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3 years ago

6/7x=42 solve for x

Mathematics

2 answers:

nordsb [41]3 years ago

5 0

6X/7 = 42

6X = 42×7 | : 6

X = 7×7

X = 49

dalvyx [7]3 years ago

5 0

Step-by-step explanation:

$\frac{6}{7x } = 42$

multiply both sides by 7 and you will get new equation as

$\frac{6}{x} = 154$

now multiply both sides with x and you will get

6 =154x ➡x = 1/49

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gulaghasi [49]

0.003 of 1-10 of is 0.03

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3 years ago

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

2 years ago

the perimeter of a rectangle is 36 feet. the length is 2 less than 3 times the width. what is the length? what is the width?

lana66690 [7]

Answer:

The width is 102

Step-by-step explanation:

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5 0

2 years ago

Calculate the slope of the line that goes through the points (2, 0) and (−4, 6)

Ivenika [448]

The answer is -1

look at the attached picture

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6 0

2 years ago

Read 2 more answers

The perimeter of a playing field for a certain sport is 174 ft. The field is a rectangle, and the length is 45 ft longer than th

Leno4ka [110]

Answer:

21 ft by 66 ft

Step-by-step explanation:

From the question,

P = 2(L+W)............... Equation 1

Where P = Perimeter of the playing Field, L = Length of the playing Field, W = width of the playing Field.

If the Length of the Field is 45 ft longer than the width,

L = 45+W............ Equation 2

Substitute Equation 2 into equation 1

P = 2(45+W+W)

P = 90+4W............. Equation 3

Given: P = 174 ft.

Substitute into equation 3

174 = 90+4W

4W = 174-90

4W = 84

W = 84/4

W = 21 ft.

Substituting the value of W into equation 2

L = 45+21

L = 66 ft.

Hence the dimensions of the playing field is 21 ft by 66 ft

3 0

2 years ago

6/7x=42 solve for x​

6/7x=42 solve for x