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2 years ago

Divisible rules what is divisible by 4518

Mathematics

1 answer:

Digiron [165]2 years ago

6 0

Answer:

where are the options

Step-by-step explanation:

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For the function: g(x)= 3 - 8 (1/4)^2-x

Alla [95]

Answer:

bye

Step-by-step explanation:

7 0

2 years ago

At the city museum, child admission is $6.10 and adult admission is $9.90. On Friday, 142 tickets were sold for a total sales of

antoniya [11.8K]

Answer:

no of child tickets sold were 58

Step-by-step explanation:

Let the number of child ticket sold be x

cost of 1 child ticket = $6.10

Total cost of x child ticket = x*cost of 1 child ticket = = $6.10x

Let the number of adult ticket sold be y

cost of 1 adult ticket = $9.90

Total cost of y adult ticket = y*cost of 1 adult ticket = = $9.90y

Given total ticket sold = 142

Thus,

number of child ticket sold + number of adult ticket sold = 142

x + y = 142 - equation 1

x = 142 -y

Total sales of ticket = $1185.40

total sales of ticket = Total cost of x child ticket + Total cost of x a dult ticket

6.10x + 9.90y = 1185.40

in this equation substituting x = 142 - y from equation 1

6.10(142 - y ) + 9.90y = 1185.40

=>866.2 - 6.10y + 9.90y = 1185.40

=> 3.80y = 1185.40 - 866.2

=> 3.80y = 319.2

=> y = 319.2/3.80 = 84

Thus, y = 84

x = 142 - y = 142 - 84 = 58

Thus, no of child tickets sold were 58

6 0

3 years ago

The Copy Shop has made 20 copies of a document for you. Since the defective rate is 0.1, you think there may be some defective c

Pepsi [2]

Answer:

Binomial

There is a 34.87% probability that you will encounter neither of the defective copies among the 10 you examine.

Step-by-step explanation:

For each copy of the document, there are only two possible outcomes. Either it is defective, or it is not. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem

Of the 20 copies, 2 are defective, so $p = \frac{2}{20} = 0.1$ .

What is the probability that you will encounter neither of the defective copies among the 10 you examine?

This is P(X = 0) when $n = 10$ .

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.1)^{0}.(0.9)^{10} = 0.3487$

There is a 34.87% probability that you will encounter neither of the defective copies among the 10 you examine.

8 0

3 years ago

Which of the following has a value of 84? <br> A. 512 <br> B. 4,096 <br> C. 32 <br> D. 83

andreev551 [17]

Answer is B. 4,096

I encountered this problem before. This is a trick question.
84 is not really the number 84. It is 8 raised to the 4th power.

So 8⁴ is equal to 4,096.

8 x 8 x 8 x 8 = 4,096

8 x 8 = 64
64 x 8 = 512
512 x 8 = 4,096

6 0

3 years ago

Select All That Apply .

makvit [3.9K]

Answer:

the 4th one ,hoped this help

7 0

2 years ago