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ira [324]
2 years ago
14

The shape of the distribution of the time required to get an oil change at a 15-minute oil-change facility is unknown. However,

records indicate that the mean time is 16.2 minutes, and the standard deviation is 3.4 minutes.(a) What is the probabilty that a random sample of n = 40 oil changes results in a sample mean time less than 15 minutes?Round to 4 decimal.(b) Suppose the manager agrees to pay each employee a $50 bonus if they meet a certain gol. On a typical Sunday, the oil-change facility will perform 40 oil changes between 10 A.M. and 12 P.M. Treating this as a random sample, what mean oil-change time would there be a 10% chance of being at or below? This will be the goal established by the manager.There is a 10% chance of being at or below a mean oil change time of ? minutes.Round to 1 decimal.
Mathematics
1 answer:
kolezko [41]2 years ago
8 0

Answer:

(a) 0.0128 to 4 decimal places

(b) The 90% confidence interval is 119.1 < μ < 121.0 to 1 decimal places

Step-by-step explanation:

See the attached documents and graphs. Cheers!

Download pdf
<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> pdf </span>
4de97e6d942fe2c9047a396b82fccba0.png
0b940014a3beda8e854506c0084127b3.png
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2 years ago
Find MO and PR <br> ............................................................
ddd [48]

MO = 12 and PR = 3

Solution:

Given \triangle M N O \sim \Delta P Q R.

Perimeter of ΔMNO = 48

Perimeter of ΔPQR = 12

MO = 12x and PR = x + 2

<em>If two triangles are similar, then the ratio of corresponding sides is equal to the ratio of perimeter of the triangles.</em>

$\Rightarrow \frac{\text{Perimeter of }\triangle MNO}{\text{Perimeter of }\triangle PQR} =\frac{MO}{PR}

$\Rightarrow \frac{48}{12} =\frac{12x}{x+2}

Do cross multiplication.

$\Rightarrow 48({x+2})= 12(12x)

$\Rightarrow 48x+96= 144x

Subtract 48x from both sides.

$\Rightarrow 48x+96-48x= 144x-48x

$\Rightarrow 96= 96x

Divide by 96 on both sides, we get

⇒ 1 = x

⇒ x = 1

Substitute x = 1 in MO an PR.

MO = 12(1) = 12

PR = 1 + 2 = 3

Therefore MO = 12 and PR = 3.

7 0
3 years ago
The owner of a bike shop that produces custom built bike frames has determined that the demand equation for bike frames is given
o-na [289]

Answer:

equilibrium point (10,340)

Step-by-step explanation:

To find the equilibrium point, equal the demand and the supply:

D(q)=S(q)\\\\-6.10q^2-5q+1000=3.2q^2+10q-80

Reorganize the terms in one side and reduce similar terms:

3.2q^2+6.1q^2+5q+10q-80-1000=0\\\\9.3q^2+15q-1080=0

that's a cuadratic equation, solve with the general formula when:

<u><em>a=9.3</em></u>, <u><em>b=15</em></u>, <u><em>c=-1080</em></u>

q_{1}=\frac{-b+\sqrt{b^{2}-4ac} }{2a}\\\\q_{2}=\frac{-b-\sqrt{b^{2}-4ac} }{2a}\\\\q_{1}=\frac{-15+\sqrt{(-15)^{2}-4(9.3)(-1080)} }{2(9.3)}\\\\q_{1}=\frac{-15+201}{18.6}\\\\q_{1}=\frac{186}{18.6}\\\\q_1=10

<em><u>q</u></em> can't be negative because it is the quantity of bike frames, so:

q_{2}=\frac{-b-\sqrt{b^{2}-4ac} }{2a}\\\\q_{2}=\frac{-15-\sqrt{(-15)^{2}-4(9.3)(-1080)} }{2(9.3)}\\\\q_{2}=\frac{-15-201}{18.6}\\\\q_{2}=\frac{-216}{18.6}\\\\

This value of <u><em>q</em></u> can't be considered.

Then substitute the value of <em><u>q</u></em> in <u><em>D(q)</em></u> to find the price <u><em>p</em></u>:

D(10) = -6.10(10)^2-5(10) + 1000\\\\D(10)=340=p

The equilibrium point (q,p) is (10,340).

6 0
2 years ago
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