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2 years ago

How can you tell if a relationship is proportiond?

1 answer:

6 0

A relationship is proportional if they represent the same kind of relationship.

One way to tell if it is proportioned is to write them as fractions and then reduce them. If the reduced fractions are the same, the relationship is proportional.

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Which inequality represents the problem?

kogti [31]

I believe A is correct.

8 0

3 years ago

Brad has recorded insect activity data for 40 days. He has recorded 1/6 of the data needed for his science fair project. How man

babymother [125]

Answer:

40/6= 6.6667(20/3)

Step-by-step explanation:

I divided the number of days into 6.

5 0

3 years ago

Read 2 more answers

Some people believe that second born subjects are more intelligent (higher IQ test). Given the data, your job is to perform the

nexus9112 [7]

Answer:

Step-by-step explanation:

When to use regression test?

This is done to find the relationship between 2 variables. One dependent and one independent.

When to use Multiple regression test?

Extension of regression test. It is used to predict value of 1 variable when it depends on 2-3 other variables. Multivariable, basically.

When to use 1-sided t-test?

We check for possibility of relationship of something from one side. 1 side effect. With a hypothesis and its alternate.

When to use 2-sided t-test?

This is same as 1-tailed but it looks for possibility from 2sides. A hypothesis and alternate, whether less or more.

Now, Options A and B are not right since we aren't comparing 2 variables here. People believe second born subjects are more intelligent. We have to perform a test to see whether this is true or not. One sided answer. So, we will use One-sided t-test.

6 0

3 years ago

Solve for x. 5(x 1) = 4(x 8)

tresset_1 [31]

Asssuming blanks are plus

5(x+1)=4(x+8)

distribute
a(b+c)=ab+ac

5(x+1)=5x+5

4(x+8)=4x+32

5x+5=4x+32
minus 4x from both sides
x+5=32
minus 5 from both sides
x=27

5 0

2 years ago

A Ferris wheel is 20 meters in diameter and boarded from a platform that is 2 meters above the ground. The six o'clock position

pantera1 [17]

Answer:

233.48s

3.84 min

Step-by-step explanation:

In order to solve this problem, we can start by drawing what the situation looks like. See attached picture.

We can model this situation by making use of a trigonometric function. Trigonometric functions have the following shape:

$y=A cos(\omega t+\phi)+C$

where:

A= amplitude =-20m because the model starts at the lowest point of the trajectory.

f= the function to use, in this case we'll use cos, since it starts at the lowest point of the trajectory.

t= time

$\omega=$ angular speed.

in this case:

$\omega=\frac{2\pi}{T}$

where T is the period, in this case 6 min or

$6min(\frac{60s}{1min})=360s$

so:

$\omega=\frac{2\pi}{360}$

$\omega = \frac{\pi}{160}$

and

$\phi$ = phase angle

C= vertical shift

in this case our vertical shift will be:

2m+20m=22m

in this case the phase angle is 0 because we are starting at the lowest point of the trajectory. So the equation for the ferris wheel will be:

$y=-20 cos(\frac{\pi}{180}t)+22$

Once we got this equation, we can figure out on what times the passenger will be higher than 13 m, so we build the following inequality:

$-20 cos(\frac{\pi}{180}t)+22>13$

so we can solve this inequality, we can start by turning it into an equation we can solve for t:

$-20 cos(\frac{\pi}{180}t)+22=13$

and solve it:

$-20 cos(\frac{\pi}{180}t)=13-22$

$-20 cos(\frac{\pi}{180}t)=-9$

$cos(\frac{\pi}{180}t)=\frac{9}{20}$

and we can take the inverse of cos to get:

$\frac{\pi}{180}t=cos^{-1}(\frac{9}{20})$

which yields two possible answers: (see attached picture)

$\frac{\pi}{180}t=1.104$ or $\frac{\pi}{180}t=5.179$

so we can solve the two equations. Let's start with the first one:

$\frac{\pi}{180}t=1.104$

$t =1.104(\frac{180}{\pi})$

t=63.25s

and the second one:

$\frac{\pi}{180}t=5.179$

$t=5.179(\frac{180}{\pi})$

t=296.73s

so now we can build our possible intervals we can use to test the inequality:

[0, 63.25] for a test value of 1

[63.25,296.73] for a test value of 70

[296.73, 360] for a test value of 300

let's test the first interval:

[0, 63.25] for a test value of 1

$-20 cos(\frac{\pi}{180}(1))+22>13$

2>13 this is false

let's now test the second interval:

[63.25,296.73] for a test value of 70

$-20 cos(\frac{\pi}{180}(70))+22>13$

15.16>13 this is true

and finally the third interval:

[296.73, 360] for a test value of 300

$-20 cos(\frac{\pi}{180}(300))+22>13$

12>13 this is false.

We only got one true outcome which belonged to the second interval:

[63.25,296.73]

so the total time spent above a height of 13m will be:

196.73-63.25=233.48s

which is the same as:

$233.48(\frac{1min}{60s})=3.84 min$

see attached picture for the graph of the situation. The shaded region represents the region where the passenger will be higher than 13 m.

3 0

3 years ago