Anneearns

The diameters of ball bearings are distributed normally. The mean diameter is 107 millimeters and the population standard deviat

tensa zangetsu [6.8K]

Answer:

0.6710

Step-by-step explanation:

The diameters of ball bearings are distributed normally. The mean diameter is 107 millimeters and the population standard deviation is 5 millimeters.

Find the probability that the diameter of a selected bearing is between 104 and 115 millimeters. Round your answer to four decimal places.

We solve using z score formula

z = (x-μ)/σ, where

x is the raw score

μ is the population mean = 107 mm

σ is the population standard deviation = 5 mm

For x = 104 mm

z = 104 - 107/5

z = -0.6

Probability value from Z-Table:

P(x = 104) = 0.27425

For x = 115 mm

z = 115 - 107/5

z = 1.6

Probability value from Z-Table:

P(x = 115) = 0.9452

The probability that the diameter of a selected bearing is between 104 and 115 millimeters is calculated as:

P(x = 115) - P(x = 104)

0.9452 - 0.27425

= 0.67095

Approximately = 0.6710

8 0

2 years ago

Which statement is not true about histograms

Eduardwww [97]

Answer:

d

Step-by-step explanation:

d is not true because width and thickness doesn't matter

8 0

3 years ago

A box company is designing a new rectangular gift container. The marketing department has designed a box with a width 2 inches s

Sladkaya [172]

Volume of the box= 56 cubic inches
let x is the length, then
width =2 inches shorter than its length = x - 2
height = 3 inches taller than its length = x+3
Volume = length x width x height
56 = x x (x-2) x (x+3)
56 = (x² -2x)(x+3)
56 = x³ +3x² -2x² - 6x
56 = x³ + x² -6x
x³+x²-6x-56 = 0
using the rational root theorem and factoring the polynomial;
(x-4)(x² +5x +14) = 0
from here;
x-4 = 0
x = 4
So, length = 4 inches
width = x - 2 = 4 -2 = 2 inches
length = x + 3 = 4 + 3 = 7 inches
volume = l x w x h = 4 x 2 x 7 = 56

3 0

2 years ago

A rock thrown vertically upward from the surface of the moon at a velocity of 36m/sec reaches a height of s = 36t - 0.8 t^2 met

Verdich [7]

Answer:

a. The rock's velocity is $v(t)=36-1.6t \:{(m/s)}$ and the acceleration is $a(t)=-1.6 \:{(m/s^2)}$

b. It takes 22.5 seconds to reach the highest point.

c. The rock goes up to 405 m.

d. It reach half its maximum height when time is 6.59 s or 38.41 s.

e. The rock is aloft for 45 seconds.

Step-by-step explanation:

Velocity is defined as the rate of change of position or the rate of displacement. $v(t)=\frac{ds}{dt}$
Acceleration is defined as the rate of change of velocity. $a(t)=\frac{dv}{dt}$

a.

The rock's velocity is the derivative of the height function $s(t) = 36t - 0.8 t^2$

$v(t)=\frac{d}{dt}(36t - 0.8 t^2) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\v(t)=\frac{d}{dt}\left(36t\right)-\frac{d}{dt}\left(0.8t^2\right)\\\\v(t)=36-1.6t$

The rock's acceleration is the derivative of the velocity function $v(t)=36-1.6t$

$a(t)=\frac{d}{dt}(36-1.6t)\\\\a(t)=-1.6$

b. The rock will reach its highest point when the velocity becomes zero.

$v(t)=36-1.6t=0\\36\cdot \:10-1.6t\cdot \:10=0\cdot \:10\\360-16t=0\\360-16t-360=0-360\\-16t=-360\\t=\frac{45}{2}=22.5$

It takes 22.5 seconds to reach the highest point.

c. The rock reach its highest point when t = 22.5 s

Thus

$s(22.5) = 36(22.5) - 0.8 (22.5)^2\\s(22.5) =405$

So the rock goes up to 405 m.

d. The maximum height is 405 m. So the half of its maximum height = $\frac{405}{2} =202.5 \:m$

To find the time it reach half its maximum height, we need to solve

$36t - 0.8 t^2=202.5\\36t\cdot \:10-0.8t^2\cdot \:10=202.5\cdot \:10\\360t-8t^2=2025\\360t-8t^2-2025=2025-2025\\-8t^2+360t-2025=0$

For a quadratic equation of the form $ax^2+bx+c=0$ the solutions are

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=-8,\:b=360,\:c=-2025:\\\\t=\frac{-360+\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2-\sqrt{2}\right)}{4}\approx 6.59\\\\t=\frac{-360-\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2+\sqrt{2}\right)}{4}\approx 38.41$

It reach half its maximum height when time is 6.59 s or 38.41 s.

e. It is aloft until s(t) = 0 again

$36t - 0.8 t^2=0\\\\\mathrm{Factor\:}36t-0.8t^2\rightarrow -t\left(0.8t-36\right)\\\\\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}\\\\t=0,\:t=45$

The rock is aloft for 45 seconds.

5 0

3 years ago

IQR: Need help asap! 30 points! What is the interquartile range of the data set?

lorasvet [3.4K]

Can you insert a photo of the question please it would greatly help me answer this

6 0

3 years ago