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2 years ago

Tyler is overdrawn at the bank by $180. His brother has $70 more than him. How much money does Tyler's brother have?

Mathematics

2 answers:

Lena [83]2 years ago

8 0

Answer:

Tyler's brother has $250

Step-by-step explanation:

180+70 =250

scoundrel [369]2 years ago

6 0

Answer: $250

Step-by-step explanation:

We Add the amount by which Tyler's account was overdrawn + $70

= $180 + 70

= $250

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A^2 + b^2 + c^2 = 2(a − b − c) − 3. (1) Calculate the value of 2a − 3b + 4c.

Verdich [7]

Answer:

$2a - 3b + 4c = 1$

Step-by-step explanation:

Given

$a^2 + b^2 + c^2 = 2(a - b - c) - 3$

Required

Determine $2a - 3b + 4c$

$a^2 + b^2 + c^2 = 2(a - b - c) - 3$

Open bracket

$a^2 + b^2 + c^2 = 2a - 2b - 2c - 3$

Equate the equation to 0

$a^2 + b^2 + c^2 - 2a + 2b + 2c + 3 = 0$

Express 3 as 1 + 1 + 1

$a^2 + b^2 + c^2 - 2a + 2b + 2c + 1 + 1 + 1 = 0$

Collect like terms

$a^2 - 2a + 1 + b^2 + 2b + 1 + c^2 + 2c + 1 = 0$

Group each terms

$(a^2 - 2a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

Factorize (starting with the first bracket)

$(a^2 - a -a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$(a(a - 1) -1(a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1) (a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b^2 + b+b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b(b + 1)+1(b + 1)) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)(b + 1)) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c^2 + c+c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c(c + 1)+1(c + 1)) = 0$

$((a - 1)^2) + ((b + 1)^2) + ((c + 1)(c + 1)) = 0$

$((a - 1)^2) + ((b + 1)^2) + ((c + 1)^2) = 0$

Express 0 as 0 + 0 + 0

$(a - 1)^2 + (b + 1)^2 + (c + 1)^2 = 0 + 0+ 0$

By comparison

$(a - 1)^2 = 0$

$(b + 1)^2 = 0$

$(c + 1)^2 = 0$

Solving for $(a - 1)^2 = 0$

Take square root of both sides

$a - 1 = 0$

Add 1 to both sides

$a - 1 + 1 = 0 + 1$

$a = 1$

Solving for $(b + 1)^2 = 0$

Take square root of both sides

$b + 1 = 0$

Subtract 1 from both sides

$b + 1 - 1 = 0 - 1$

$b = -1$

Solving for $(c + 1)^2 = 0$

Take square root of both sides

$c + 1 = 0$

Subtract 1 from both sides

$c + 1 - 1 = 0 - 1$

$c = -1$

Substitute the values of a, b and c in $2a - 3b + 4c$

$2a - 3b + 4c = 2(1) - 3(-1) + 4(-1)$

$2a - 3b + 4c = 2 +3 -4$

$2a - 3b + 4c = 1$

7 0

3 years ago

A rope has knots every 47 feet 3 inches. what length is this in feet?

nadezda [96]

12 inches = 1 foot

thus 47 feet 3 inches = (47+3/12) feet =47.25 feet

3 0

2 years ago

Read 2 more answers

I'm in need of help. The file should be attached to this. Thanks for your help!

Amanda [17]

company a: 2mins

company b: 7mins

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6 0

3 years ago

QUESTION 20 PLEASE HELPPPP ASAPPPP

WARRIOR [948]

Answer: x=-1, y=-8

Step-by-step explanation: solve for the first variable in one of the equations then subsitutte the result into the other equation

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2 years ago

Using a proportion, what can you infer about the number of households in her town that have more than three children?

eduard

Answer:

About 296 households have more than three children.

Step-by-step explanation:

The complete question is:

Carmen used a random number generator to simulate a survey of how many children live in the households in her town. There are 1,346 unique addresses in her town with numbers ranging from zero to five children. The results of 50 randomly generated households are shown below. Children in 50 Households Number of Children Number of Households 0 5 1 11 2 13 3 10 4 9 5 2 Using a proportion, what can you infer about the number of households in her town that have more than three children? About 242 households have more than three children. About 269 households have more than three children. About 296 households have more than three children. About 565 households have more than three children.

Solution:

The data provided for the number of children and number of households is:

Number of Children (X) Number of Households (f (X))

0 5

1 11

2 13

3 10

4 9

5 2

TOTAL 50

Compute the probability of households having more than three children as follows:

P (More than 3 children) = P (4 children) + P (5 children)

$=\frac{9}{50}+\frac{2}{50}$

$=\frac{11}{50}$

$=0.22$

Let the random variable Y be defined as the number of households having more than three children.

The probability of the random variable Y is, p = 0.22.

The entire population, i.e. total number of addresses in Carmen's town with numbers ranging from zero to five children, consists of N = 1,346 households.

Compute the expected number of households having more than three children as follows:

$E(Y) =N\times p$

$=1346\times 0.22\\=296.12\\\approx 296$

Thus, about 296 households have more than three children.

4 0

2 years ago