All categories

3 years ago

X+3menor que 8 qual o resultado

Mathematics

1 answer:

elixir [45]3 years ago

7 0

x+3<8

8-3 =5

X<5

any number less than 5 will work

You might be interested in

What is fourteen times one thousand ?*

ioda

The trick with answering your 10s, 100s, 1000s, etc, is just adding zeroes onto the ends of the first number. So, since this is 14 times 1000, you would add 3 zeroes onto 14. So your answer is 14,000. Hope this helped!

7 0

3 years ago

Read 2 more answers

The U.S. Census Bureau announced that the median sales price in 2014 for new homes was $282,800 and the mean sales price was $34

victus00 [196]

Answer:

a) 0.9964

b) 0.3040

Step-by-step explanation:

Given data:

standard deviation = $90,000

Mean sales price =$345,800

sample mean = $370,000

Total number of sample = 100

calculate z score for [/tex](\bar x = 370000)[/tex]

$z = \frac{\bar x - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{370000 - 345800}{\frac{90000}{\sqrt{100}}}$

z = 2.689

P(x<370000) = P(Z<2.689)

FROM STANDARD NORMAL DISTRIBUTION TABLE FOR Z P(Z<2.689) = 0.9964

calculate z score for (\bar x = 350000)

$z = \frac{\bar x - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{350000 - 345800}{\frac{90000}{\sqrt{100}}}$

z = 2.133

$P(350000 \leq \bar x \leq 365000) = P(0.467 \leq Z\leq 2.133) = P(Z\leq 2.133) - P(Z\leq 0.467)$

FROM NORMAL DISTRIBUTION TABLE Z VALUE FOR

$P(Z\leq 2.133) = 0.9836$

$P(Z\leq 0.467) = 0.6796$

SO, = 0.9836 - 0.6796 = 0.3040

4 0

3 years ago

Find f(1/3) if f(n)=3n

tia_tia [17]

F(n)=3n
f(1/3)=3(1/3)
f(1/3)=1

Hope I didn't mess up for your sake

8 0

3 years ago

What is the formulaa to fund % of selling overheads(Tender sheet)

Lina20 [59]

I might not know the answer to this question but I hope you find one

4 0

3 years ago

Prove the theorem (AB )^T= B^T. A^T

Lisa [10]

Answer:

$(AB)^T = B^T.A^T (Proved)$

Step-by-step explanation:

Given (AB )^T= B^T. A^T;

To prove this expression, we need to apply multiplication law, power law and division law of indices respectively, as shown below.

$(AB)^T = B^T.A^T\\\\Start, from \ Right \ hand \ side\\\\B^T.A^T = \frac{B^T.A^T}{A^T}.\frac{B^T.A^T}{B^T} (multiply \ through) \\\\ = \frac{A^{2T}.B^{2T}}{A^T.B^T} \\\\=\frac{(AB)^{2T}}{(AB)^T} \ \ (factor \ out \ the power)\\\\= (AB)^{2T-T} \ (apply \ division \ law \ of \ indices; \ \frac{x^a}{x^b} = x^{a-b})\\\\= (AB)^T \ (Proved)$

3 0

3 years ago