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Ratling [72]
2 years ago
9

(09.02)

Mathematics
1 answer:
dmitriy555 [2]2 years ago
4 0
X^2 - 7x + 12 = 0
x^2 - 4x - 3x + 12 = 0
x (x - 4) -3 ( x - 4) = 0
(x - 4) (x - 3) = 0
x = 3 , 4

Check the answer by plugging in 3 and 4 for x, if the equation equals zero then you have your answer.
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18 divided by 3-7+2x5<br> Please help!!!
chubhunter [2.5K]
The answer is to 18/3-7+2 x 5 is 3
6 0
3 years ago
Please help me with these 19 and 20 and tell me what you use to get the answer
ss7ja [257]
36 + h = 75...subtract 36 from both sides
h = 75 - 36
h = 39

-176 = h + (-219)
-176 = h - 219....add 219 to both sides
219 - 176 = h
43 = h
6 0
3 years ago
Read 2 more answers
Assume that the weight loss for the first month of a diet program varies between 6 pounds and 14 pounds, and is spread evenly ov
Llana [10]

Hello!

For this uniform distribution, the shape of the density function is a line. We can begin by identifying key elements of the distribution:

Range: 6 to 14 pounds

Formula of density function:

y = 1/Range

1 - (14 - 6) = 1/8 = 0.125

f(x) = 0.125

1)

P(x = 7)

A uniform distribution is a continuous distribution, so the probability of x being an EXACT value is approximately 0.

<u>This is the same as if we took the following integral. (Finding the area under a curve with the same start and end point)</u>
\int\limits^a_a {f(x)} \, dx  = 0

Therefore, P(x = 7) = 0.

2)
We can think of this as finding the area of a rectangle.

The width would be the difference between 8.25 pounds and 12 pounds:
W = 12 - 8.25 = 3.75

The height would be the function (y = .125), or:
h = .125

Using the equation for the area of a rectangle: (A = h * w)

A = 3.75 * .125 = 0.469

Thus, P(8.25 < x < 12) = 0.469

3)
To find the probability GREATER than 10.50 pounds, we can subtract this value from the max value for the width:
W = 14 - 10.50 = 3.50

The height is the same as above:
h = .125

Solve for the area:
A = 3.50 * .125 = 0.4375

P(x > 10.5) = 0.4375

5 0
2 years ago
Find x in this 45°-45°-90° triangle.<br><br> x=
Blababa [14]

Answer:

24

Step-by-step explanation:

This is the same question as before just flipped around

6 0
3 years ago
Suppose we want to choose a value of x within 3 units of 11. [This means a value of x that is less than 3 units away from 11.] T
Dahasolnce [82]

Answer:

(a) [8, 14]

(b)  8 \leq x \leq 14

(c)See attachment

Step-by-step explanation:

We want to choose a value of x within 3 units of 11.

(a)Now, 11-3=8 and 11+3=14

The possible values of x ranges is in the closed interval [8,14]

(b) Since x is within 3 units of 11., we have:

|11-x|\leq3

Solving the absolute inequality

-3 \leq 11-x \leq 3\\$In $ -3 \leq 11-x\\ x \leq 11+3\\x \leq 14\\\\$In $ 11-x \leq 3\\ 11-3 \leq x\\8 \leq x\\$Therefore,an inequality that represents all values of x that meet this constraint is:$\\8 \leq x \leq 14

(c)To draw the number line, we use a closed dot since we have the less than or equal to sign.

5 0
3 years ago
Read 2 more answers
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