A square has an area of 9 cm2 what is its side length​

Answers

Answer 1
Answer:

Value of side length is 3 cm

Give that;

Area of square = 9 cm²

FInd:

Value of side length

Computation:

We know that,

Area of square = Side²

BY putting value into formula

⇒ Side² = 9

⇒ Side² = 3²

Side = 3

So,

Value of side length = 3 cm

Learn more about square;

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Answer 2
Answer:

Answer:

Length side is 3 cm

Step-by-step explanation:

the length in square is called side.

A= sides X sides

9 = s^2

S= 3

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there is a bag filled with 3 blue and 4 red marbles. A marble is taken at random from the bag, the colour is noted and then it is not replaced. another marble is taken at random. what is the probability of getting 2 reds ?
F (x)=2x-1 Find f (-2)​

Let (Ω,F,P) be a probability space. Recall the R.V. X is called simple, if there exist A₁, …., Aₙ ∈ F and b₁, …, bₙ ∈ R such that X = ᵢ₌ₗ∑ⁿ bᵢ 1{A ᵢ }. Note that A; need not be disjoint. ...) ... a) Show that if R.V. X,Y are simple, then so is the linear combination aX + bY. b) Is it true that if X₁, X₂ are all simple then so are X₁ + X₂, and X₁ ∧ X₂ = min X1, X₂? c) Is the representation of X (i.e. the sets A₁,.., Aₙ and the scalars b₁, ..., bₙ) unique? d) What if Aᵢ are known to be disjoint? How many values can X take then? e) Show that R.V. X is simple if and only it takes finitely many values. f) Show that E[X] is well defined for X simple R.V.. In other words, if X has two representations: X = ᵢ₌ₗ∑ⁿ bᵢ 1{Aᵢ } and X = ᵢ₌ₗ∑ᵐ Cᵢ 1{Bᵢ), such that B₁, ..., Bₘ ∈ F and C₁, ..., Cₘ ∈ R, then we must have ᵢ₌ₗ∑ⁿ Bᵢ P(Aᵢ) = ᵢ₌ₗ∑ᵐ cᵢP(Bᵢ).

Answers

Answer:

see below

Step-by-step explanation:

a) If random variables X and Y are simple, then their linear combination aX + bY is also simple.

b) If X₁ and X₂ are simple, then X₁ + X₂ and X₁ ∧ X₂ are also simple.

c) The representation of a simple random variable X is not unique. There can be different sets of {Aᵢ} and corresponding scalars {bᵢ} that represent the same X.

d) If the sets Aᵢ are known to be disjoint, then X can take as many unique values as there are disjoint sets {Aᵢ}.

e) A random variable X is simple if and only if it takes a finite number of values.

f) The expected value E[X] is well-defined for a simple random variable X, regardless of its representation. It can be calculated using the formula E[X] = ∑ᵢ bᵢ P(Aᵢ) or E[X] = ∑ᵢ cᵢ P(Bᵢ), where {Aᵢ} and {Bᵢ} are sets and {bᵢ} and {cᵢ} are scalars representing X.

Simplify your answer and write it as a proper fraction,improper fraction,or interger

Answers

Answer:

2

Step-by-step explanation:

Take any 2 points on the graph,

(50, 10) & (90, 90)

Use the slope formula:

(y2-y1)/(x2-x1)

= (90-10)/(90-50)

= (80)/(40)

= 2

Hence, the slope is 2.

Feel free to mark this as brainliest! :D

Point (1,-5) 6. The slope is 5 , and the line passes through the (-1,-3).

Answers

Step-by-step explanation:

You have given a point (1, -5), a slope of 5, and the line passes through the point (-1, -3). You can use this information to find the equation of the line.

The equation of a line in point-slope form is:

y - y1 = m(x - x1)

Where (x1, y1) is a point on the line, and m is the slope.

Using the given point (1, -5) and the slope m = 5, you can plug these values into the equation:

y - (-5) = 5(x - 1)

Now, simplify:

y + 5 = 5(x - 1)

To put it in slope-intercept form (y = mx + b), expand and solve for y:

y + 5 = 5x - 5

Subtract 5 from both sides:

y = 5x - 5 - 5

y = 5x - 10

So, the equation of the line with a slope of 5 that passes through the point (1, -5) is:

y = 5x - 10

ACT mathematics score for a particular year are normally distributed with a mean of 28 and a standard deviation of 2.4 pointsA. What is the probability a randomly selected score is greater than 30.4?

B. what is the probability a randomly selected score is less than 32.8?

C. What is the probability a randomly selected score is Between 25.6 and 32.8?

Answers

Answer:

a.   0.1587

b.  0.8849

c.  0.1814

Step-by-step explanation:

a. Given that \mu=28, \ \ \sigma=2.4

-The probability a randomly selected score is greater than 30.4 is calculated as:

P(X>30.4)=1-P(X<30.4)\n\nz=(\bar X-\mu)/(\sigma)\n\n=(30.4-28)/(2.4)=1\n\n\therefore P(X>30.4)=1-P(z<1)\n\n=1-0.84134\n\n=0.1587

Hence, the probability of a score greater than 30.4 is 0.1587

b. Given that \mu=28, \ \ \ \sigma=2.4

The probability a randomly selected score is less than 32.8 is calculated as:

P(X<32.8)=P(z<(\bar X-\mu)/(\sigma))\n\nz=(\bar X-\mu)/(\sigma)\n\n=(32.8-28)/(2.4)=1.2\n\nP(X<32.8)=P(z<1.2)\n\n=0.88493

Hence, the probability that a randomly selected score is less than 32.8 is 0.8849

c. The probability that a score is between 25.6 and 32.8 is calculated as follows:

P(25.6<X<32.8)=P((\bar X-\mu)/(\sigma)<z<(\bar X-\mu)/(\sigma))\n\n=P((25.6-28)/(2.4)<z<(32.8-28)/(2.4))\n\n=P(-1<z<2.0)\n\n=0.15866+(1-0.97725)\n\n=0.1814

Hence, the probability is 0.1814

Convert the fraction to higher terms. Enter the value of the missing numerator 3/10=?/80

Answers

3/10 multiply with 8 on both numerator and denominator

therefore, 3/10 = 3x8/10x8 =  24/80

I think the answer is 24/80 because you have multiplied the denominator by 8 to get to ?/80 from 3/10. Therefore, you need to multiply 3 by 8 also, which when you do, you get 24/80

Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth. −2y2 + 6y = −2

Answers

-2y^2 + 6y + 2 = 0
a = -2, b = 6, c = 2
x = (- b + - sqrt(b^2-4ac))/(2a)
x=(-6+-sqrt(36-4*-2*2))/-4
x=(-6+-sqrt(36+16))/-4
x=(-6+-sqrt(52))/-4
x = (-6 +- 2sqrt13)/-4
x = (3 + - sqrt13)/2

Answer:

-0.3 Or 3.3

Step-by-step explanation:

-2y^2 + 6y + 2 = 0

X = -6 + 7.21/-4 or -7 - 7.21/-4

X = -0.3 or X = 3.3