All categories

3 years ago

Please help explanation if possible

Mathematics

1 answer:

Eva8 [605]3 years ago

5 0

Answer:

612 adults

361 students

Step-by-step explanation:

To solve this question, set two equations:

Let x be number of adults and y be number of students.

As there are in total 937 people, the equation would be the sum of both adults and children:

$x+y=937$

$x=937-y$ ...... ( 1 )

As the total sale amount is $1109, the equation would be to add up the ticket fee:

$2x+0.75y=1,109$ ...... ( 2 )

Put ( 1 ) into ( 2 ):

$2(937-y)+0.75y=1,109$

$1874-2y+0.75y=1,109$

$-1.25y=-765$

$y=763/1.25$

$y=612$

Put y into ( 1 ):

$x=973-612$

$x=361$

Therefore there are 612 adults and 361 students.

You might be interested in

Based on historical data, your manager believes that 41% of the company's orders come from first-time customers. A random sample

TEA [102]

Answer:

The probability that the sample proportion is between 0.35 and 0.5 is 0.7895

Step-by-step explanation:

To calculate the probability that the sample proportion is between 0.35 and 0.5 we need to know the z-scores of the sample proportions 0.35 and 0.5.

z-score of the sample proportion is calculated as

z= $\frac{p(s)-p}{\sqrt{\frac{p*(1-p)}{N} } }$ where

p(s) is the sample proportion of first time customers
p is the proportion of first time customers based on historical data

N is the sample size

For the sample proportion 0.35:

z(0.35)= $\frac{0,35-0.41}{\sqrt{\frac{0.41*0.59}{72} } }$ ≈ -1.035

For the sample proportion 0.5:

z(0.5)= $\frac{0,5-0.41}{\sqrt{\frac{0.41*0.59}{72} } }$ ≈ 1.553

The probabilities for z of being smaller than these z-scores are:

P(z<z(0.35))= 0.1503

P(z<z(0.5))= 0.9398

Then the probability that the sample proportion is between 0.35 and 0.5 is

P(z(0.35)<z<z(0.5))= 0.9398 - 0.1503 =0.7895

6 0

3 years ago

Please see attachment and SHOW WORK. Thank You. Will mark brainliest!!

Fofino [41]

a. Let c and b represent the numbers of capped and broken bottles, respectively. Here is the system of equations.

... c + b = 15 . . . . 15 bottles have come Erica's way

... 4c -2b = 6 . . . .Erica's net pay for the day is 6 cents

b. By substitution, we can write an expression for b in terms of c:

... b = 15 - c

Then put it into the second equation:

... 4c - 2(15-c) = 6

... 6c - 30 = 6 . . . . eliminate parentheses

... 6c = 36 . . . . . . .add 30

... c = 6 . . . . . . . . . divide by 6

... b = 15 - 6 = 9 . . find the value of b

Erica has capped 6 bottles and broken 9.

_____

Here's one way to do it by elimination. We can divide the second equation by 2 to put it into standard form:

... 2c - b = 3

Now, adding it to the first equation gives

... (2c -b) + (c +b) = (3) + (15)

... 3c = 18 . . . . . simplify. Notice the b variable has been eliminated

... c = 6 . . . . . . . divide by 3

... 6 + b = 15

... b = 9 . . . . . . substitute into the first equation and solve for b by subtracting 6

Erica has capped 6 bottles and broken 9. (same solution as above)

4 0

3 years ago

What is the total cost for a $100 dinner plus 20% tip plus 8% sales tax

hoa [83]

$129.6

you do 100 times .08 and get 8 so 100 + 8 = 108
then you do 108 times .20 and get 21.6 so 108 + 21.6 = 129.6

6 0

2 years ago

Let a be a rational number and b be an irrational number. Which of the following are true statements?(there is more than 1 answe

kramer

A.) the sum of a and b is never rational.
This is a true statement. Since an irrational umber has a decimal part that is infinite and non-periodical, when you add a rational number to an irrational number, the result will have the same infinite non periodical decimal part, so the new number will be irrational as well.

B.) The product of a and b is rational
This one is false. Zero is a rational number, and when you multiply an irrational number by zero, the result is always zero.

C.) b^2 is sometimes rational
This one is true. When you square an irrational number that comes from a square root like $\sqrt{2}$ , you will end with a rational number: $( \sqrt{2} )^{2}=2$ , but, if you square rationals from different roots than square root like $\sqrt[3]{2}$ , you will end with an irrational number: $\sqrt[3]{2^{2} } = \sqrt[3]{2}$ .

D.) a^2 is always rational
This one is false. If you square a rational number, you will always end with another rational number.

E.) square root of a is never rational
This one is false. The square root of perfect squares are always rational numbers: $\sqrt{64} =8$ , $\sqrt{16} =4$ ,...

F.) square root of b is never rational
This one is true. Since the square root of any non-perfect square number is irrational, and all the irrational numbers are non-perfect squares, the square root of an irrational number is always irrational.

We can conclude that given that a is a rational number and b be an irrational number, A, C, D, and F are true statements.

4 0

3 years ago

Solve for x. 5(3x + 9) = - 135

tekilochka [14]

Answer:

x=-12

tha last one

7 0

4 years ago

Read 2 more answers