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3 years ago

What is the value of x in the equation 5.5x = 17.6? 3.2 12.1 23.1 96.8

Mathematics

2 answers:

Fudgin [204]3 years ago

5 0

3.2

Because:
5.5x=17.6
—— ——
5.5x 5.5x

X=3.2

Yakvenalex [24]3 years ago

4 0

Answer:

3.2

Step-by-step explanation:

rephrase question to isolate the variable x

can do this is by dividing both sides by 5.5

5.5 ÷ 5.5 = 1 so:

x = 17.6 ÷ 5.5

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What is 3x squared-12x=0

lubasha [3.4K]

x=0 or x=4

just factorize basically

cancels the x’s so it’d be 3x-12=0 and from that x=4

3 0

4 years ago

Find the value of the expression. 7! ÷ 3! 5,040 5,034 840 24

koban [17]

First, let's review what the "factorial operator" ! means.

5! = 5*4*3*2*1.

3! = 3*2*1

7! = 7*6*5*4*3*2*1

Then:

7! 7*6*5*4*3*2*1 7*6*5*4 5040
----- = ----------------------- = ------------- = ----------- = 5040 (answer)
3! 3*2*1 1 1

3 0

3 years ago

Suppose a normal distribution has a mean of 38 and a standard deviation of 2. What is the probability that a data value is betwe

Dimas [21]

Answer:

P ( 37 < x < 41) = P(-0.5 < Z < 1.5) = 0.6247

Step-by-step explanation:

We know mean u = 38 standard dev. s = 2

We want P ( 37 < x < 41)

P( (37 - 38) / 2 < Z) = P(-0.5 < Z)

P( Z < (41 - 38)/2 ) = P( Z < 1.5)

Find P(Z < -0.5) = 0.3085

Find P(Z > 1.5) = 0.0668

so P(-0.5 < Z < 1.5) = 1 - P(Z < -0.5) - P(Z > 1.5)

P(-0.5 < Z < 1.5) = 1 - 0.3085 - 0.0668

P(-0.5 < Z < 1.5) = 0.6247

P ( 37 < x < 41) = P(-0.5 < Z < 1.5) = 0.6247

3 0

3 years ago

Solve the equation for x.<br> -=7<br> Answer here

Leviafan [203]

Complete Question:

Solve the equation for x

$x-7=7$

Answer:

$x = 14$

Step-by-step explanation:

Given

$x-7=7$

Required

Solve for x

$x-7=7$

Add 7 to both sides

$x-7+7=7+7$

$x=7+7$

$x = 14$

Hence:

For $x-7=7$ , $x = 14$

5 0

3 years ago

Identify the radius and center.<br><br> x^2 + y^2 - 2x + 4y - 11 = 0

miss Akunina [59]

<h2>Hello!</h2>

The answer is:

Center: (1,-2)

Radius: 4 units.

To solve the problem, using the given formula of a circle, we need to find its standard equation form which is equal to:

$(x-h)^{2}+(y-k)^{2}=r^{2}$

Where,

"h" and "k"are the coordinates of the center of the circle and "r" is its radius.

So, we need to complete the square for both variable "x" and "y".

The given equation is:

$x^2+y^2-2x+4y-11=0$

So, solving we have:

$x^2+y^2-2x+4y=11$

$(x^2-2x+(\frac{2}{2})^{2} )+(y^2+4y+(\frac{4}{2})^{2})=11+(\frac{2}{2})^{2} +(\frac{4}{2})^{2}\\\\(x^2-2x+1)+(y^2+4y+4)=11+1+4\\\\(x^2-1)+(y^2+2)=16$

$(x^2-1)+(y^2-(-2))=16$

Now, we have that:

$h=1\\k=-2\\r=\sqrt{16}=4$

So,

Center: (1,-2)

Radius: 4 units.

Have a nice day!

Note: I have attached a picture for better understanding.

5 0

3 years ago