Help ASAP! Will name brainliest!

Lostsunrise [7]

easy 214.7 because if it was rounded it would be 214.700 an this is way bigger than 214.275

8 0

3 years ago

A quadratic equation ax^2+bx+c=0 has -11 and 4 as solutions. Find the values of b and c if the value of a is 1 (hint, use zero f

Alina [70]

Answer:

b = 7, c = -44

Step-by-step explanation:

If the quadratic equation has the solutions -11 and 4, the two factors are:

$(x+11)(x-4)=0$

Since when we use the zero factor property we get

x+11=0 ⇒ x= -11

x-4=0 ⇒ x=4

Thus, we have used the zero factor property in reverse to find the factorization of the quadratic equation.

Now we develop the multiplications between parenthesis:

$(x+11)(x-4)=0\\x^2-4x+11x-44=0\\x^2+7x-44=0$

So b is the number that accompanies the x: b = 7

and c is the independent number: c = -44

3 0

3 years ago

The shapes in this question are drawn on a centimetre grid.

xenn [34]

Answer:

A. 12 centimeters B. 4 centimeters C. 20 centimeters

Step-by-step explanation:

To find area you need to do the Height time the Length.

For A it is 3x4

For B it is 4x1

And for C it is 4x5

You get this equation by multiplying the amount of squares that the Height and Length measured!

Hope this helped!

:DD

8 0

3 years ago

Read 2 more answers

What is the reference angle for 120°?

Drupady [299]

Since the angle 240° 240 ° is in the third quadrant, the reference angle formula is Ar=Ac−180°<span> A r = A c </span>- 180 °<span> .</span>

4 0

3 years ago

Read 2 more answers

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

#SPJ4

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