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2 months ago

What is the value of x in the equation 8x – 2y = 48, when y = 4?

Mathematics

1 answer:

tester [12.3K]2 months ago

8 0

Respuesta:

Explicación paso a paso:

8x-2y=48, sustituir y por 4, 8x-2(4)=48, multiplicamos, 8x-8=48, sumamos 8 a ambos lados para deshacernos del -8, 8x=56 y finalmente, dividimos entre 8 para eliminar el 8 de x, x=7.

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On a coordinate plane, a curved line with a minimum value of (0.8, negative 11.4) and maximum values of (negative 1.6, 56) and (

Leona [12618]

Answer:

2,4

I completed it and got it correct

7 1

2 months ago

In a circle, an arc 60 centimeters long subtends an angle of 5 radians. What is the radius of the circle?

Inessa [12570]

Got it!
There are 2π radians in a complete circle.

Now, let's calculate the circumference.
5/2π = 60/circumference.
Next, solve for the circumference.
By multiplying both sides by 2π, we have: 5 * circumference = 120π.
Now divide both sides by 5, and we find: circumference = 24π.

Using the formula c = 2πr,
we set 24π = 2πr.
Dividing both sides by 2π gives us r = 12. Thus, the radius measures 12cm.

6 0

3 months ago

Read 2 more answers

In Gallup's Annual Consumption Habits Poll, telephone interviews were conducted for a random sample of 1,014 adults aged 18 and

babunello [11817]

Answer:

(b)E(x)=1.3087

(c)Variance of x =1.7119

(d)E(y)=2.0447

Step-by-step explanation:

The random variable x refers to the average number of cups of coffee consumed daily.

The total number of respondents equals 1014

(a) Probability distribution for x.

$\left|\begin{array}{c|cccccc}x&0&1&2&3&4\\\\P(x)&\dfrac{365}{1014}&\dfrac{264}{1014}&\dfrac{193}{1014}&\dfrac{91}{1014}&\dfrac{101}{1014} \end{array}\right|$

(b) Expected value for x

$E(x)=\left(0\times\dfrac{365}{1014}\left)+\left(1\times\dfrac{264}{1014}\left)+\left(2\times\dfrac{193}{1014}\left)+\left(3\times\dfrac{91}{1014}\left)+\left(4\times\dfrac{101}{1014}\right)$

E(x)=1.3087

Variance $=\sum (x-\mu)^2P(x)$

$\left|\begin{array}{c|cccccc}x&0&1&2&3&4\\(x-\mu)^2&1.7167&0.0953&0.4779&2.8605&7.2431\\\\P(x)&\dfrac{365}{1014}&\dfrac{264}{1014}&\dfrac{193}{1014}&\dfrac{91}{1014}&\dfrac{101}{1014} \\\\(x-\mu)^2P(x)&0.6179&0.0248&0.0910&0.2567&0.7215\end{array}\right|$

$Variance, \sum (x-\mu)^2P(x)=1.7119$

(d)

$\left|\begin{array}{c|cccccc}y&1&2&3&4\\\\P(y)&\dfrac{264}{649}&\dfrac{193}{649}&\dfrac{91}{649}&\dfrac{101}{649} \end{array}\right|$

$E(y)=\left(1\times\dfrac{264}{649}\left)+\left(2\times\dfrac{193}{649}\left)+\left(3\times\dfrac{91}{649}\left)+\left(4\times\dfrac{101}{649}\right)$

E(y)=2.0447

The expected value for y surpasses that of x.

6 0

2 months ago