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2 years ago

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Foster’s answer to the question he wrote for Ranger was 7x + (–3), while Ranger’s answer was 7x – 3. They both checked their wor

k by letting x = 5, and they both claim to be correct. Are they both correct? Explain.

2 answers:

gladu [14]2 years ago

6 0

Both Foster and Ranger are correct because subtracting 3 is the same as adding –3. If you let x = 5 in either of the expressions, you get the answer 32.

yan [13]2 years ago

4 0

Answer:

Yes, they are both correct.

Step-by-step explanation:

Subtracting is the same as adding a negative number.

Ranger's answer was 7x-3; this is the same as 7x+(-3). This means they are both correct.

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Mrs. Matthews wants to have $18,000 in the bank in 2 years. If she deposits $9000 today at 6% compounded quarterly for 2 years,

OverLord2011 [107]

Depends if you want
1. find how much he will earn, find the differnce between that and 18000
2. see how much to invest till he will get 18000

A= $P(1+ \frac{r}{n})^{nt}$

A=futre amount
P=present amout
r=rate in decimal
n=number of times per year ccompounded
t=time in years

1.
A=?
P=9000
r=0.06
n=4 (quarter means 4 times per year)
t=2
?= $9000(1+ \frac{0.06}{4})^{(4)(2)}$
?= $9000(1+ 0.015)^{8}$
?= $9000(1.015)^{8}$
?=10138.4 will be earned
18000-10138.4=7861.6 needed

2.
A=18000
P=9000+x
r=0.06
n=4 (quarter means 4 times per year)
t=2
18000= $(9000+x)(1+ \frac{0.06}{4})^{(4)(2)}$
18000= $(9000+x)(1+ 0.015)^{8}$
18000= $(9000+x)(1.015)^{8}$
divide both sides by 1.015^8
15978.8=9000+x
minus 9000 both sides
6978.8 needed

if he willnot be investing any more, he needs $7861.6 more
if he will invest more he will need to invest $6978.8 more

6 0

2 years ago

Read 2 more answers

Assume that two marbles are drawn without replacement from a box with 1 blue, 3 white, 2 green, and 2 red marbles. find the prob

navik [9.2K]

Before I answer the question i am going to start with the blue Marble you would have a 1/8 chance of drawing the blue marble from the box and after that you would have a 3/7 of drawing a white marble because you did not replace the blue marble and I hope this help.

8 0

2 years ago

Which function has a range of y < 3?

Setler79 [48]

Using a graphing tool

Let's graph each of the cases to determine the solution of the problem

<u>case A)</u> $y=3(2^{x})$

see the attached figure N $1$

The range is the interval--------> (0,∞)

$y> 0$

therefore

the function $y=3(2^{x})$ is not the solution

<u>case B)</u> $y=2(3^{x})$

see the attached figure N $2$

The range is the interval--------> (0,∞)

$y> 0$

therefore

the function $y=2(3^{x})$ is not the solution

<u>case C)</u> $y=-2^{x}+3$

see the attached figure N $3$

The range is the interval--------> (-∞,3)

$y< 3$

therefore

the function $y=-2^{x}+3$ is the solution

<u>case D)</u> $y=2^{x}-3$

see the attached figure N $4$

The range is the interval--------> (-3,∞)

$y>-3$

therefore

the function $y=2^{x}-3$ is not the solution

<u>the answer is</u>

$y=-2^{x}+3$

5 0

2 years ago

Read 2 more answers

Uma urna contém 10 bolas identificadas pelas letras A, B, ..., J. Uma bola é extraída ao acaso da urna e sua letra é observada.

arlik [135]

Answer:

a) Probability that a letter of the drawn ball is vowel = (3/10) = 0.30

b) Probability that a letter of the drawn ball is consonant = (7/10) = 0.70

a) Probabilidade de uma letra da bola sacada ser vogal = (3/10) = 0.30

b) Probabilidade de uma letra da bola sacada ser consoante = (7/10) = 0,70

Step-by-step explanation:

English Translation

A ballot box contains 10 balls identified by the letters A, B, ..., J. A ball is drawn at random from the ballot box and its letter is observed. (Make the sample space and all events explicit). What is the probability that a letter of the drawn ball is: a) Vowel b) Consonant

Solution

The 10 balls are identified by A, B, C, D, E, F, G, H, I and J

Note that the probability of an event is given as the number of elements in that event divided by the number of elements in the sample space.

a) Probability of drawing a ball that has a vowel letter = P(v)

P(v) = n(v) ÷ n(S)

n(v) = Number of balls with vowel letters = 3 (that is, A, E and I)

n(S) = Total number of balls = 10

P(v) = (3/10) = 0.30

b) Probability of drawing a ball that has a consonant letter = P(c)

P(c) = n(c) ÷ n(S)

n(c) = Number of balls with consonant letters = 7 (that is, B, C, D, F, G, H and J)

n(S) = Total number of balls = 10

P(c) = (7/10) = 0.70

In Portugese/Em português

As 10 bolas são identificadas por A, B, C, D, E, F, G, H, I e J.

Observe que a probabilidade de um evento é fornecida como o número de elementos nesse evento dividido pelo número de elementos no espaço de amostra.

a) Probabilidade de desenhar uma bola com uma letra de vogal = P (v)

P (v) = n (v) ÷ n (S)

n (v) = Número de bolas com letras de vogal = 3 (ou seja, A, E e I)

n (S) = Número total de bolas = 10

P (v) = (3/10) = 0,30

b) Probabilidade de desenhar uma bola com uma letra consoante = P (c)

P (c) = n (c) ÷ n (S)

n (c) = Número de bolas com letras consoantes = 7 (ou seja, B, C, D, F, G, H e J)

n (S) = Número total de bolas = 10

P (c) = (7/10) = 0,70

Hope this Helps!!!!

Espero que isto ajude!!!!

3 0

2 years ago

Solve the recurrence relation: hn = 5hn−1 − 6hn−2 − 4hn−3 + 8hn−4 with initial values h0 = 0, h1 = 1, h2 = 1, and h3 = 2 using (

musickatia [10]

(a) Suppose $h_n=r^n$ is a solution for this recurrence, with $r\neq0$ . Then

$r^n=5r^{n-1}-6r^{n-2}-4r^{n-3}+8r^{n-4}$
$\implies1=\dfrac5r-\dfrac6{r^2}-\dfrac4{r^3}+\dfrac8{r^4}$
$\implies r^4-5r^3+6r^2+4r-8=0$
$\implies (r-2)^3(r+1)=0\implies r=2,r=-1$

So we expect a general solution of the form

$h_n=c_1(-1)^n+(c_2+c_3n+c_4n^2)2^n$

With $h_0=0,h_1=1,h_2=1,h_3=2$ , we get four equations in four unknowns:

$\begin{cases}c_1+c_2=0\\-c_1+2c_2+2c_3+2c_4=1\\c_1+4c_2+8c_3+16c_4=1\\-c_1+8c_2+24c_3+72c_4=2\end{cases}\implies c_1=-\dfrac8{27},c_2=\dfrac8{27},c_3=\dfrac7{72},c_4=-\dfrac1{24}$

So the particular solution to the recurrence is

$h_n=-\dfrac8{27}(-1)^n+\left(\dfrac8{27}+\dfrac{7n}{72}-\dfrac{n^2}{24}\right)2^n$

(b) Let $G(x)=\displaystyle\sum_{n\ge0}h_nx^n$ be the generating function for $h_n$ . Multiply both sides of the recurrence by $x^n$ and sum over all $n\ge4$ .

$\displaystyle\sum_{n\ge4}h_nx^n=5\sum_{n\ge4}h_{n-1}x^n-6\sum_{n\ge4}h_{n-2}x^n-4\sum_{n\ge4}h_{n-3}x^n+8\sum_{n\ge4}h_{n-4}x^n$
$\displaystyle\sum_{n\ge4}h_nx^n=5x\sum_{n\ge3}h_nx^n-6x^2\sum_{n\ge2}h_nx^n-4x^3\sum_{n\ge1}h_nx^n+8x^4\sum_{n\ge0}h_nx^n$
$G(x)-h_0-h_1x-h_2x^2-h_3x^3=5x(G(x)-h_0-h_1x-h_2x^2)-6x^2(G(x)-h_0-h_1x)-4x^3(G(x)-h_0)+8x^4G(x)$
$G(x)-x-x^2-2x^3=5x(G(x)-x-x^2)-6x^2(G(x)-x)-4x^3G(x)+8x^4G(x)$
$(1-5x+6x^2+4x^3-8x^4)G(x)=x-4x^2+3x^3$
$G(x)=\dfrac{x-4x^2+3x^3}{1-5x+6x^2+4x^3-8x^4}$
$G(x)=\dfrac{17}{108}\dfrac1{1-2x}+\dfrac29\dfrac1{(1-2x)^2}-\dfrac1{12}\dfrac1{(1-2x)^3}-\dfrac8{27}\dfrac1{1+x}$

From here you would write each term as a power series (easy enough, since they're all geometric or derived from a geometric series), combine the series into one, and the solution to the recurrence will be the coefficient of $x^n$ , ideally matching the solution found in part (a).

3 0

2 years ago