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elixir [45]
3 years ago
12

I’m the triangle shown below what is the approximate value of x

Mathematics
1 answer:
Dafna1 [17]3 years ago
8 0

Answer:

The answer is choice B: 22.52 units.

Step-by-step explanation:

We use the Pythagoras's theorem which says for a right triangle with sides x,y and hypotenuse z:

x^2+y^2=z^2.

In our case

z=26

y=13

So we solve for x:

x=\sqrt{z^2-y^2}=\sqrt{26^2-13^2}  =\boxed{22.52\: units.}

Which is choice B.

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Multiply.<br><br> 1.25(−3.6)<br><br> Enter your answer as a decimal in the box.
Darina [25.2K]

-4.5 is your answer, hope this helped



5 0
3 years ago
What is the probability of rolling an even number on a standard number cube and then rolling a one
expeople1 [14]

Answer:

multiply the chances

1/2 (every other number is even right)

1/2 x 1/6 (if it is a normal dice) = 1/12

1/12 chance

Step-by-step explanation:

4 0
3 years ago
if some one did 35% of their hw over 7 days if that person reads the same amount of pages each night , how many more days will i
Llana [10]

Answer:

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Step-by-step explanation:

6 0
3 years ago
Customers arrive at a bank drive-up window every 6 minutes based on a Poisson distribution. Once they arrive at the teller, serv
Jet001 [13]

Answer:

(a) 0.01029 (b) 4.167 customers (c) 0.4167 hours or 25 minutes (d) 0.5 hours or 30 minutes

Step-by-step explanation:

With an arrival time of 6 minutes, λ=10 clients/hour.

With a service time of 5 minutes per transaction, μ=12 transactions/hour.

(a) The probability of 3 or fewer customers arriving in one hour is

P(C<=3)=P(1)+P(2)+P(3)

P(X)=\lambda^{X}*e^{-\lambda}  /X!

P(1)=10^{1}*e^{-10}  /1!=0.00045\\P(3)=10^{3}*e^{-10}  /3!=0,00227\\P(2)=10^{2}*e^{-10}  /2!=0,00757\\P(x\leq3)=0.00045+0.00227+0.00757 = 0.01029

(b) The average number of customers waiting at any point in time (Lq) can be calculated as

L_{q}=p*L=\frac{\lambda}{\mu}*\frac{\lambda}{\mu-\lambda}\\L_{q}=\frac{10}{12}*\frac{10}{12-10}\\\\L_{q}=100/24=4.167

(c) The average time waiting in the system (Wq) can be calculated as

W_{q}=p*W=\frac{\lambda}{\mu}*\frac{1}{\mu-\lambda}\\W_{q}=(10/12)*(1/(12-10))\\W_{q}=(10/24)=0.4167

(d) The average time in the system (W), waiting and service, can be calculated as

W=\frac{1}{\mu-\lambda}\\W=\frac{1}{12-10}=0.5

6 0
3 years ago
Find the area of this semi-circle with diameter, d = 49cm.<br> Give your answer rounded to 2 DP.
lilavasa [31]
The area of a semicircle is (πr^2)/2, with r being the radius
the radius is going to be the diameter diced by two: 49/2 = 24.5
now, plug in and solve
(π24.5^2)/2 = 600.25π/2 = 300.125 π
π is roughly equivalent to 3.14, so we can multiply to simplify
area = 942.87

8 0
2 years ago
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