If the probability of getting a particular result in an experiment is 75.3%. What is the probability of not getting that result?

boeing plans to increase its price for jetliners. with a selling price $201.5 million and a cost of $190.1 million, what was the

scoray [572]

so from 190.1 to 201.5 is 201.5 - 190.1 = 11.4.

now, if we take 190.1 as the 100%, what is 11.4 off of it in percentage?

$\bf \begin{array}{ccll} amount&\%\\ \cline{1-2} 190.1&100\\ 11.4&x \end{array}\implies \cfrac{190.1}{11.4}=\cfrac{100}{x}\implies 190.1x=1140 \\\\\\ x=\cfrac{1140}{190.1}\implies x\approx 5.99684$

3 0

2 years ago

Read 2 more answers

Write an expression that represents the volume of a rectangular-shaped box with a length 3 inches less than its height and a wid

AfilCa [17]

The expression that represents the volume of a rectangular-shaped box is: A. x³ + x² - 12x.

<h3>What is the Volume of a Rectangular Prism?</h3>

A rectangular-shaped box is a rectangular prism. To find the volume of a rectangular-shaped box, we would apply the formula of the volume of a rectangular prism, which is expressed as: V = (length)(width)(height).

We are given the following dimensions of the rectangular-shaped box:

Let the height be represented as x

Height of the rectangular-shaped box = x in.

Length of the rectangular-shaped box = (x - 3) in.

Width of the rectangular-shaped box = (x + 4) in.

Plug in the values into the volume formula:

Volume of the rectangular-shaped box = (x - 3)(x)(x + 4)

Expand (x - 3)(x)(x + 4)

Volume of the rectangular-shaped box = x³ + x² - 12x.

Therefore, the expression that represents the volume of a rectangular-shaped box with the given dimensions is: A. x³ + x² - 12x.

Learn more about the volume of rectangular prism on:

brainly.com/question/12917973

#SPJ1

5 0

1 year ago

8th grade teachers and students from M.I.T went on a field trip to an zoo. Tickets

alukav5142 [94]

As there are multiple answers going with trial and error seems good in this case.

A

5.00 + 5.00 + 5.00 + 5.00 = 20.00

3.50 + 3.50 + 3.50 + 3.50 + 3.50 + 3.50 + 3.50 + 3.50 + 3.50 = 31.5

20.00 + 31.5 = 51.5

B

5.00 + 5.00 + 5.00 + 5.00 = 20.00

3.50 + 3.50 + 3.50 + 3.50 + 3.50 + 3.50 + 3.50 + 3.50 + 3.50 + 3.50 = 35.00

20.00 + 35.00 = 55.00

C

5.00 + 5.00 + 5.00 = 15.00

3.50 + 3.50 + 3.50 + 3.50 + 3.50 + 3.50 + 3.50 + 3.50 + 3.50 = 31.5

15.00 + 31.5 =46.5

Knowing that for the first one all A,B, and C were wrong we can determine that D is the answer. But wait if I made one mistake and didn't understand if $156.5 is equivalent to $156.50 that would mean A is correct.

Always recheck your answers and good luck,

Jon

6 0

2 years ago

Does anyone know the answer to number 37?

kotegsom [21]

Answer: 37) 170

Step-by-step explanation:

85 - (-85) = 85 + 85 = 170

$\textit{\textbf{Spymore}}$

7 0

3 years ago

Determine whether the sequences converge.

Alik [6]

$a_n=\sqrt{\dfrac{(2n-1)!}{(2n+1)!}}$

Notice that

$\dfrac{(2n-1)!}{(2n+1)!}=\dfrac{(2n-1)!}{(2n+1)(2n)(2n-1)!}=\dfrac1{2n(2n+1)}$

So as $n\to\infty$ you have $a_n\to0$ . Clearly $a_n$ must converge.

The second sequence requires a bit more work.

$\begin{cases}a_1=\sqrt2\\a_n=\sqrt{2a_{n-1}}&\text{for }n\ge2\end{cases}$

The monotone convergence theorem will help here; if we can show that the sequence is monotonic and bounded, then $a_n$ will converge.

Monotonicity is often easier to establish IMO. You can do so by induction. When $n=2$ , you have

$a_2=\sqrt{2a_1}=\sqrt{2\sqrt2}=2^{3/4}>2^{1/2}=a_1$

Assume $a_k\ge a_{k-1}$ , i.e. that $a_k=\sqrt{2a_{k-1}}\ge a_{k-1}$ . Then for $n=k+1$ , you have

$a_{k+1}=\sqrt{2a_k}=\sqrt{2\sqrt{2a_{k-1}}\ge\sqrt{2a_{k-1}}=a_k$

which suggests that for all $n$ , you have $a_n\ge a_{n-1}$ , so the sequence is increasing monotonically.

Next, based on the fact that both $a_1=\sqrt2=2^{1/2}$ and $a_2=2^{3/4}$ , a reasonable guess for an upper bound may be 2. Let's convince ourselves that this is the case first by example, then by proof.

We have

$a_3=\sqrt{2\times2^{3/4}}=\sqrt{2^{7/4}}=2^{7/8}$
$a_4=\sqrt{2\times2^{7/8}}=\sqrt{2^{15/8}}=2^{15/16}$

and so on. We're getting an inkling that the explicit closed form for the sequence may be $a_n=2^{(2^n-1)/2^n}$ , but that's not what's asked for here. At any rate, it appears reasonable that the exponent will steadily approach 1. Let's prove this.

Clearly, $a_1=2^{1/2}$ . Let's assume this is the case for $n=k$ , i.e. that $a_k$ . Now for $n=k+1$ , we have

$a_{k+1}=\sqrt{2a_k}$

and so by induction, it follows that $a_n$ for all $n\ge1$ .

Therefore the second sequence must also converge (to 2).

4 0

3 years ago