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solniwko [45]
3 years ago
13

If five less than y is six more than x + 1, then by how much is x less than y? ​

Mathematics
1 answer:
Pie3 years ago
6 0

Answer:

12 more

Step-by-step explanation:

y-5= 6+ (x+1)

5+6+1=12

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Solve: |4x + 5| + 8 >(or equal to)45
KatRina [158]
This is pretty simple once you get used to this


First set it equal to the original and the negative one
(>_) is the bigger and equal to part

4x+5>_45 or 4x+5>_-45

Then solve for x for both


4x>_40. 4x>_-50
X>_10 X>_-25/2
8 0
3 years ago
I need to solve this problem using the substitution method. I keep getting crazy numbers so I’m definitely doing something wrong
Oksanka [162]

Answer:

S=3 R=1

Step-by-step explanation:

3x1+3=6

1+(3x3)=10

3x3=9

9+1=10.

4 0
3 years ago
ABC Auto Insurance classifies drivers as good, medium, or poor risks. Drivers who apply to them for insurance fall into these th
Misha Larkins [42]

Answer:

a.P(E_1/A)=0.0789

b.P(E_2/A)=0.395\

c.P(E_3/A)=0.526

Step-by-step explanation:

Let E_1,E_2,E_3 are the events that denotes the good drive, medium drive and poor risk driver.

P(E_1)=0.30,P(E_2)=0.50,P(E_3)=0.20

Let A be the event that denotes an accident.

P(A/E_1)=0.01

P(A/E_2=0.03

P(A/E_3)=0.10

The company sells Mr. Brophyan insurance policy and he has an accident.

a.We have to find the probability Mr.Brophy is a good driver

Bayes theorem,P(E_i/A)=\frac{P(A/E_i)\cdot P(E_1)}{\sum_{i=1}^{i=n}P(A/E_i)\cdot P(E_i)}

We have to find P(E_1/A)

Using the Bayes theorem

P(E_1/A)=\frac{P(A/E_1)\cdot P(E_1)}{P(E_1)\cdot P(A/E_1)+P(E_2)P(A/E_2)+P(E_3)P(A/E_3)}

Substitute the values then we get

P(E_1/A)=\frac{0.30\times 0.01}{0.01\times 0.30+0.50\times 0.03+0.20\times 0.10}

P(E_1/A)=0.0789

b.We have to find the probability Mr.Brophy is a medium driver

P(E_2/A)=\frac{0.03\times 0.50}{0.038}=0.395

c.We have to find the probability Mr.Brophy is a poor driver

P(E_3/A)=\frac{0.20\times 0.10}{0.038}=0.526

7 0
3 years ago
Harry is trying to solve the equation 0 = 2x2 − x − 6 using the quadratic formula. He has made an error in one of the steps belo
bija089 [108]
<h3><u>Given</u> - </h3>

➙ a quadratic equation in which Harry lagged due to an error made by him, 2x² - x - 6= 0

<h3><u>To solve</u> - </h3>

➙ the given quadratic equation.

<h3><u>Concept applied</u> - </h3>

➙ We will apply the quadratic formula as given in the question. So, let's study about quadratic equation first because we are supposed to apply the formula in equation.

What is quadratic equation?

➙ A quadratic equation in the variable x is an equation of the form ax² + bx + c = 0, where a, b, c are real numbers, a ≠ 0.

Now, what is quadratic formula?

➙The roots of a quadratic equation ax + bx + c = 0 are given by \sf{\:\frac{-b \pm\: \sqrt {b ^ 2 - 4ac}}{2a}} provided b - 4ac ≥ 0.

<h3><u>Solution</u> - </h3>

here as per the given quadratic equation,

a = 2, b = -1 and c = -6

putting in the formula,

\implies\sf{x=\frac{-(-1) \pm\: \sqrt {(-1)^2 - 4(2)(-6)}}{2(2)}}

\implies\sf{x=\frac{1 \pm\: \sqrt {1+48}}{4}}

\implies\sf{x=\frac{1 \pm\: \sqrt {49}}{4}}

\implies\sf{x=\frac{1 \pm\: 7}{4}}

Solving one by one,

\implies\sf{x=\frac{1 + \: 7}{4}}

\implies\sf{x=\frac{8}{4}}

\implies{\boxed{\bf{x=2}}}

________________

\implies\sf{x=\frac{1 - \: 7}{4}}

\implies\sf{x=\frac{-6}{4}}

\implies{\boxed{\bf{x=\frac{-3}{2}}}}

________________________________

<em><u>Note</u> - Hey dear user!! You haven't provided the solution which was solved by Harry (A.T.Q). Please go through the solution as it will help you to find the error done by Harry.</em>

<em>________________________________</em>

Hope it helps!! (:

4 0
2 years ago
What is the value of x?
denis-greek [22]

Answer:

x = 17

Step-by-step explanation:

Hi there!

Because this is an isosceles triangle, we know that the angles measuring (x+17°) and (4x+34°) are congruent. Knowing this, we can set up the following equation and solve for <em>x</em>:

x+17 = 4x-34

17 = 3x-34

0 = 3x-51

3x = 51

x = 17

Therefore, the value of <em>x</em> is 17.

I hope this helps!

8 0
2 years ago
Read 2 more answers
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