Answer A B C D and E

The average weight of a particular box of crackers is 26.0 ounces with a standard deviation of 0.5 ounce. The weights of the

Olenka [21]

Answer:

(a) 99.865%

(b) 0.135%

Explanation:

Given that the weight of the boxes are normally distributed.

The average weight of the particular box,

$\mu=26.0$ ounce

The standard deviation of weight,

$\sigma=0.5$ ounce.

Let $z$ be the standard normal variable,

$z=\frac{x-\mu}{\sigma}$

And, the probability of the boxes having weight x ounces is

$P(z)=\frac{1}{2\pi}e^{-\frac{z^2}{2}}$

For $x=24.5$ ,

$z=\frac{x-\mu}{\sigma}=\frac{24.5-26}{0.5}=-3$

(a) For the boxes having weight more than 24.5 ounces: $z>-3$

So, the probability of boxes for $z>-3$ is

$P(z>-3)=\int_{-3}^{\infty}\left(\frac{1}{2\pi}e^{-\frac{z^2}{2}}\right)dx$

=0.99865

So, the percent of the boxes weigh more than 24.5 ounces is 99.865%.

(b) For the boxes having weight less than 24.5 ounces: z<-3

So, the probability of boxes for z<-3 is

$P(z-3}=1-0.99865$

=0.00135

So, the percent of the boxes weigh more than 24.5 ounces is 0.135%.

8 0

4 years ago

A 4 kg billiard ball moving on a horizontal surface has a speed of 16 m/s when it strikes a horizontal coiled spring is brought

ruslelena [56]

Answer:

spring constant of the spring is 1820.44 N/m

Explanation:

given data

ball mass = 4 kg

speed = 16 m/s

distance = 0.75 m

to find out

spring constant of the spring

solution

we know that kinetic energy of ball = energy store in spring as compression

so we can express it as

0.5 × m × v² = 0.5 × k × x² ....................1

so put here value we get spring constant k

m × v² = k × x²

4 × 16² = k × 0.75²

solve it we get

k = 1820.44 N/m

so spring constant of the spring is 1820.44 N/m

6 0

4 years ago

Read 2 more answers

a 40 kg block is being pulled at constant velocity across a horizontal surface by a 150 N force at an angle 60 above the horizon

k0ka [10]

Answer:

0.19

Explanation:

mass of block, m = 40 kg

F = 150 N

Angle make with the horizontal, θ = 60 degree

Let μ be the coefficient of kinetic friction

The component of force along horizontal direction is F Cos θ

= 150 cos 60 = 75 N

As it is moving with constant velocity it mean the acceleration of the block is zero.

Applied force in horizontal direction = friction force

75 = μ x Normal reaction

75 = μ x m x g

75 = μ x 40 x 9.8

μ = 0.19

Thus, the coefficient of kinetic friction is 0.19.

8 0

3 years ago

Imagine a sunny day at the pool. The sun is out and you are thinking about how the light travels from the sun and then hits the

Dovator [93]

Answer:

it evaporats

Explanation:

because the sun is so hot that the water will turn into gas hope i helped

5 0

4 years ago

Read 2 more answers

A man with a weight of 100 N is located

valentina_108 [34]

Answer:

Given,

mass of man = 100 N = 10 kg

height = h = 25m

since the man does not move anything with his force, work done by him is zero

work done on the man = gain in potential energy

P.E=mgh

P.E=10×9.8×25

P.E=2.45KJ

Explanation:

so, potential energy gained by man is 2.45 KJ

5 0

3 years ago