)=0.80

Percentage of audience at first & second concert P(X \cap Y)=0.10

Percentage of audience in first & third concert P(X \cap Z)=0.08

Percentage of audience in second & third concert P(Y\cap Z)=0.07

 

Generally the equation for probability of attending all concerts P(X\cap Y\cap Z)is mathematically given by

P(X \cup Y \cup Z)=P(X)+P(Y)+P(Z)-P(X \cap Y)-P(X \cap Z)-P(Y\cap Z)+P(X\cap Y\cap Z)

P(X\cap Y\cap Z)=P(X \cup Y \cup Z)-P(X)-P(Y)-P(Z)+P(X \cap Y)+P(X \cap Z)+P(Y\cap Z)

P(X\cap Y\cap Z)=0.80-0.28-0.42-0.30+0.10+0.80+0.70

P(X\cap Y\cap Z)=0.05

Therefore the probability that a randomly selected audient attended all the concerts

P(X\cap Y\cap Z)=0.05

8 0
3 years ago
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