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2 years ago

A=1/2 (a+b)h work out A if a=2 b=7 h=3

Mathematics

1 answer:

BartSMP [9]2 years ago

6 0

Step-by-step explanation:

A =1/2(a+b)h

1/2(2+7)3
1/2×27
13.5

hope it helps.

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Help please! i don’t understand how to do this!

atroni [7]

Answer:

x=c+b/a

Step-by-step explanation:

ax-b=c

+b +b

ax=c+b

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a=c+b/a

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2 years ago

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I need help finding the value of n

Degger [83]

8n+3+9n+7=180
17n+10=180
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Answer
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How do I do 3 x 10^5

Ilya [14]

Answer: 3(10^5)

=3*10^5

=3*(10*10*10*10*10)

=300000

This is how you should do It, just when it is squared you should times the number like up here you times 10 times 10 5 times so if you see it you don't times it by 5 you do 10 times 10 times 10 times 10 times 10 like this

You should get it now

Have a good day

Yours,

Nathaniel

Step-by-step explanation:

4 0

2 years ago

Part Sophia's brother, Stanley, is cooking carrots for the family dinner. He needs to make less than twice as many servings of c

Masteriza [31]

Answer:

Step-by-step explanation:

Twice the serving of mashed potatoes is 2 x 10 = 20

If y is the number then y < 20

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2 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

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

$\displaystyle \cdots = \int_0^\infty e^{-\frac y4} \, dy$

$\displaystyle \cdots = -4 \left(\lim_{y\to\infty} e^{-\frac y4} - e^0\right) = \boxed{4}$

8 0

1 year ago