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3 years ago

Weights of cars passing over a bridge part 2

Mathematics

1 answer:

DaniilM [7]3 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)

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The lengths of pregnancies in a small rural village are normally distributed with a mean of 261 days and a standard deviation of

valina [46]

Answer:

(221.39, 300.61) and (255.2223, 266.7777)

Step-by-step explanation:

Given that X, the lengths of pregnancies in a small rural village are normally distributed with a mean of 261 days and a standard deviation of 17 days

Middle 98% would lie on either side of the mean with probability ±2.33 in the std normal distribution on either side of 0

Corresponding we have x scores as

Between $261-2.33*17$ and $261+2.33*17$

i.e. in the interval = (221.39, 300.61)

If sample size = 47, then std error of sample would be

$\frac{17}{\sqrt{47} }$

So 98% of pregnancies would lie between

$261-2.33*\frac{17}{\sqrt{47} }$ and $262+2.33*\frac{17}{\sqrt{47} }$

= (255.2223, 266.7777)

8 0

3 years ago

Which is equivalent to 3 square root of 8 1/4 x

Slav-nsk [51]

Answer:

√

Step-by-step explanation:

Simplify

(Something is wrong with my keyboard hopefully you can put this together I'm sorry)

4 0

3 years ago

Read 2 more answers

The expression f(x) = 12(1.035)* models the monthly growth of membership in the new drama club at a school. According to the fun

olga nikolaevna [1]

Answer:

The monthly growth rate is 3.5%.

Step-by-step explanation:

The exponential growth function is given as follows:

$y=a(1+r)^{x}$

Here,

y = final value

a = initial value

r = growth rate

x = time taken

The provided expression for the monthly growth of membership in the new drama club at a school is:

$f(x) = 12\cdot(1.035)^{x}$

Comparing this function with the exponential growth function:

$a(1+r)^{x}=12(1.035)^{x}\\\\a(1+r)^{x}=12(1+0.035)^{x}$

Then value of r is 0.035 or 3.5%.

Thus, the monthly growth rate is 3.5%.

7 0

3 years ago

Find each value.Use a calculator is needed P(9,2) .P(5,5). P(7,7)

Andrews [41]

The answers are:
P(9, 2) = 72
P(5, 5) = 120
P(7, 7) = 5040

These are examples of permutations. The problems are asking us to find the total number of ways an event can happen.

In the first case, P(9, 2), we are asked to find the ways that 2 things can be chosen out of a group of 9.
9 x 8 = 72

P(5, 5) = 5 x 4 x 3 x 2 x 1
P(7, 7) = 7 x 6 x 5 x 4 x 3 x 2 x 1

3 0

3 years ago

what is the mean of the set : 95,45,37,82,90,100,91,78,67,84,85,85,82,91,93,92,76,84,100,59,92,77,68,88

zmey [24]

Answer:

The mean is 80.875

Step-by-step explanation:

8 0

3 years ago

Weights of cars passing over a bridge part 2​

Weights of cars passing over a bridge part 2