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2 years ago

Find the distance between the

Mathematics

1 answer:

Dovator [93]2 years ago

7 0

Answer:

$\displaystyle d = \sqrt{29}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Coordinates (x, y)

Algebra II

Distance Formula: $\displaystyle d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Step-by-step explanation:

*Note:

The distance formula is derived from the Pythagorean Theorem.

Step 1: Define

Identify

Point (5, 10)

Point (10, 12)

Step 2: Find distance d

Substitute in points [Distance Formula]: $\displaystyle d = \sqrt{(10 - 5)^2 + (12 -10)^2}$
[√Radical] (Parenthesis) Subtract: $\displaystyle d = \sqrt{5^2 + 2^2}$
[√Radical] Evaluate exponents: $\displaystyle d = \sqrt{25 + 4}$
[√Radical] Add: $\displaystyle d = \sqrt{29}$

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suppose X and Y are independent random variables, both with normal distributions. If X has a mean of 45 with a standard deviatio

djyliett [7]

Answer:

0.9772 = 97.72% probability that a randomly generated value of X is greater than a randomly generated value of Y

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question:

$\mu_X = 45, \sigma_X = 4, \mu_Y = 35, \sigma_Y = 3$

What is the probability that a randomly generated value of X is greater than a randomly generated value of Y

This means that the subtraction of X by Y has to be positive.

When we subtract two normal variables, the mean is the subtraction of their means, and the standard deviation is the square root of the sum of their variances. So

$\mu = \mu_X - \mu_Y = 45 - 35 = 0$

$\sigma = \sqrt{\sigma_X^2+\sigma_Y^2} = \sqrt{25} = 5$

We want to find P(X > 0), that is, 1 subtracted by the pvalue of Z when X = 0. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0 - 10}{5}$

$Z = -2$

$Z = -2$ has a pvalue of 0.0228

1 - 0.0228 = 0.9772

0.9772 = 97.72% probability that a randomly generated value of X is greater than a randomly generated value of Y

4 0

3 years ago

Solving an equation involving complementary or supplementary angles! Please help I really would appreciate it

Schach [20]

2x + x + 33 = 90
3x + 33 = 90
3x = 57
x = 19
Angle 1 = 38°
Angle 2 = 52°

8 0

3 years ago

ASHA 777 [7]

Answer:

Step-by-step explanation:

Vertex A of the triangle ABC when rotated by 90° counterclockwise about the origin,

Rule to be followed,

A(x, y) → P(-y, x)

Therefore, A(1, 1) → P(-1, 1)

Similarly, B(3, 2) → Q(-2, 3)

C(2, 5) → R(-5, 2)

Triangle given in second quadrant will be the triangle PQR.

If the point P of triangle PQR is reflected across a line y = x,

Rule to be followed,

P(x, y) → X(y, x)

P(-1, 1) → X(1, -1)

Similarly, Q(-2, 3) → Y(3, -2)

R(-5, 2) → Z(2, -5)

Therefore, triangle given in fourth quadrant is triangle XYZ.

6 0

3 years ago

Help me, Thanks!!

Marrrta [24]

2.) dT/dt = -t^2 + 3t

dT = (-t^2 + 3t)dt

T = -t^3/3 + 3t^2/2 + C

at 4:00 AM

80 = -4^3/3 + 3*4^2/2 + C

C = 80 + 4^3/3 - 3*4^2/2

C = 77.33

T = -t^3/3 + 3t^2/2 + 77.33

at 8:00 AM

T = -8^3/3 + 3*8^2/2 + 77.33

T = 2.67 deg

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3 years ago

Finding the geometric mean

grigory [225]

Answer:

To find the geometric mean of two numbers, just find the product of those numbers and take the square root!

Step-by-step explanation:

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2 years ago