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3 years ago

Explain why 4:6 and 8: 10 are not equivalent ratios.

Mathematics

2 answers:

brilliants [131]3 years ago

8 0

Answer:

Step-by-step explanation:

You have to find a common denominator for both ratios

if you were to multiply the 4:6 ratio by ten it would give you 40:60

if you were to multiply 8:10 by 6 you would get 48:60

48:60 is greater than 40:60

andrew11 [14]3 years ago

4 0

Answer:

Because when 4:6 is simplified, you get 2:3, but when 8:10 is simplified, you get 4:5

So, they are not equivalent ratios because their simplified ratios are not the same.

Step-by-step explanation:

Hope this is helpful.

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Solve for C<br><br> C<br> — - 4.3 = 11.5 <br> 2

Maslowich

Answer:

The correct answer is,

$c = 31.6$

7 0

3 years ago

A company selling light bulbs claims in its advertisements that its light bulbs’ average life is 1000 hours. In fact, the life s

Ket [755]

Answer:

The probability that a randomly chosen light bulb will last less than 900 hours is 0.1587.

Step-by-step explanation:

The life span of these light bulbs is normally distributed with a mean of 1000 hours and a standard deviation of 100 hours

Mean = $\mu = 1000 hours$

Standard deviation = $\sigma = 100 hours$

We are supposed to find the probability that a randomly chosen light bulb will last less than 900 hours.i.e. P(x<900)

So, $Z=\frac{x-\mu}{\sigma}$

$Z=\frac{900-1000}{100}$

Z=-1

P(x<900)=P(z<-1)=0.1587

Hence the probability that a randomly chosen light bulb will last less than 900 hours is 0.1587.

4 0

3 years ago

The center of a circle is the point (3, 2). if the point (6, −2) lies on this circle, find the standard equation for the circle.

garri49 [273]

Given
centre of circle O(3,2)=O(x0,y0)
point on circle P(6,-2)

Standard equation of circle:
(x-x0)^2+(y-y0)^2=r^2

r=radius of circle
= (distance OP)
= sqrt((6-3)^2+(-2-2)^2)
=sqrt(3^2+(-4)^2)
=sqrt(25)

=>
r^2=(sqrt(25))^2=25

Equation of circle
(x-x0)^2+(y-y0)^2=r^2
(x-3)^2+(y-2)^2=25 ............... standard equation of circle

5 0

4 years ago

Solve 3y^{2} - (y + 2) (y - 2)

hram777 [196]

Answer:

3y^2 - (y + 2) (y - 2) = 0

<=> 3y^2 - (y^2 - 4) = 0

<=> 2y^2 + 4 =0

<=>y^2 + 2 = 0

=> Because y^2 is always equal or larger than 0, there is no real solution.

Hope this helps!

8 0