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4 years ago

The distance around a meteor crater is 9,753 ft. Find the diameter of the crater. Use 22 Over 7 for piπ.

1 answer:

4 0

So distance areound=circumference=pi times diameter
so
circumference=9753
pi=22/7

subsitute
9753=22/7 time diameter
multipliy both sides by 7/22 to clear fraction
68271/22=diamter
3103 and 5/22=3103.23 ft

answer is 3103.23 feet

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Suppose the expected tensile strength of type-A steel is 103 ksi and the standard deviation of tensile strength is 7 ksi. For ty

ExtremeBDS [4]

Answer:

i So the approximate distribution of $\= X$ is $\mu_{\= X} =103$ and $\sigma_{\= X} = 0.783$

ii So the approximate distribution of $\= Y$ is $\mu_{\= Y} =105$ and $\sigma_{\= Y} = 0.645$

the approximate distribution of $\=X - \= Y$ is $E (\= X - \= Y) = -2$ and $\sigma_{\= X - \=Y}=1.029$

Here we can see that the mean of the approximate distribution is negative which tell us that this negative value of the data for $\=X - \= Y$ sample are more and their frequency occurrence is higher than the positive values

the value of $P(-1 \le \=X - \= Y \le 1)$ is $= -0.1639$

Step-by-step explanation:

From the question we are given that

The expected tensile strength of the type A steel is $\mu_A = 103 ksi$

The standard deviation of type A steel is $\sigma_A = 7ksi$

The expected tensile strength of the type B steel is $\mu_B = 105\ ksi$

The standard deviation of type B steel is $\sigma_B = 5 \ ksi$

Also the assumptions are

Let $\= X$ be the sample average tensile strength of a random sample of 80 type-A specimens

Here $n_a =80$

Let $\= Y$ be the sample average tensile strength of a random sample of 60 type-B specimens.

Here $n_b = 60$

Let the sampling distribution of the mean be

$\mu _ {\= X} = \mu$

$=103$

Let the sampling distribution of the standard deviation be

$\sigma _{\= X} = \frac{\sigma }{\sqrt{n_a} }$

$= \frac{7}{\sqrt{80} }$

$=0.783$

So What this mean is that the approximate distribution of $\= X$ is $\mu_{\= X} =103$ and $\sigma_{\= X} = 0.783$

For $\= Y$

The sampling distribution of the sample mean is

$\mu_{\= Y} = \mu$

$= 105$

The sampling distribution of the standard deviation is

$\sigma _{\= Y} = \frac{\sigma }{\sqrt{n_b} }$

$= \frac{5}{\sqrt{60} }$

$= 0.645$

So What this mean is that the approximate distribution of $\= Y$ is $\mu_{\= Y} =105$ and $\sigma_{\= Y} = 0.645$

Now to obtain the approximate distribution for $\=X - \= Y$

$E (\= X - \= Y) = E (\= X) - E(\= Y)$

$= \mu_{\= X} - \mu_{\= Y}$

$= 103 -105$

$= -2$

The standard deviation of $\=X - \= Y$ is

$\sigma_{\= X - \=Y} = \sqrt{\sigma_{\= X}^2 - \sigma_{\= Y}^2}$

$= \sqrt{(0.783)^2 + (0.645)^2}$

$=1.029$

Now to find the value of $P(-1 \le \=X - \= Y \le 1)$

Let us assume that $F = \= X - \= Y$

$P(-1 \le F \le 1)$ $= P [\frac{-1 -E (F)}{\sigma_F} \le Z \le \frac{1-E(F)}{\sigma_F} ]$

$= P[\frac{-1-(-2)}{1.029} \le Z \le \frac{1-(-2)}{1.029} ]$

$= P[0.972 \le Z \le 2.95]$

$= P(Z \le 0.972) - P(Z \le 2.95)$

Using the z-table to obtain their z-score

$= 0.8345 - 0.9984$

$= -0.1639$

3 0

3 years ago

What is the LCM and GCF of 13 and 15?!

AlladinOne [14]

$LCM\\\\
13:13\\1:1\\\\
15:3\\
5:5\\1:1\\\\LCM(13,15)=1*3*5*13=195\\\\
GCF:\\\\GCF(13,15)=1.$

3 0

4 years ago

Read 2 more answers

What is the answer plz?

Studentka2010 [4]

Answer:

read the explanation

Step-by-step explanation:

the first one is positive 24 and the second one is -2

8 0

3 years ago

Read 2 more answers

What is an equation of the line that passes through the points (-6, 3) and (-6, -3)?

Harlamova29_29 [7]

9514 1404 393

Answer:

x = -6

Step-by-step explanation:

A plot of the points tells you one is above the other on the same vertical line. The equation for a vertical line is ...

x = constant

In order for the line to go through points that have x-coordinates of -6, the constant must be -6. The equation of the line is ...

x = -6

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3 years ago

If the probability of an event is , and there are 28 total outcomes, how many favorable outcomes are there?

Aleks [24]

14 would be our answer

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3 years ago