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2 years ago

Someone please help me!?

Mathematics

2 answers:

lisabon 2012 [21]2 years ago

6 0

Answer:

5.5

Step-by-step explanation:

Line is passing through the points (1, 0) and (3, 11)

Slope of line $=\frac{11 -0}{3-1}$

Slope of line $=\frac{11}{2}$

Slope of line $=5.5$

Ksenya-84 [330]2 years ago

3 0

Answer:

rise is 11 and run is 2

Step-by-step explanation:

i believe but not fully sure

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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

Eugene walked all the way to school at 3 mph then realized she forgot her math book (how could she?!), so she ran back at 7 mph.

Fynjy0 [20]

$\text{Answer: }1\frac{23}{40}\ miles\text{ far she live from school.}$

Explanation:

Since we have given that

Speed at which Eugene walked all the way to school = 3 mph

After realizing that she forgot her math book, she ran back

Speed at which Eugene walked back to her house = 7 mph

Let the distance between her house and her school be x

According to question,

$\frac{x}{3}+\frac{x}{7}=\frac{45}{60}\\\\\frac{7x+3x}{21}=\frac{3}{4}\\\\\frac{10x}{21}=\frac{3}{4}\\\\x=\frac{3\times 63}{4\times 10}=\frac{63}{40}=1\frac{23}{40}\ miles.$

Hence,

$1\frac{23}{40}\ miles\text{ far she live from school.}$

8 0

3 years ago

What addition expression is equivalent to<br> 3-(-7)?

Lesechka [4]

Answer:

3 - (-7)

= 3 + 7.

Step-by-step explanation:

A double negative is equivalent to a plus.

4 0

3 years ago

Read 2 more answers

Paige’s income statement for the month of December is shown. Paige monthly income statement for December. Total wages are 1,850

DedPeter [7]

Answer:

Paige’s net income for December is $950

Step-by-step explanation:

Here in this question, we are interested in calculating Paige’s net income for December

Mathematically;

Net income = Monthly wage - Total Expenses

From the question, monthly wage = $1,850 while the total expenses = $900

Thus, the net income will be ; $1,850 - $900 = $950

6 0

3 years ago

Helpmeeejjrjjrjrjtktktktii

Artist 52 [7]

I would first convert the mixed fraction into an improper fraction. To put it lightly, the reciprocal is the numerator and denominator are switched. For this instance it would be 4/37.

7 0

3 years ago