Nad podaną liczba wpisz liczbę 2 razy większa a pod nią 2 razy mniejsza

Paraphin [41]

Answer:

If you want to use the Rational Zeros Theorem, as instructed, you need to use synthetic division to find zeros until you get a quadratic remainder.

P: ±1, ±2, ±3, ±6 (all prime factors of constant term)

Q: ±1, ±7 (all prime factors of the leading coefficient)

P/Q: ±1, ±2, ±3, ±6, ±1/7, ±2/7, ±3/7, ±6/7 (all possible values of P/Q)

Now, start testing your values of P/Q in your polynomial:

f(x)=7x4-9x3-41x2+13x+6

You can tell f(1) and f(-1) are not zeros since they're not = 0Now try f(2) and f(-2):

f(2)=7(16)-9(8)-41(4)+13(2)+6

112-72-164+26+6 ≠ 0

f(-2)=7(16)-9(-8)-41(4)+13(-2)+6

112+72-164-26+6 = 0 OK!! There is a zero at x=-2

This means (x+2) is a factor of the polynomial.

Now, do synthetic division to find the polynomial that results from

(7x4-9x3-41x2+13x+6)÷(x+2):

-2⊥ 7 -9 -41 13 6

-14 46 -10 -6

7 -23 5 3 0 The remainder is 0, as expected

The quotient is a polynomial of degree 3:

7x3-23x2+5x+3

Now, continue testing the P/Q values with this new polynomial. Try f(3):

f(3)=7(27)-23(9)+5(3)+3

189-207+15+3 = 0 OK!! we found another zero at x=3

Now, another synthetic division:

3⊥ 7 -23 5 3

21 -6 -3_

7 -2 -1 0

The quotient is a quadratic polynomial:

7x2-2x-1 This is not factorable, you need to apply the quadratic formula to find the 3rd and 4th zeros:

x= (1±2√2)÷7

The polynomial has 4 zeros at x=-2, 3, (1±2√2)÷7

Step-by-step explanation:

7 0

3 years ago

X and y are normal random variables with e(x) = 2, v(x) = 5, e(y) = 6, v(y) = 8 and cov(x,y)=2. determine the following: e(3x 2y

andriy [413]

The result for the given normal random variables are as follows;

a. E(3X + 2Y) = 18

b. V(3X + 2Y) = 77

c. P(3X + 2Y < 18) = 0.5

d. P(3X + 2Y < 28) = 0.8729

<h3>What is normal random variables?</h3>

Any normally distributed random variable having mean = 0 and standard deviation = 1 is referred to as a standard normal random variable. The letter Z will always be used to represent it.

Now, according to the question;

The given normal random variables are;

E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8.

Part a.

Consider E(3X + 2Y)

$\begin{aligned}E(3 X+2 Y) &=3 E(X)+2 E(Y) \\&=(3) (2)+(2)(6 )\\&=18\end{aligned}$

Part b.

Consider V(3X + 2Y)

$\begin{aligned}V(3 X+2 Y) &=3^{2} V(X)+2^{2} V(Y) \\&=(9)(5)+(4)(8) \\&=77\end{aligned}$

Part c.

Consider P(3X + 2Y < 18)

A normal random variable is also linear combination of two independent normal random variables.

$3 X+2 Y \sim N(18,77)$

Thus,

$P(3 X+2 Y < 18)=0.5$

Part d.

Consider P(3X + 2Y < 28)

$Z=\frac{(3 X+2 Y-18)}{\sqrt{77}}$

$\begin{aligned} P(3X + 2Y < 28)&=P\left(\frac{3 X+2 Y-18}{\sqrt{77}} < \frac{28-18}{\sqrt{77}}\right) \\&=P(Z < 1.14) \\&=0.8729\end{aligned}$

Therefore, the values for the given normal random variables are found.

To know more about the normal random variables, here

brainly.com/question/23836881

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The correct question is-

X and Y are independent, normal random variables with E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8. Determine the following:

a. E(3X + 2Y)

b. V(3X + 2Y)

c. P(3X + 2Y < 18)

d. P(3X + 2Y < 28)

8 0

2 years ago

Is 8,749,628 divisible by 8?<br><br> Help me plz

CaHeK987 [17]

Answer:

so yes

Step-by-step explanation:

If you can divide two numbers without a remainder, then the first number is divisible by the second.

For example, 12 is divisible by 2.

12÷2=6

But 12 is not divisible by 5. When you divide 12 by 5, you get a remainder.

12÷5=2 R2

8 0

2 years ago

Read 2 more answers

Jason takes 2 centiliters of medicine. How many milliliters is this?

spin [16.1K]

Jason has taken 20ml

7 0

3 years ago

Read 2 more answers

Mr. lira opened an account with a deposit of $2,000.

NNADVOKAT [17]

Answer:

A) 25%

Step-by-step explanation:

Simple interest rate = $3,500

Principal, p = $2,000

Time, t = 7 years

Rate, r = ?

Simple interest rate = (principal × rate × time) / 100

3,500 = (2,000 × r × 7) / 100

3,500 * 100 = 14,000r

350,000 = 14,000r

r = 350,000/14,000

r = 25%

8 0

2 years ago