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4 years ago

Tickets to a baseball game cost $12 for adults and $7 for children. How much will tickets cost for 3 adults and 4 children?

Mathematics

1 answer:

ludmilkaskok [199]4 years ago

4 0

Answer:

So total cost of tickets = $64

Step-by-step explanation:

Given:

Cost of adult = $12

Cost of Children = $7

Total Adults = 3

Total children = 4

To Find:

Total Cost = ?

Solution:

We are given per person price and no of persons too now

Total cost = Cost of children ticket + Cost of Adults ticket

Now we will find the values

Cost of 3 adults ticket = total person * cost per person

= 3 * 12

=$36

Cost of 4 Children ticket = total person * cost per person

= 4 * 7

=$28

Now

Total cost = Cost of children ticket + Cost of Adults ticket

putting value

Total cost = 28 + 36

= $ 64

So total cost of tickets = $64

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Identify all sets to which the number belongs.

ra1l [238]

Option B

-0.249851765 is a irrational number

<em><u>Solution:</u></em>

Given number is -0.249851765

We have to classify the number

Let us first understand about irrational, real, whole, integer and rational numbers

<h3><u>Integers:</u></h3>

An integer is a whole number (not a fractional number) that can be positive, negative, or zero. Examples of integers are: -5, 1, 5, 8, 97, and 3,043

Every integer can be expressed as a decimal, but most numbers that can be expressed as a decimal are not integers. If all the digits after the decimal point are zeroes, the number is an integer. If there are any non-zero digits after the decimal point, the number is not an integer.

Thus -0.249851765 has non zero digits after decimal point. So it is not a integer

<h3><u>Whole numbers:</u></h3>

Whole numbers are positive numbers, including zero, without any decimal or fractional parts. Negative numbers are not considered "whole numbers."

But the given number -0.249851765 is negative number. So it is not a whole number

<h3><u>Natural numbers:</u></h3>

A natural number is an integer greater than 0. Natural numbers begin at 1 and increment to infinity: 1, 2, 3, 4, 5, etc. Natural numbers are also called "counting numbers" because they are used for counting.

A decimal is a natural number if it is non-negative and the only digits after its decimal points are zero

So the given number -0.249851765 is a negative number and so it is not a natural number

<h3><u>Rational numbers:</u></h3>

A rational number is a number that can be expressed as a fraction (ratio) in the form $\frac{p}{q}$ where p and q are integers and q is not zero.

The rational numbers includes all positive numbers, negative numbers and zero that can be written as a ratio (fraction) of one number over another.

When a rational number fraction is divided to form a decimal value, it becomes a terminating or repeating decimal.

So -0.249851765 is not a rational number

<h3><u>Irrational number:</u></h3>

An irrational number is real number that cannot be expressed as a ratio of two integers.

When an irrational number is expressed in decimal form, it goes on forever without repeating

So the given number -0.249851765 is irrational number

We can conclude that:

-0.249851765 is a irrational number, So Option B is correct

3 0

3 years ago

What is the value of the lower quartile? <br><br><br><br><br> 0<br><br> 8<br><br> 6<br><br> 2

Elanso [62]

the value of the lower quartile is 2 because that’s where the box ends

5 0

4 years ago

The enrollment at a local college increased 4% over last year’s enrollment of 2100. Find the current enrollment.

sveticcg [70]

Answer:

Current enrollment 2940

Step-by-step explanation:4% of 2100 is 840. You can find this by multiplying .4 by 2100. You would add 840 to 2100 to get 2940.

3 0

4 years ago

Read 2 more answers

In a binomial distribution, n = 8 and π=0.36. Find the probabilities of the following events. (Round your answers to 4 decimal p

skelet666 [1.2K]

Answer:

$\mathbf{P(X=5) =0.0888}$

P(x ≤ 5 ) = 0.9707

P ( x ≥ 6) = 0.0293

Step-by-step explanation:

The probability of a binomial mass distribution can be expressed with the formula:

$\mathtt{P(X=x) =(^{n}_{x} ) \ \pi^x \ (1-\pi)^{n-x}}$

$\mathtt{P(X=x) =(\dfrac{n!}{x!(n-x)!} ) \ \pi^x \ (1-\pi)^{n-x}}$

where;

n = 8 and π = 0.36

For x = 5

The probability $\mathtt{P(X=5) =(\dfrac{8!}{5!(8-5)!} ) \ 0.36^5 \ (1-0.36)^{8-5}}$

$\mathtt{P(X=5) =(\dfrac{8!}{5!(3)!} ) \ 0.36^5 \ (0.64)^{3}}$

$\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 \times 5!}{5!(3)!} ) \times \ 0.0060466 \ \times 0.262144}$

$\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 }{3 \times 2 \times 1} ) \times \ 0.0060466 \ \times 0.262144}$

$\mathtt{P(X=5) =({8 \times 7 } ) \times \ 0.0060466 \ \times 0.262144}$

$\mathtt{P(X=5) =0.0887645}$

$\mathbf{P(X=5) =0.0888}$ to 4 decimal places

b. x ≤ 5

The probability of P ( x ≤ 5) $\mathtt{P(x \leq 5) = P(x = 0)+ P(x = 1)+ P(x = 2)+ P(x = 3)+ P(x = 4)+ P(x = 5})$

${P(x \leq 5) = ( \dfrac{8!}{0!(8!)} \times (0.36)^0 \times (1-0.36)^8 \ ) + \dfrac{8!}{1!(7!)} \times (0.36)^1 \times (1-0.36)^7 \ +$ $\dfrac{8!}{2!(6!)} \times (0.36)^2 \times (1-0.36)^6 \ + \dfrac{8!}{3!(5!)} \times (0.36)^3 \times (1-0.36)^5 + \dfrac{8!}{4!(4!)} \times (0.36)^4 \times (1-0.36)^4 \ + \dfrac{8!}{5!(3!)} \times (0.36)^5 \times (1-0.36)^3 \ )$