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4 years ago

HELPPPPP MEEEEE PLEASEEEEE.

1 answer:

8 0

You can make the first equation equal to x: x=0.5y+5 and substitute it into the second equation: 2(0.5x+5)-2y=4 -> -y+10=4 so y=6.

Then plug y into either equation: 2x-(6)=10 -> x=8

The correct ordered pair is (8,6)

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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

WILL MARK YOU BRAINLIEST!!!!

Tema [17]

Answer:

A is not 3, 7

Step-by-step explanation:because it isint

4 0

3 years ago

Casey is six times as old as Ralph. In two years, she will be four times as old as Ralph. How old is each of them now?

lisov135 [29]

Ralph is 3 years old

Casey is 18 years old

Step-by-step explanation:

Let the age of Ralph be x

Then Casey is 6 times as old as Ralph will be 6x

In two years

Ralph =x+2

Casey=6x+2 ⇒ 4(x+2)

Equate the two equations for Casey as

6x+2 = 4x+8

6x-4x=8-2

2x=6

x=6/2= 3

Current age;

Ralph= x= 3 years old

Casey=6x = 6*3= 18 years

Learn More

Forming expressions and solving equations;brainly.com/question/1280754

Keywords :six, times, years

#LearnwithBrainly

4 0

3 years ago

Please please please help me

Montano1993 [528]

the answer is c i used a algebra calculator

7 0

3 years ago

PLEASE HELP QUICK WILL MARK BRAINLIEST!!!!!!!!!

Helga [31]

Answer:s>_18

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers