Ask question

Ask question

All categories

3 years ago

13

A part manufactured by plastic injection molding has a historical mean of 100 and a historical standard deviation of 8. Find the

value of the mean thickness required to make the probability of exceeding 101 less than 8%.

1 answer:

damaskus [11]3 years ago

8 0

Answer:

The probability of thickness exceeding 101 is 0.4483.

Step-by-step explanation:

Let <em>X</em> denote the thickness of the part manufactured by plastic injection molding.

Assume that <em>X</em> follows a normal distribution with mean, <em>μ</em> = 100 and standard deviation, <em>σ</em> = 8.

Compute the probability of thickness exceeding 101 as follows:

$P(X>101)=P(\frac{X-\mu}{\sigma}>\frac{101-100}{8})$

$=P(Z>0.125)\\\\=1-P(Z$

Thus, the probability of thickness exceeding 101 is 0.4483.

You might be interested in

X=7; 4x + 8/2=??? Explain

AlladinOne [14]

Answer:

Step-by-step explanation:

4(7)+8/2

28+4

=32

7 0

2 years ago

E price of a large pizza increased from $21 to $26. what is the percent of increase in the cost of pizza? round to the nearest p

rodikova [14]

Percent increase=increase/original times 100

increase=26-21=5
original=21

so
5/21 times 100=0.23809523809523809523809523809524 times 100=23.809523809523809523809523809524
round
24%

so 24% increase

7 0

3 years ago

Can someone help me with this, please?<br>No links or false answers, please

VARVARA [1.3K]

The exact value is found by making use of order of operations. The

functions can be resolved using the characteristics of quadratic functions.

Correct responses:

$\displaystyle 1\frac{4}{7} \div \frac{2}{3} - 1\frac{5}{7} =\frac{9}{14}$

x = -4, y = 12

When P = 1, V = 6

$\displaystyle x = -3 \ or \ x = \frac{1}{2}$

$\displaystyle i. \hspace{0.1 cm} \underline{ f(x) = 2 \cdot \left(x - 1.25 \right)^2 + 4.875 }$

ii. The function has a minimum point

iii. The value of <em>x</em> at the minimum point, is <u>1.25</u>

iv. The equation of the axis of symmetry is <u>x = 1.25</u>

<h3>Methods by which the above responses are found</h3>

First part:

The given expression, $\displaystyle \mathbf{ 1\frac{4}{7} \div \frac{2}{3} -1\frac{5}{7}}$ , can be simplified using the algorithm for arithmetic operations as follows;

$\displaystyle 1\frac{4}{7} \div \frac{2}{3} - 1\frac{5}{7} = \frac{11}{7} \div \frac{2}{3} - \frac{12}{7} = \frac{11}{7} \times \frac{3}{2} - \frac{12}{7} = \frac{33 - 24}{14} =\underline{\frac{9}{14}}$

Second part:

y = 8 - x

2·x² + x·y = -16

Therefore;

2·x² + x·(8 - x) = -16

2·x² + 8·x - x² + 16 = 0

x² + 8·x + 16 = 0

(x + 4)·(x + 4) = 0

<u>x = -4</u>

y = 8 - (-4) = 12

<u>y = 12</u>

Third part:

(i) P varies inversely as the square of <em>V</em>

Therefore;

$\displaystyle P \propto \mathbf{\frac{1}{V^2}}$

$\displaystyle P = \frac{K}{V^2}$

V = 3, when P = 4

Therefore;

$\displaystyle 4 = \frac{K}{3^2}$

K = 3² × 4 = 36

$\displaystyle V = \sqrt{\frac{K}{P}$

When P = 1, we have;

$\displaystyle V =\sqrt{ \frac{36}{1} } = 6$

When P = 1, V =<u> 6</u>

Fourth Part:

Required:

Solving for <em>x</em> in the equation; 2·x² + 5·x - 3 = 0

Solution:

The equation can be simplified by rewriting the equation as follows;

2·x² + 5·x - 3 = 2·x² + 6·x - x - 3 = 0

2·x·(x + 3) - (x + 3) = 0

(x + 3)·(2·x - 1) = 0

$\displaystyle \underline{x = -3 \ or\ x = \frac{1}{2}}$

Fifth part:

The given function is; f(x) = 2·x² - 5·x + 8

i. Required; To write the function in the form a·(x + b)² + c

The vertex form of a quadratic equation is f(x) = a·(x - h)² + k, which is similar to the required form

Where;

(h, k) = The coordinate of the vertex

Therefore, the coordinates of the vertex of the quadratic equation is (b, c)

The x-coordinate of the vertex of a quadratic equation f(x) = a·x² + b·x + c, is given as follows;

$\displaystyle h = \mathbf{ \frac{-b}{2 \cdot a}}$

Therefore, for the given equation, we have;

$\displaystyle h = \frac{-(-5)}{2 \times 2} = \mathbf{ \frac{5}{4}} = 1.25$

Therefore, at the vertex, we have;

$k = \displaystyle f\left(1.25\right) = 2 \times \left(1.25\right)^2 - 5 \times 1.25 + 8 = \frac{39}{8} = 4.875$

a = The leading coefficient = 2

b = -h

c = k

Which gives;

$\displaystyle f(x) \ in \ the \ form \ a \cdot (x + b)^2 + c \ is \ f(x) = 2 \cdot \left(x + \left(-1.25 \right) \right)^2 +4.875$

Therefore;

$\displaystyle \underline{ f(x) = 2 \cdot \left(x -1.25\right)^2 + 4.875}$

ii. The coefficient of the quadratic function is <em>2</em> which is positive, therefore;

<u>The function has a minimum point</u>.

iii. The value of <em>x</em> for which the minimum value occurs is -b = h which is therefore;

The x-coordinate of the vertex = h = -b =<u> 1.25 </u>

iv. The axis of symmetry is the vertical line that passes through the vertex.

Therefore;

The axis of symmetry is the line <u>x = 1.25</u>.

Learn more about quadratic functions here:

brainly.com/question/11631534

6 0

2 years ago

Read 2 more answers

a) A group of 50 people are going to run a race. The top three runners earn gold, silver, and bronze medals. b) Creating a nine

rjkz [21]

a) n (No. of peoples) = 50

r (No. of prizes) = 3

p(n,r) = p (50,3)

$\begin{gathered} P(n,r)=\frac{n!}{(n-r)!} \\ P(50,3)=\frac{50!}{(50-3)!} \\ =117600 \end{gathered}$

b) n(No. of players)=14

r (No. of players for Batting)=9

$\begin{gathered} P(14,9)=\frac{14!}{(14-9)!} \\ =726485760 \end{gathered}$

c)n(Total applicants)=30

r(tester positions)=4

$\begin{gathered} P(30,4)=\frac{30!}{(30-4)!} \\ \text{ =}\frac{30!}{26!} \\ \text{ =657720} \end{gathered}$

7 0

6 months ago

Which reason completes the proof below?

sergeinik [125]

Observe the figure given.

Given: ABCD is a parallelogram.

To prove: ABCD is a rhombus

Statement

1. ABCD is a parallelogram. AC bisects $\angle BCD, \angle BAD$ .

Reason: Given

2. $\angle 1 \cong \angle 2, \angle 3 \cong \angle 4$

Reason: Definition of an angle bisector

3. $AC \cong AC$

Reason: Reflexive property of congruence

4. $\Delta ABC \cong \Delta ADC$

Reason: ASA Postulate

5. $AB \cong AD, BC \cong DC$

Reason: corresponding parts of congruent triangles are equal.

6. $AB \cong CD, BC \cong AD$

Reason: Opposite sides of the parallelogram are congruent.

7. $AB \cong AD \cong BC \cong CD$

Reason: Transitive property of congruence

8. ABCD is a rhombus

Reason: Definition of rhombus

4 0

3 years ago