10x is greater than or equal to 120 cm^2

Mathematics

1 answer:

Agata [3.3K]2 years ago

8 0

Answer:

x greater than or equal to 12cm^2

Step-by-step explanation:

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a group of five people go out to eat and buy appetizers and main dishes they decide to split the bill so each person pays 20% of

alexandr1967 [171]

5 people x $5 appetizers = $25
5 people x $10 main dish = $50
Total cost for the group of 5 = $75

$75 - 20% (.2) = $15 each person has to pay

Check ur work
$15 x 5 = $75

3 0

3 years ago

13) John and James live 0.7km apart. If John takes 25cm steps, then how many steps would it take

aleksandrvk [35]

It will take John 2800 steps to walk from his house to James' house

Step-by-step explanation:

First of all we have to convert all the measurements in same unit.

Distance = 0.7 km

As 1 km = 1000 m

0.7*100 = 700m

And

1 m = 100 cm

700*100 = 70000cm

Now,

25 cm = 1 step

$So,\\70000cm = \frac{70000}{25}\\=2800\ steps$

It will take John 2800 steps to walk from his house to James' house

Keywords: Conversions, Lengths

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7 0

3 years ago

Hi need help for this maths question

Sauron [17]

a) If f(y) is a probability density function, then both f(y) ≥ 0 for all y in its support, and the integral of f(y) over its entire support should be 1. eˣ > 0 for all real x, so the first condition is met. We have

$\displaystyle \int_{-\infty}^\infty f(y) \, dy = \frac14 \int_0^\infty e^{-\frac y4} \, dy = -\left(\lim_{y\to\infty}e^{-\frac y4} - e^0\right) = \boxed{1}$

so both conditions are met and f(y) is indeed a PDF.

b) The probability P(Y > 4) is given by the integral,

$\displaystyle \int_{-\infty}^4 f(y) \, dy = \frac14 \int_0^4 e^{-\frac y4} \, dy = -\left(e^{-1} - e^0\right) = \frac{e - 1}{e} \approx \boxed{0.632}$

c) The mean is given by the integral,

$\displaystyle \int_{-\infty}^\infty y f(y) \, dy = \frac14 \int_0^\infty y e^{-\frac y4} \, dy$

Integrate by parts, with

$u = y \implies du = dy$

$dv = e^{-\frac y4} \, dy \implies v = -4 e^{-\frac y4}$

Then

$\displaystyle \int_{-\infty}^\infty y f(y) \, dy = \frac14 \left(\left(\lim_{y\to\infty}\left(-4y e^{-\frac y4}\right) - \left(-4\cdot0\cdot e^0\right)\right) + 4 \int_0^\infty e^{-\frac y4} \, dy\right)$