PLEASE HELP ASAP Z ⋅ 3 1/3

B) Then solve for x given that the area of the shaded region is 48 square units. Show all your work

nasty-shy [4]

Answer:

10

Step-by-step explanation:

Here it's given that the area of the shaded region is 48 u² and we need to find out the value of x .

The given figure is made up of two rectangles where the smaller one is inscribed into the bigger one .

The area of shaded region will be equal to the difference of the areas of the two rectangles , that is equal to 48 u² .
The area of rectangle is length × breadth .
The dimensions of bigger rectangle is (x+3)(x-4) and that of smaller one is 3 × x .

So that ,

$\longrightarrow$ ar(bigger) - ar(smaller) = 48

$\longrightarrow$ [ (x+3) (x-4)] - 3(x) = 48

$\longrightarrow$ [ x² - 4x +3x -12 - 3x ] = 48

$\longrightarrow$ x² -4x -12 -48 = 0

$\longrightarrow$ x² - 4x - 60 = 0

$\longrightarrow$ x² - 10x + 4x - 60 = 0

$\longrightarrow$ x( x -10) +4(x -10) = 0

$\longrightarrow$ (x +4)(x-10) = 0

$\longrightarrow$ x = -4,10

Since sides can't be negative ,

$\longrightarrow$ x = 10

<h3>Hence the required answer is 10.</h3>

8 0

3 years ago

Find the value of x in the equation below. x/10 = 8

steposvetlana [31]

The answer is 80 because 80 divided by 10 = 8

7 0

3 years ago

Confidence Interval Mistakes and Misunderstandingsâ€”Suppose that 500 randomly selected recent graduates of a university were as

kvv77 [185]

Answer:

The correct 95% confidence interval is (8.4, 8.8).

Step-by-step explanation:

The information provided is:

$n=500\\\bar x=8.6\\\sigma=2.2$

(a)

The (1 - α)% confidence interval for population mean (μ) is:

$CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

The 95% confidence interval for the average satisfaction score is computed as:

8.6 ± 1.96 (2.2)

This confidence interval is incorrect.

Because the critical value is multiplied directly by the standard deviation.

The correct interval is:

$8.6\pm 1.96 (\frac{2.2}{\sqrt{500}})=8.6\pm 0.20=(8.4,\ 8.6)$

(b)

The (1 - α)% confidence interval for the parameter implies that there is (1 - α)% confidence or certainty that the true parameter value is contained in the interval.

The 95% confidence interval for the mean rating, (8.4, 8.8) implies that the true there is a 95% confidence that the true parameter value is contained in this interval.

The mistake is that the student concluded that the sample mean is contained in between the interval. This is incorrect because the population is predicted to be contained in the interval.

(c)

The (1 - α)% confidence interval for population parameter implies that there is a (1 - α) probability that the true value of the parameter is included in the interval.

The 95% confidence interval for the mean rating, (8.4, 8.8) implies that the true mean satisfaction score is contained between 8.4 and 8.8 with probability 0.95 or 95%.

Thus, the students is not making any misinterpretation.

(d)

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

In this case the sample size is,

n = 500 > 30

Thus, a Normal distribution can be applied to approximate the distribution of the alumni ratings.

7 0

3 years ago

What is the equation of the line that passes through the point (3,5) and is perpendicular to the line x=4

Mamont248 [21]

Hello :
an equation is : y-5 = 0(x-3).... perpendicular to the line x=4 ( parrallel to x-axis)
so ;
y=5

7 0

3 years ago

A metallurgist has an alloy with 5% titanium and an alloy with 25% titanium. He needs 100 grams of an alloy with 13% titanium

Nimfa-mama [501]

So... the alloy with 5% titanium is ...some alloy :), so is a mixed metal and among its mix it has 5% of titanium, it has other metals, but only 5% is titanium.

that means that if the amount of it is say "x", then the amount of titanium in it will be 5% or (5/100) * x, or 0.05x.

now, the 25% titanium, is the same issue, if say a quantity of "y" will only have 25% of titanium, regardless of other metals, so in "y" there is 25% or (25/100) * y, or 0.25y.

so, we need a mixture that yields only a 13% alloy, or 13/100, and will be 100 grams, so the amount of titanium in it will then be (13/100) * 100 or 13.

$\bf \begin{array}{lccclll}
&\stackrel{grams}{amount}&\stackrel{\%}{quantity}&\stackrel{concentrated}{quantity}\\
\textit{5\% alloy}&x&0.05&0.05x\\
\textit{25\% alloy}&y&0.25&0.25y\\
------&------&------&------\\
mixture&100&0.13&13
\end{array}
\\\\\\
\begin{cases}
x+y=100\implies \boxed{y}=100-x\\
0.05x+0.25y=13\\
----------\\
0.05x+0.25\left( \boxed{100-x} \right)=13
\end{cases}
\\\\\\
0.05x-0.25x+25=13\implies -0.20x=-12\implies x=\cfrac{-12}{=0.20}
\\\\\\
x=\stackrel{grams}{60}$

how much of the 25% alloy will be needed? well, y = 100 - x.

4 0

3 years ago