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Elanso [62]
2 years ago
13

A drive in theater charges $3.50 per car, the drive in already admitted 100 cars, solve an inequality to find the number of cars

the drive in needs to admit to make atleast 500$
Mathematics
1 answer:
Ilia_Sergeevich [38]2 years ago
4 0

They need to admit 43 more cars to make at least $500

<u>Solution:</u>

Given, A drive in theater charges $3.50 per car,  

The drive in already admitted 100 cars,  

Then, money collected until now is $ 3.50 x 100 = $ 350  

We have to solve an inequality to find the number of cars the drive in needs to admit to make atleast 500$

Now, let the number of cars need to be admitted be "n"

Then, inequality is:

collected amount + needed amount ≥ $ 500  

$ 350 + $ 3.50 x n ≥ $ 500

350 + 3.5n ≥ 500

3.5n ≥ 500 – 350  

On solving, we get

3.5n ≥ 150  

n ≥ 42.857

As number of cars can’t be fractions, n = 43

Hence, they need to admit 43 more cars to make at least $500

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Answer:

Step-by-step explanation:

x^3 - 8x² - x + 8 =0

x^3 - 8x² - (x - 8 )=0

x²(x-8) - (x-8) = 0

(x-8)(x² -1) =0

x-8 = 0  or x² - 1 = 0

x-8=0 or (x-1)(x+1)=0

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8 0
3 years ago
Read 2 more answers
Given a minimum usual value of 135.8 and a maximum usual value of 155.9, determine which (1 point) of the following values would
maxonik [38]

Answer:  b. 134

Step-by-step explanation:

Given : A minimum usual value of 135.8 and a maximum usual value of 155.9.

Let x denotes a usual value.

i.e.  135.8< x < 155.9

Therefore , the interval for the usual values is [135.8, 155.9] .

If interval for any usual value is [135.8, 155.9] , then any value should lie in this otherwise we call it unusual.

Let's check all options

a. 137  ,

since  135.8< 137 < 155.9

So , it is usual.

b. 134

since 134<135.8 (Minimum value)

So , it is unusual.

c. 146  

since  135.8< 146 < 155.9

So , it is usual.

d. 155  

since 135.8< 1155 < 155.9

So , it is usual.

Hence, the correct answer is b. 134 .

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3 years ago
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denis23 [38]

Answer:

{HH, HT, TH, TT}

Step-by-step explanation:

The set of all possible outcomes in tossing a coin twice is;

{HH, HT, TH, TT}

In the first toss the coin may land Heads. In the second toss the coin may land Heads or Tails. This can be represented as;

HH, HT

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In the first toss the coin is also likely to land Tails. In the second toss the coin may land Heads or Tails. This can be represented as;

TH, TT

Tails in the first toss followed by a Head in the second toss. Tails in the first and second tosses.

Combining these two possibilities will give us the set of all possible outcomes in tossing a coin twice is;

{HH, HT, TH, TT}

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