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maksim [4K]
3 years ago
11

In a pacemaker factory, all pacemakers must go through a rigorous quality assurance check before being shipped. The table shows

the results
of the quality assurance checks. Find the probability that a non-defective pacemaker "fails." Round your answer to the nearest tenth of a percent. NEED ASAP
Mathematics
1 answer:
Len [333]3 years ago
5 0

Answer:

16.1%

Step-by-step explanation:

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MATH HELP!!! 100PTS!!!
Lostsunrise [7]

Answer:

x =\dfrac{20 +\sqrt{454}}{18}, \quad \dfrac{20 -\sqrt{454}}{18}

Step-by-step explanation:

<u>Given functions</u>:

  p(x)=2x-4

  r(x)=\dfrac{6x-1}{9x+1}

Solve for p(x) = r(x):

\begin{aligned}p(x) & = r(x)\\\implies 2x-4 & = \dfrac{6x-1}{9x+1}\\(2x-4)(9x+1)&=6x-1\\18x^2+2x-36x-4&=6x-1\\18x^2-40x-3&=0\end{aligned}

As the found quadratic equation cannot be factored, use the Quadratic Formula to solve for x:

<u>Quadratic Formula</u>

x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0

Therefore:

a=18, \quad b=-40, \quad c=-3

Substitute the values of a, b and c into the <u>quadratic formula</u> and solve for x:

\begin{aligned}\implies x & =\dfrac{-(-40) \pm \sqrt{(-40)^2-4(18)(-3)} }{2(18)}\\\\& =\dfrac{40 \pm \sqrt{1816}}{36}\\\\& =\dfrac{40 \pm \sqrt{4 \cdot 454}}{36}\\\\& =\dfrac{40 \pm \sqrt{4}\sqrt{454}}{36}\\\\& =\dfrac{40 \pm2\sqrt{454}}{36}\\\\& =\dfrac{20 \pm\sqrt{454}}{18}\end{aligned}

Therefore, the solutions are:

x =\dfrac{20 +\sqrt{454}}{18}, \quad \dfrac{20 -\sqrt{454}}{18}

Learn more about quadratic equations here:

brainly.com/question/27750885

brainly.com/question/27739892

6 0
2 years ago
Elizabeth’s aunt started a college savings account for her with $6,000. What is the balance of the account after 5 years if the
vaieri [72.5K]

Answer:

6750

Step-by-step explanation:

simple interest = P×r%×t

P = $6000

r = 2.5%

t = 5 years

I = 6000 × 2.5/100 × 5

= 750

6000 + 750 = $6750

7 0
3 years ago
PLS SEE ATTACHMENT! NEED HELP!!!
pshichka [43]

Answer:idk

Step-by-step explanation:

idk

6 0
3 years ago
Read 2 more answers
Jackson sells toolboxes for farming tools. He plans on selling them during a fair this weekend. Jackson estimates he will sell t
Nataly [62]

The linear equation for the given situation is represented as:

20x + 12y = 1,300

correct answer option: b

<h3>What is a linear equation with two variables?</h3>

A general linear equation in two variables is defined as an equation of the form ax + by + c = 0, where x and y are the two variables and a, b, and c are real numbers and a and b are non-zero (x and y).

Given:

We use variables to represent the cost of the big and the smaller boxes.

According to the question, let x be the selling price of the big boxes.

The selling price of the bigger boxes is y.

The total selling price of the smaller boxes is 20x.

The total selling price of the bigger ones is 12y.

Now we know he is making a profit of $1,300 on the total transaction.

Hence the linear equation in two variables for the given situation is represented as:

20x + 12y = 1,300

To learn more about linear equations in two variables visit:

brainly.com/question/24085666

#SPJ4

I understand that the question you are looking for is:

Jackson sells two types of toolboxes. he plans on selling them during a farming fair this weekend. He estimates he will sell 20 of the smaller boxes (X) and 12 of the larger boxes (Y). if he'd like the profit to be $1300, which of the following best displays the equation that represents this information?

A-  y+12=20(x-1300)

B-  y=- \frac{5}{3}x+ \frac{325}{3}

C-  20x+12y=1300

D-  12y=20x-1300

6 0
1 year ago
Suppose that a function​ f(x) is defined for all real values of x except at xequals=c. can anything be said about the existence
Margaret [11]

we are given that

f(x) is defined for all values of x except at x=c

Limit may or may not exist

case-1:

If there is hole at x=c , then limit exist

case-2:

If there is vertical asymptote at x=c , then limit does not exist

Examples:

case-1:

\lim_{x \to c} \frac{x^2-cx}{(x-c)}

We can simplify it

\lim_{x \to c} \frac{x(x-c)}{(x-c)}

=\lim_{x \to c} x

=c

so, we can see that limit exist and it's value defined

case-2:

\lim_{x \to c} \frac{1}{(x-c)}

Left limit is

\lim_{x \to c-} \frac{1}{(x-c)}

=-\infty

Right Limit is

\lim_{x \to c+} \frac{1}{(x-c)}

=+\infty

so, we can see that left limit is not equal to right limit

so, limit does not exist

4 0
3 years ago
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