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2 years ago

ONE HUNDRED LOTTERY TICKETS ARE SOLD FOR $5.00 EACH. ONE PRIZE OF $300

Mathematics

1 answer:

olga55 [171]2 years ago

5 0

1/100 tickets = 1% chance = 0.01
$5 spent on 1 ticket

0.01(300) - 5
3-5
-$2

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-UIVIT UUNNU INTEREST

Jobisdone [24]

Answer:

Step-by-step explanation:

Principal, P = $4,000

Interest rate, r = 3% = 0.03

Period, t = 20 years

Number of times compounded in a year = 4

Amount , A = P( 1 + r/n)^tn

A = 4000( 1 + 0.03/4)^4*20

A = 4000( 1 + 0.0075) ^80

A =4000( 1.0075) ^80

A = 4000* 1.818

A = $7272.18

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3 years ago

One-stray Problem Solving

anygoal [31]

A) 5000 m²
b) A(x) = x(200 -2x)
c) 0 < x < 100
Step-by-step explanation:
b) The remaining fence, after the two sides of length x are fenced, is 200-2x. That is the length of the side parallel to the building. The product of the lengths parallel and perpendicular to the building is the area of the playground:
A(x) = x(200 -2x)
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a) A(50) = 50(200 -2·50) = 50·100 = 5000 . . . . m²
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c) The equation makes no sense if either length (x or 200-2x) is negative, so a reasonable domain is (0, 100). For x=0 or x=100, the playground area is zero, so we're not concerned with those cases, either. Those endpoints could be included in the domain if you like.

3 0

2 years ago

Jill makes 40 hours per week. She makes $20 per hour. What is his gross pay

krek1111 [17]

Answer:

Weekly Gross Income: $800

Anual Gross Income: $41,600

Monthly Gross Income: $3,466.67

Step-by-step explanation:

Weekly: 40 x 20 = 800

Anual: 40 x 20= 800 800x52= 41,600

Monthly:41,600 / 12= 3,466.6666666666666666666666

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3 years ago

Can you help me 5x+9=29

Artyom0805 [142]

It actually equals 49

3 0

3 years ago

Read 2 more answers

Two problems here I need solved! I need every step, so please have that with your answers!!

Free_Kalibri [48]

QUESTION 1

We want to solve,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}$

We factor the denominator of the fraction on the right hand side to get,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.$

This implies
$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.$

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{(x-4)(x - 2)}$

We multiply through by LCM of
$(x-4)(x - 2)$

$(x - 2) + x(x-4) = 2$

We expand to get,

$x - 2 + {x}^{2} - 4x= 2$

We group like terms and equate everything to zero,

${x}^{2} + x - 4x - 2 - 2 = 0$

We split the middle term,

${x}^{2} + - 3x - 4 = 0$

We factor to get,

${x}^{2} + x - 4x- 4 = 0$

$x(x + 1) - 4(x + 1) = 0$

$(x + 1)(x - 4) = 0$

$x + 1 = 0 \: or \: x - 4 = 0$

$x = - 1 \: or \: x = 4$

But
$x = 4$
is not in the domain of the given equation.

It is an extraneous solution.

$\therefore \: x = - 1$
is the only solution.

QUESTION 2

$\sqrt{x+11} -x=-1$

We add x to both sides,

$\sqrt{x+11} =x-1$

We square both sides,

$x + 11 = (x - 1)^{2}$

We expand to get,

$x + 11 = {x}^{2} - 2x + 1$

This implies,

${x}^{2} - 3x - 10 = 0$

We solve this quadratic equation by factorization,

${x}^{2} - 5x + 2x - 10 = 0$

$x(x - 5) + 2(x - 5) = 0$

$(x + 2)(x - 5) = 0$

$x + 2 = 0 \: or \: x - 5 = 0$

$x = - 2 \: or \: x = 5$

But
$x = - 2$
is an extraneous solution

$\therefore \: x = 5$

7 0

3 years ago