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3 years ago

Solve 30x -12,000 less than or equal to 980

Mathematics

1 answer:

Lyrx [107]3 years ago

6 0

Answer:

$\large\boxed{x\leq432\dfrac{2}{3}\to x\in\left(-\infty,\ 432\dfrac{2}{3}\right]}$

Step-by-step explanation:

$30x-12,000\leq980\qquad\text{add 12,000 to both sides}\\\\30x-12,000+12,000\leq980+12,000\\\\30x\leq12,980\qquad\text{divide both sides by 30}\\\\x\leq\dfrac{12,980}{30}\\\\x\leq\dfrac{1,298}{3}\\\\x\leq432\dfrac{2}{3}$

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How many elements are in the union of five sets if the sets contain 10,000 elements each, each pair of sets has 1000 common elem

masya89 [10]

Answer:

40951

Step-by-step explanation:

Using the principles of inclusion - Exclusion

Where C(n,r)=n!/(n-r)!r!

Total elements in the five sets including number repetition is given as (10000)×C(5, 1) =10000× 5!/(5-1)!1!=10000×5=50000

Total Number of elements in each pair including number repetition of sets is given as

=(1000) × C(5, 2) =10000

Number of elements in each triple of sets is given as

=(100) × C(5, 3) =1000

Number of elements in every four sets

=(10) × C(5, 4)=50

Number of elements in every one set

(1) × C(5, 5)=1

Therefore total number of unique elements=50000-10000+1000-50+1

=40951

5 0

3 years ago

Find the oth term of the geometric sequence 9, -18, 36, ...

Sphinxa [80]

Answer:

2304

Step-by-step explanation:

Given :-

A geometric sequence is given to us which is 9 , -18 , 36.

And we need to find out the 9th term of the sequence. Here firstly we should find the Common Ratio and then we can substitute the respective values in the formula to find the nth term of a geometric sequence .

Common Ratio :-

$:\implies$ CR = -18÷ 9 = -2

The 9 th term :-

$:\implies$ T_n = arⁿ - ¹

$:\implies$ T_9 = 9× (-2) ⁹ - ¹

$:\implies$ T_9 = 9 × (-2)⁸

$:\implies$ T_9 = 9 × 256

$:\implies$ T_9 = 2304

Hence the 10th term is 2304.

3 0

3 years ago

The amount of time Ricardo spends brushing his teeth follows a Normal distribution with unknown mean and standard deviation. Ric

Papessa [141]

Let $X$ denote the number of minutes spent brushing his teeth, and let $\mu$ be the mean and $\sigma$ the standard deviation for this distribution.

$\mathbb P(X$

The z-score corresponding to this probability is approximately $z=-0.2533$ , which means

$\dfrac{1-\mu}\sigma=-0.2533\iff\mu-0.2533\sigma=1$

Next, (note the sign change)

$\mathbb P(X>2)=0.02\implies\mathbb P(X\le2)=\mathbb P\left(\dfrac{X-\mu}\sigma\le\dfrac{2-\mu}\sigma\right)=0.98$

The corresponding z-score is approximately $z=2.0538$ , so you have

$\dfrac{2-\mu}\sigma=2.0538\iff\mu+2.0538\sigma=2$

Solving the two equations for $\mu$ and $\sigma$ , you'll find that the mean is approximately $\mu=1.1098$ and the standard deviation is approximately $\sigma=0.4334$ .