All categories

3 years ago

PLEASE HELP 4 QUESTIONS 90 POINTS

Mathematics

2 answers:

kondor19780726 [428]3 years ago

4 0

1. n^2 -8n +16 = 25

Subtract 25 from both sides

n^2 - 8n + 16 - 25 = 0

Simplify

n^2 - 8n - 9 =0

Factor

(n-9)(n+1) = 0

Solve for n

n-9 = 0, n = 9

n+1 = 0, n = -1

Solution: 9,-1

2. C = b^2/25

Multiply both sides by 25:

25c = b^2

Take square root of both sides

b = +/-√25c

Simplify:

b = 5√C, -5√C

3. d = 16t^2 +12t

subtract d from both side:

16t^2 + 12t -d =0

Use quadratic formula to solve:

t = (3 +/-√(9-4d))/8

4. 5w^2 +10w =40

Subtract 40 from both side:

5w^2 + 10w -40 = 0

Factor:

5(w-2)(w+4)=0

Divide both sides by 5:

(w-2)(w+4)=0

Solve for w:

w-2 = 0, w = 2

w+4=0, w = -4

Solution: 2,-4

spayn [35]3 years ago

3 0

Answer:

n^2 − 8n + 16=25

C = b^2/25

d = −16t^2 + 12t

5w^2 + 10w = 40

Step-by-step explanation:

You might be interested in

In Exercises 51−56, the letters a, b, and c represent nonzero constants. Solve the equation for x

OverLord2011 [107]

Answer:

x = ¾-a

Step-by-step explanation:

x + a = ¾

Subtract a from each side

x + a -a= ¾-a

x = ¾-a

6 0

2 years ago

Read 2 more answers

Suppose we roll a fair die and let X represent the number on the die. (a) Find the moment generating function of X. (b) Use the

Likurg_2 [28]

Answer:

(a) moment generating function for X is $\frac{1}{6}\left(e^{t}+e^{2 t}+e^{2 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$

(b) $\mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}$

Step-by step explanation:

Given X represents the number on die.

The possible outcomes of X are 1, 2, 3, 4, 5, 6.

For a fair die, $P(X)=\frac{1}{6}$

(a) Moment generating function can be written as $M_{x}(t)$ .

$M_x(t)=\sum_{x=1}^{6} P(X=x)$

$M_{x}(t)=\frac{1}{6} e^{t}+\frac{1}{6} e^{2 t}+\frac{1}{6} e^{3 t}+\frac{1}{6} e^{4 t}+\frac{1}{6} e^{5 t}+\frac{1}{6} e^{6 t}$

$M_x(t)=\frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$

(b) Now, find $E(X) \text { and } E\((X^{2})$ using moment generating function

$M^{\prime}(t)=\frac{1}{6}\left(e^{t}+2 e^{2 t}+3 e^{3 t}+4 e^{4 t}+5 e^{5 t}+6 e^{6 t}\right)$

$M^{\prime}(0)=E(X)=\frac{1}{6}(1+2+3+4+5+6)$

$\Rightarrow E(X)=\frac{21}{6}$

$M^{\prime \prime}(t)=\frac{1}{6}\left(e^{t}+4 e^{2 t}+9 e^{3 t}+16 e^{4 t}+25 e^{5 t}+36 e^{6 t}\right)$

$M^{\prime \prime}(0)=E(X)=\frac{1}{6}(1+4+9+16+25+36)$

$\Rightarrow E\left(X^{2}\right)=\frac{91}{6}$

Hence, (a) moment generating function for X is $\frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$ .

(b) $\mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}$

6 0

3 years ago

Help me pls ........................

Ymorist [56]

Answer:

The answer is

$y = 3x + 5$

Step-by-step explanation:

❃Incase you forgot what the linear equation formula is ↶

$y = mx + b$

❃Since we already have the slope, we don't need to solve for that.

➊ First: We are going to find the y-intercept.

$y = mx + b \\ 2 = 3( - 1) + b \\ 2 = - 3 + b \\ \frac{ + 3 = + 3 \: \: \: \: \: }{5 = b}$

➋Second: Plug in.

$y = 3x + 5$

7 0

2 years ago

Melissa has reached her credit card limit of $6,500. Since Melissa has poor credit history, the annual interest rate on her cred

GrogVix [38]

You my 220+235 and then you get the Nader

4 0

2 years ago

How many possible outcomes are there when flipping a coin 9 times?

otez555 [7]

Each time the coin is tossed, there are two possible outcomes. The total number of possible outcomes when 9 tosses are made is given by:
$2^{9}=512\ outcomes$

3 0

3 years ago