All categories

2 years ago

Weights of cars passing over a bridge part 2

Mathematics

1 answer:

DaniilM [7]2 years ago

6 0

Step-by-step explanation:

The solution to this problem is very much similar to your previous ones, already answered by Sqdancefan.

Given:

mean, mu = 3550 lbs (hope I read the first five correctly, and it's not a six)

standard deviation, sigma = 870 lbs

weights are normally distributed, and assume large samples.

Probability to be estimated between W1=2800 and W2=4500 lbs.

Solution:

We calculate Z-scores for each of the limits in order to estimate probabilities from tables.

For W1 (lower limit),

Z1=(W1-mu)/sigma = (2800 - 3550)/870 = -.862069

From tables, P(Z<Z1) = 0.194325

For W2 (upper limit):

Z2=(W2-mu)/sigma = (4500-3550)/879 = 1.091954

From tables, P(Z<Z2) = 0.862573

Therefore probability that weight is between W1 and W2 is

P( W1 < W < W2 )

= P(Z1 < Z < Z2)

= P(Z<Z2) - P(Z<Z1)

= 0.862573 - 0.194325

= 0.668248

= 0.67 (to the hundredth)

You might be interested in

Can you help me find the volume base area latera area and total area please to this problem

krek1111 [17]

Answer:

BA = 25π,

LA = 25√2π,

TA = 25π + 25√2π,

V = 41 and 2 / 3π

Step-by-step explanation:

We need to determine the height here, as it is not given, and is quite important to us. The height is a perpendicular line segment to the radius, hence forming a 45 - 45 - 90 degree triangle as you can see. Therefore, by " Converse to Base Angles Theorem " the height should be equal in length to the radius,

( Height = 5 inches = Radius

______

Now knowing the height, let's begin by calculating the base area. By it's name, we have to find the area of the base. As it is a circle, let us apply the formula " πr^2 "

$\pi r^2\\= \pi ( 5 )^2\\= 25\pi$ - Base Area = 25π

______

The lateral area is simply the surface area excluding the base area, the surface area having a formula of " πr^2 + πrl. " Thus, the lateral area can be calculated through the formula " πrl, " but as we are not given the slant height ( l ) we have to use another formula, $l= \sqrt{r^2+h^2}$ -

$\pi r( \sqrt{r^2+h^2} )\\= \pi( 5 )( \sqrt{5^2 + 5^2} )\\= 25\sqrt{2} \pi$ - Lateral Area = 25√2π

______

And the surface area is the base area + lateral area -

$25\pi + 25\sqrt{2} \pi$ - Surface Area

______

The volume of a cone is 1 / 3rd that of a cylinder, with a simple formula of Base * height. Therefore, we can conclude the following -

$1 / 3( 25\pi )( 5 )\\= 25 / 3( 5 )( \pi )\\= 41 \frac{2}{3}\pi$ - Volume = 41 and 2 / 3π

8 0

2 years ago

IfA=(-2,-4) and B=(-8,4), what is the length of AB?

lisabon 2012 [21]

Answer:

<h2>10 units </h2>

solution,

A(-2,-4)--->(X1,y1)

B(-8,4)----->(x2,y2)

now,

$ab = \sqrt{(x2 - x1) ^{2} + {(y2 - y1)}^{2} } \\ \: \: \: \: = \sqrt{ (- 8 - ( - 2)) ^{2} + (4 - ( - 4)) ^{2} } \\ \: \: \: = \sqrt{( - 8 + 2) ^{2} + {(4 + 4)}^{2} } \\ \: \: \: = \sqrt{( - 6) ^{2} + {(8)}^{2} } \\ \: \: = \sqrt{36 + 64} \\ \: \: \: = \sqrt{100} \\ \: \: \: = \sqrt{ {(10)}^{2} } \\ \: \: \: \: = 10 \: units$