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1 year ago

In Exercises 8-10, sketch the figure described.

Mathematics

1 answer:

Lunna [17]1 year ago

6 0

The sketch answers to question 8, 9 and 10 is given in the image attached.

<h3>What is an intersecting lines?</h3>

A link is known to be intersecting if two or more lines are said to have cross one another in a given plane.

Note that the intersecting lines are known to be one that often share a common point, and it is one that can be seen on all the intersecting lines, and it is known to be the point of intersection.

Looking at the image attached, you can see how plane A and line c intersecting at all points on line c and also GM and GH and line CD and plane X as they are not intersecting

Therefore, The sketch answers to question 8, 9 and 10 is given in the image attached.

Learn more about intersecting lines from

brainly.com/question/2065148

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please help me understand tell me what the answer is and explain I will give all my points. Thank You

Aleksandr-060686 [28]

Ok so Jesse is 10-15%
Margie is <10%
Pietro is >15%
Susana is 10%-15%
And I think Raul maybe >15%

Basically you can determine how far off they are by doing the number they guessed divided by the actual number (132) so for Jesse I did 120/132 and I got .909 and so on. Just take the two numbers after the decimal and that’s your percent. So the .909 is about 90% so now do 100% (of the 132) - 90 and you get 10. So I put 10%-15%

4 0

3 years ago

A sample of 100 cars driving on a freeway during a morning commute was drawn, and the number of occupants in each car was record

makkiz [27]

Answer:

$E(X)=1*0.74 +2*0.1 +3*0.11+ 4*0.03 +5*0.02=1.49$

$Var(X)=E(X^2)-[E(X)]^2 =3.11-(1.49)^2 =0.8899$

$Sd(X)=\sqrt{Var(X)}=\sqrt{0.8899}=0.943$

Step-by-step explanation:

For this case we have the following data given:

X 1 2 3 4 5

F 74 10 11 3 2

The total number of values are 100, so then we can find the empirical probability dividing the frequency by 100 and we got the followin distribution:

X 1 2 3 4 5

P(X) 0.74 0.10 0.11 0.03 0.02

Previous concepts

In statistics and probability analysis, the expected value "is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values".

The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).

And the standard deviation of a random variable X is just the square root of the variance.

Solution to the problem

In order to calculate the expected value we can use the following formula:

$E(X)=\sum_{i=1}^n X_i P(X_i)$

And if we use the values obtained we got:

$E(X)=1*0.74 +2*0.1 +3*0.11+ 4*0.03 +5*0.02=1.49$

In order to find the standard deviation we need to find first the second moment, given by :

$E(X^2)=\sum_{i=1}^n X^2_i P(X_i)$

And using the formula we got:

$E(X^2)=1^2 *0.74 +2^2 *0.1 +3^2 *0.11 +4^2 0.03 +5^2 *0.02=3.11$

Then we can find the variance with the following formula:

$Var(X)=E(X^2)-[E(X)]^2 =3.11-(1.49)^2 =0.8899$

And then the standard deviation would be given by:

$Sd(X)=\sqrt{Var(X)}=\sqrt{0.8899}=0.943$

7 0

3 years ago

400,000,000,000 90,000,000,000 6,000,000,000 60,000,000 300,000 40,000 2,000 800 10 1 in standard form

Greeley [361]

Perhaps you want the sum: 496,060,342,811.

7 0

3 years ago

Two solutions to y'' – 2y' – 35y = 0 are yı = e, Y2 = e -5t a) Find the Wronskian. W = 0 Preview b) Find the solution satisfying

pashok25 [27]

Answer:

a. $w(t)=-12e^{2t}$

b. $y(t)=-\frac{9}{2}e^{7t}-\frac{5}{2}e^{-5t}$

Step-by-step explanation:

We have a differential equation

y''-2 y'-35 y=0

Auxillary equation

$(D^2-2D-35)=0$

By factorization method we are finding the solution

$D^2-7D+5D-35=0$

$(D-7)(D+5)=0$

Substitute each factor equal to zero

D-7=0 and D+5=0

D=7 and D=-5

Therefore ,

General solution is

$y(x)=C_1e^{7t}+C_2e^{-5t}$

Let $y_1=e^{7t} \;and \;y_2=e^{-5t}$

We have to find Wronskian

$w(t)=\begin{vmatrix}y_1&y_2\\y'_1&y'_2\end{vmatrix}$

Substitute values then we get

$w(t)=\begin{vmatrix}e^{7t}&e^{-5t}\\7e^{7t}&-5e^{-5t}\end{vmatrix}$

$w(t)=-5e^{7t}\cdot e^{-5t}-7e^{7t}\cdot e^{-5t}=-5e^{7t-5t}-7e^[7t-5t}$

$w(t)=-5e^{2t}-7e^{2t}=-12e^{2t}$

a. $w(t)=-12e^{2t}$

We are given that y(0)=-7 and y'(0)=23

Substitute the value in general solution the we get

$y(0)=C_1+C_2$

$C_1+C_2=-7$ ....(equation I)

$y'(t)=7C_1e^{7t}-5C_2e^{-5t}$

$y'(0)=7C_1-5C_2$

$7C_1-5C_2=23$ ......(equation II)

Equation I is multiply by 5 then we subtract equation II from equation I

Using elimination method we eliminate $C_1$

Then we get $C_2=-\frac{5}{2}$

Substitute the value of $C_2$ in I equation then we get

$C_1-\frac{5}{2}=-7$

$C_1=-7+\frac{5}{2}=\frac{-14+5}{2}=-\frac{9}{2}$

Hence, the general solution is

b. $y(t)=-\frac{9}{2}e^{7t}-\frac{5}{2}e^{-5t}$

7 0

3 years ago

Suppose that it rains in Spain an average of once every 10 days, and when it does, hurricanes have a 8% chance of happening in H

Law Incorporation [45]

Answer:

The probability that it rains in Spain when hurricanes happen in Hartford is 0.1127

Step-by-step explanation:

This is a question where you use must use Bayes' Theorem.

The easiest way to do Bayes' type questions is to carefully define your terms.

Let R be the event that it is raining in Spain. R' is the event it isn't.

Let H be the event that it is hurricane in Hartford. H' is the event it isn't.

We know

P(R) = 1/10,

P(H | R) = 0.08,

P(H | R') = 0.07 and we want P(R | H).

Bayes Theorem says P(R | H) = [P(H | R)×P(R)] / P(H)

where

P(H) = P(H | R)×P(R) + P(H | R')×P(R')

Therefore,

P(R | H) = [P(H | R)×P(R)] / [P(H | R)×P(R) + P(H | R')×P(R')]

P(R | H) = [0.08 × 1/10] / [(0.08 × 1/10) + (0.07 × (1 - 1/10)]

P(R | H) = 8 / 71

P(R | H) = 0.1127

Therefore, the probability that it rains in Spain when hurricanes happen in Hartford is 0.1127.

7 0

3 years ago