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3 years ago

11

Sam built a ramp to a loading dock. The ramp has a vertical support 2 m from the base of the loading dock and 3m from the base o

f the ramp. If the vertical support is 1.2 m in height, what is the height of the loading dock?

1 answer:

blsea [12.9K]3 years ago

5 0

Answer:

tan angle = x/5.

Step-by-step explanation:

If I understand your description,

tan angle at bottom of ramp = 1.2/3.

Then tan angle = x/5.

Check my thinking.

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slava [35]

Answer is <span>16y + 7z

</span><span><span><span><span>9y</span>+<span>11z</span></span>+<span>7y</span></span>−<span>4z

</span></span><span>=<span><span><span><span><span>9y</span>+<span>11z</span></span>+<span>7y</span></span>+</span>−<span>4z

</span></span></span>Combine Like Terms:

<span>=<span><span><span><span>9y</span>+<span>11z</span></span>+<span>7y</span></span>+<span>−<span>4z

</span></span></span></span><span>=<span><span>(<span><span>9y</span>+<span>7y</span></span>)</span>+<span>(<span><span>11z</span>+<span>−<span>4z</span></span></span>)

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7 0

2 years ago

A gambler has a coin which is either fair (equal probability heads or tails) or is biased with a probability of heads equal to 0

yawa3891 [41]

Answer:

(a) 0.1719

(b) 0.3504

Step-by-step explanation:

For every coin the number of heads follows a Binomial distribution and the probability that x of the 10 times are heads is equal to:

$P(x)=\frac{n!}{x!(n-x)!}*p^x*(1-p)^{10-x}$

Where n is 10 and p is the probability to get head. it means that p is equal to 0.5 for the fair coin and 0.3 for the biased coin

So, for the fair coin, the probability that the number of heads is less than 4 is:

$P(x$

Where, for example, P(0) and P(1) are calculated as:

$P(0)=\frac{10!}{0!(10-0)!}*0.5^0*(1-0.5)^{10-0}=0.0009\\P(1)=\frac{10!}{1!(10-1)!}*0.5^1*(1-0.5)^{10-1}=0.0098$

Then, $P(x$ , so there is a probability of 0.1719 that you conclude that the coin is biased given that the coin is fair.

At the same way, for the biased coin, the probability that the number of heads is at least 4 is:

$P(x\geq4 )=P(4)+P(5)+P(6)+...+P(10)$

Where, for example, P(4) is calculated as:

$P(4)=\frac{10!}{4!(10-4)!}*0.3^4*(1-0.3)^{10-4}=0.2001$

Then, $P(x\geq4 )=0.3504$ , so there is a probability of 0.3504 that you conclude that the coin is fair given that the coin is biased.

7 0

3 years ago

!!Plz urgent help!!

Firdavs [7]

Answer:

890 beads can be fitted in the triangular prism.

Step-by-step explanation:

If we can fill the spherical beads completely in the triangular prism,

Volume of the triangular prism = Volume of the spherical beads

Volume of triangular prism = Area of the triangular base × Height

From the picture attached,

Area of the triangular base = $\frac{1}{2}(\text{Base})(\text{Height})$

= $\frac{1}{2}(AB)(BC)$

By applying Pythagoras theorem in the given triangle,

AC² = AB² + BC²

(13)² = 5² + BC²

169 = 25 + BC²

BC² = 144

BC = 12

Area of the triangular base = $\frac{1}{2}(5)(12)$

= 30 cm²

Height of the triangular prism = 18 cm

Volume of the triangular prism = 30 × 18

= 540 cm³

Volume of one spherical bead = $\frac{4}{3}\pi r^{3}$

= $\frac{4}{3}\pi (0.525)^{3}$

= 0.606 cm³

Let there are 'n' beads in the triangular prism,

Volume of 'n' beads = Volume of the prism

540 = 0.606n

n = 890.90

n ≈ 890

Therefore, 890 beads can be fitted in the triangular prism.

3 0

2 years ago

Solve using the box method

Lena [83]

$\huge\text{Hey there!}$

$\large\text{Just SIMPLIFY the given EQUATION or find the DIFFERENCE}\\\large\text{OF the SQUARES... Here is the formula: }\mathsf{\bf a^2 - b^2 =(a+b)(a-b)}$

$\large\text{Equation: }\mathsf{\dfrac{(4x^2-9)}{(2x + 3)}}$

$\large\text{Rewrite }\mathsf{ 4x^9 - 9}\large\text{ in the formation of }\mathsf{a^2 - b^2}\large\text{ whereas}\mathsf{a = 2x \ \&\ b = 3.}$

$\large\text{Equation: }\mathsf{\dfrac{(2x)^2-3^2}{2x + 3}}$

$\large\text{This is where you try to do the DIFFERENCE OF its SQUARES}$

$\mathsf{\dfrac{(2x + 3)(2x - 3)}{2x + 3}}$

$\large\text{CANCEL out: }\mathsf{(2x + 3)\ - (2x +3)}\large\text{ because it gives you 0}$

$\large\text{This leaves us with }\mathsf{\bf 2x - 3}\large\text{ as your POSSIBLE ANSWER}$

$\boxed{\boxed{\large\text{Answer: \huge \bf 2x - 3}}}\huge\checkmark$

$\text{Good luck on your assignment and enjoy your day!}$

~ $\frak{Amphitrite1040:)}$

$\large\text{Note: There is/are many ways to solve for equations like this.... this was just}\\\large\text{the quickest and easiest way to understand it!}$

3 0

2 years ago

Combine Like Terms : 7x^3+4x^2-x^3+6x+5x-4x^3

larisa86 [58]

Answer:

2x^2+4x^2+11x

Step-by-step explanation:

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2 years ago