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Rudiy27
3 years ago
13

Solve 18n-7p-15n=5p for n

Mathematics
2 answers:
umka2103 [35]3 years ago
6 0
Answer is D:

18n - 7p -15n = 5p
18n- 15n = 5p +7p
3n= 12p
n= 12p ÷ 3
n= 4p
Elden [556K]3 years ago
6 0

Answer:

Step-by-step explanation:

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The answer is √3. This is because √3 is already an irrational number and if you multiply it but either -5/9 or add it by 4, the product or sum will not terminate, nor repeat, or is able to turn in to a fraction
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There are 18 female students and 16 male students in a class. Which of the following expresses the ratio of female students to m
fiasKO [112]

Hey!

-------------------------------------------------

Solution:

The ratio is FEMALES to MALES.

So its 18/16, 18:16, or 18 to 16.

~Now we need to simplify.

18:16 → 18/2:16/2 → 9:8

*Note* When you have a ratio you need to know which comes first. If you put males to females that incorrect. It has to be correct according to the question.

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Answer:

\large\boxed{B)~9~to~8}

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Hope This Helped! Good Luck!

4 0
4 years ago
What is the equation of the circle?
boyakko [2]

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5 0
3 years ago
A segment with endpoints A (2, 1) and C (4, 7) is partitioned by a point B such that AB and BC form a 3:2 ratio. Find B.
sdas [7]

\bf ~~~~~~~~~~~~\textit{internal division of a line segment} \\\\\\ A(2,1)\qquad C(4,7)\qquad \qquad \stackrel{\textit{ratio from A to C}}{3:2} \\\\\\ \cfrac{A\underline{B}}{\underline{B} C} = \cfrac{3}{2}\implies \cfrac{A}{C} = \cfrac{3}{2}\implies 2A=3C\implies 2(2,1)=3(4,7)\\\\[-0.35em] ~\dotfill\\\\ B=\left(\frac{\textit{sum of "x" values}}{\textit{sum of ratios}}\quad ,\quad \frac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)\\\\[-0.35em] ~\dotfill

\bf B=\left(\cfrac{(2\cdot 2)+(3\cdot 4)}{3+2}\quad ,\quad \cfrac{(2\cdot 1)+(3\cdot 7)}{3+2}\right)\implies B=\left( \cfrac{4+12}{5}~,~\cfrac{2+21}{5} \right) \\\\\\ B=\left(\cfrac{16}{5}~~,~~\cfrac{23}{5} \right)\implies B=\left( 3\frac{1}{5}~~,~~4\frac{3}{5} \right)

8 0
4 years ago
Read 2 more answers
The probability density function of the time you arrive at a terminal (in minutes after 8:00 A.M.) is f(x) = 0.1 exp(−0.1x) for
Blababa [14]

f_X(x)=\begin{cases}0.1e^{-0.1x}&\text{for }x>0\\0&\text{otherwise}\end{cases}

a. 9:00 AM is the 60 minute mark:

f_X(60)=0.1e^{-0.1\cdot60}\approx0.000248

b. 8:15 and 8:30 AM are the 15 and 30 minute marks, respectively. The probability of arriving at some point between them is

\displaystyle\int_{15}^{30}f_X(x)\,\mathrm dx\approx0.173

c. The probability of arriving on any given day before 8:40 AM (the 40 minute mark) is

\displaystyle\int_0^{40}f_X(x)\,\mathrm dx\approx0.982

The probability of doing so for at least 2 of 5 days is

\displaystyle\sum_{n=2}^5\binom5n(0.982)^n(1-0.982)^{5-n}\approx1

i.e. you're virtually guaranteed to arrive within the first 40 minutes at least twice.

d. Integrate the PDF to obtain the CDF:

F_X(x)=\displaystyle\int_{-\infty}^xf_X(t)\,\mathrm dt=\begin{cases}0&\text{for }x

Then the desired probability is

F_X(30)-F_X(15)\approx0.950-0.777=0.173

7 0
4 years ago
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