The tape diagram represents an equation.

Evelyn has $524.96 in her checking account. She must maintain a $500 balance to avoid a fee. She wrote a check for $32.50 today.

cupoosta [38]

A linear inequality to represent the algebraic expression is given as 492.46 - x ≥ 500

<h3>Linear Inequality</h3>

Linear inequalities are inequalities that involve at least one linear algebraic expression, that is, a polynomial of degree 1 is compared with another algebraic expression of degree less than or equal to 1.

In this problem, her minimum balance must not decrease beyond $500 or she will pay a fee.

where

x = withdrawals

The inequality to represent this can be written as

524.96 - 32.50 - x ≥ 500

Simplifying this;

492.46 - x ≥ 500

The linear inequality is 492.46 - x ≥ 500

Learn more on linear inequality here;

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4 0

1 year ago

You have 42 ft. of fencing (1 ft. segments) to make a rectangular garden. How much should each side be to maximize your total ar

Verizon [17]

Answer:

$Length=10.5\ ft$

$Width=10.5\ ft$

$Area=110.25\ ft^{2}$

Step-by-step explanation:

Let

x----> the length of the rectangular garden

y---> the width of the rectangular garden

we know that

The perimeter of the rectangle is equal to

$P=2(x+y)$

we have

$P=42\ ft$

so

$42=2(x+y)$

simplify

$21=(x+y)$

$y=21-x$ ------> equation A

Remember that the area of rectangle is equal to

$A=xy$ ----> equation B

substitute equation A in equation B

$A=x(21-x)$

$A=21x-x^{2}$ ----> this is a vertical parabola open downward

The vertex is a maximum

The y-coordinate of the vertex is the maximum area

The x-coordinate of the vertex is the length side of the rectangle that maximize the area

using a graphing tool

The vertex is the point $(10.5,110.25)$

see the attached figure

so

$x=10.5\ ft$

Find the value of y

$y=21-10.5=10.5\ ft$

The garden is a square

the area is equal to

$A=(10.5)(10.5)=110.25\ ft^{2}$ ----> is equal to the y-coordinate of the vertex is correct

6 0

3 years ago

17 __ a solution od d - 6 = 23<br><br> is it or is it not

anygoal [31]

Answer:

29 is the answer.

:-)

;-)

3 0

2 years ago

Read 2 more answers

Is multiplying 10 to the power of 3 the same as multiplying by 10 factors of 3

andrew-mc [135]

No because since multiplying by 10³ is multiplying by 10 times 10 times 10 which equals 1000 and multiplying by 10 factors of 3 is multiplying by 3 times 3 times 3 times 3…….for 10 times equals 59049

7 0

3 years ago

A fair die is rolled 12 times. the number of times an even number occurs on the 12 rolls has

bonufazy [111]

Answer:

Step-by-step explanation:

For a fair die, there are six likely options; 1, 2, 3, 4, 5, and 6

the probability of a even number is 3/6 = 0.5

Since the results of the die roll is independent and each trial is mutually exclusive, the distribution to explain the probability of occurrence will follow a binomial distribution such that n is the number of trials

x = number of successful throws

therefore for a Binomial distribution where

P(X =x) = nCx . P^x . (1-P)^ (n-x)

since p = 0.5, and n = 12, the distribution follows

P(X = x) = 12Cx . 0.5^x . (1 - 0.5)^(12- x)

= 12Cx . 0.5^x . 0.5)^(12- x)

where x = (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)

since we are interested in the probability of the number of times an even number occurs

it can occur either as P(X = 0), P(X =1), P(X =2), P(X =3), P(X =4), P(X =5), P(X =6), P(X =7), P(X =8), P(X =9), P(X =10), P(X =11), and P(X =12)

For no even number in 12 rolls,

P(X = 0) = 12C0 . 0.5^0 . 0.5^(12- 0) = 0.000244

For one even number in 12 rolls,

P(X = 1) = 12C1 . 0.5^1 . 0.5^(12- 1) = 0.002930

For two even number in 12 rolls,

P(X = 2) = 12C2 . 0.5^2 . 0.5^(12- 2) = 0.016113

For three even number in 12 rolls,

P(X = 3) = 12C3 . 0.5^3 . 0.5^(12- 3) = 0.053711

For four even number in 12 rolls,

P(X = 4) = 12C4 . 0.5^4 . 0.5^(12- 4) = 0.120850

For five even number in 12 rolls,

P(X = 5) = 12C5 . 0.5^5 . 0.5^(12- 5) = 0.193359

For six even number in 12 rolls,

P(X = 6) = 12C6 . 0.5^6 . 0.5^(12- 6) = 0.225586

For seven even number in 12 rolls,

P(X = 7) = 12C7 . 0.5^7 . 0.5^(12- 7) = 0.193359

For eight even number in 12 rolls,

P(X = 8) = 12C8 . 0.5^8 . 0.5^(12- 8) = 0.120850

For nine even number in 12 rolls,

P(X = 9) = 12C9 . 0.5^9 . 0.5^(12- 9) = 0.053711

For ten even number in 12 rolls,

P(X = 10) = 12C10 . 0.5^10 . 0.5^(12- 10) = 0.016113

For eleven even number in 12 rolls,

P(X = 11) = 12C11 . 0.5^11 . 0.5^(12- 11) = 0.002930

For twelve even number in 12 rolls,

P(X = 12) = 12C12 . 0.5^12 . 0.5^(12- 12) = 0.000244

Final test summation[P(X)] = 1

i.e.

P(X = 0) + P(X =1) + P(X =2) + P(X =3) + P(X =4) + P(X =5) + P(X =6) + P(X =7) + P(X =8) + P(X =9) + P(X =10) + P(X =11) + P(X =12) = 1

Hence since 0.000244 + 0.002930 + 0.016113 + 0.053711 + 0.120850 + 0.193359 + 0.225586 + 0.193359 + 0.120850 + 0.053711 + 0.016113 + 0.002930 + 0.000244 = 1.000000,

the probability value stands

7 0

3 years ago