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3 years ago

5

A total of 503 tickets were sold for the school play. They were either adult tickets or student tickets. There were 53 more stud

ent tickets sold than adult tickets. How many adult tickets were sold?

1 answer:

Sloan [31]3 years ago

6 0

Answer:

225 adult tickets were sold

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The weights (to the nearest pound) of some boxes to be shipped are found to be:. Weight 65 68 69 70 71 72 90 95 frequency 1 2 5

Lena [83]

So the mean is 72.97

We need to subtract the mean from each value and square it.
(65-72.97)^2= 63.5209
(68-72.97)^2=24.7009
(69-72.97)^2=15.7609
(70-72.97)^2=8.8209
(71-72.97)^2= 3.8809
(72-72.97)^2=0.9409
(90-72.97)^2=290.0209
(95-72.97)^2=485.3209

Now we add up the new values ( also consider their frequency) and find their mean.
Add the values
63.5209+(2 •24.7009=49.4018)+(5•15.7609=78.8045)+(8•8.8209=70.5672)+(7•3.8809=27.1663)+(3•0.9409=2.8227)+(2•290.0209=580.0418)+(2•485.3209=970.6418)= 1,842.967
Divide by total numburs to find the mean
1,842.967/ 30=61.43223333

The standar deviation is the square root of the mean so is
Square root of 61.43223333=7.837871735
Round to the nearest tenth
Standard Deviation is 7.8

7 0

3 years ago

. What are the exact solutions of x2 - x - 4 = 0?

Dimas [21]

X ² - x - 4 = 0
$x_{1} = \frac{1+ \sqrt{1+16} }{2}= \frac{1+ \sqrt{17} }{2} = 2.56$
$x_{2} = \frac{1- \sqrt{17} }{2}= - 1.56$
Exact values are x 1 = 2.56 and x 2 = -1.56

7 0

2 years ago

Read 2 more answers

Abel is making brownies for a bake sale.

IgorLugansk [536]

Answer:

You did not post an image.

Step-by-step explanation:

8 0

2 years ago

HELP ASAP PLZZZZ

Tcecarenko [31]

QUESTION 1

The given system of equations is

$3d - e = 7...eqn(1)$
$d + e = 5...eqn(2)$

To solve by linear combination, we add equation (1) to equation (2) to get,

$3d + d= 7 + 5$

$4d = 12$

We divide through by 4 to obtain,

$d = \frac{12}{4}$

$d = 3$

We put d=3 into equation (2) to get,

$3+ e = 5$

$e = 5 - 3$

$e = 2$

$\boxed {The \: solution \: is \: (3, 2)}$

QUESTION 2

The given system is

4x + y = 5 ...eqn(1)

3x + y = 3 ...eqn(2)

To solve by linear combination, we subtract equation (2) from equation (1) to eliminate y from the equation.

This will give us,

$4x - 3x = 5 - 3$

This implies that,

$x = 2$

Put x=3 into equation (1) to get,

$4(2) + y = 5$

$8+ y = 5$

$y = 5 - 8$

$y = - 3$

The solution is

$(2,-3)$

QUESTION 3

We want to solve the system;

a – 2b = –2 ....eqn(1)

2a + 2b = 14...eqn(2)

by linear combination.

We need to add equation (1) to equation (2) to eliminate b.

This implies that,

$2a + a = 14 + - 2$

Simplify,

$3a = 12$

Divide both sides by 3 to get,

$a = 4$
Put a=4 into equation (2) to obtain,

$2(4) + 2b = 14$

$8 + 2b = 14$
$2b = 14 - 8$

$2b = 6$

$b = 3$

The ordered pair in the form (a, b) is

$(4,3)$

QUESTION 4

The given system of equations is

11x + 4y = 18 ...eqn(1)

3x + 4y = 2 ...eqn(2)

We subtract equation (2) from equation (1) to get,

$11x - 3x = 18 - 2$

$8x = 16$

$x = 2$

Put x=2 into equation (2) to obtain,

$3(2) + 4y = 2$

This implies that,

$6 + 4y = 2$

$4y = 2 - 6$

$4y = - 4$

$y=-1$

The correct answer is (2,-1).

QUESTION 5

The given system is ;

2d + e = 8...eqn1

d – e = 4...eqn2

We add the two equations to eliminate e.

This implies that,

$2d + d = 8 + 4$

$3d = 12$

We divide both sides by 3 to get,

$d = 4$

We put d=4 into equation (2) to get,

$4 - e = 4$

$- e = 4 - 4$

$- e = 0$

$e = 0$

The solution is

$(4,0)$

7 0

3 years ago

32 and 2.56 what is the unit rate

Serggg [28]

12.5

I hope this helped.

5 0

2 years ago