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dlinn [17]
2 years ago
12

What is the area of the shape below

Mathematics
1 answer:
Tems11 [23]2 years ago
7 0

8cm, since 6 full squares and 4 halves

hope it helps...!!!

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Which ordered pairs are solutions to the inequality y−4x≤−6 ?
amid [387]

I think it is A, (-2,-14)

7 0
3 years ago
Which function grows the fastest for large values of x? f(x)=8x f(x)=3x f(x)=4x2+3 f(x)=1.5x 20 points
Aleonysh [2.5K]
The 4 functions are:
f_1 (x) = 8x
f_2(x)=3x
f_3(x)=4x^2+3
f_4(x)=1.5 x

Let's keep in mind that for large values of x, a quadratic function grows faster than a linear function:
ax^2 \ \textgreater \  kx for large values of x

In this problem, we can see that the only quadratic function is f_3(x), while all the others are linear functions, so the function that grows faster for large values of x is
f_3(x) = 4x^2 +3
7 0
3 years ago
1. Reduce the ratio to the simplest form: <br> a)5:10=_____. <br> b)60:54=____.
Nikitich [7]

Answer:

1:2 by division of 5

10:9 by division of 6

6 0
2 years ago
Read 2 more answers
A line segment on a number line has its endpoints at -9 and 6. what is midpoint
ivolga24 [154]
Midpoint formula = (-9 + 6) / 2 = -3/2
8 0
3 years ago
The number of people arriving for treatment at an emergency room can be modeled by aPoisson process with a rate parameter of 5/h
saveliy_v [14]

Answer:

(a) The probability of having exactly four arrivals during a particular hour is 0.1754.

(b) The probability that at least 3 people arriving during a particular hour is 0.7350.

(c) The expected arrivals in a 45 minute period (0.75 hours) is 3.75 arrivals.

Step-by-step explanation:

(a) If the arrivals can be modeled by a Poisson process, with λ = 5/hr, the probability of having exactly four arrivals during a particular hour is:

P(X=4)=\frac{\lambda^{X}*e^{-\lambda}  }{X!} =\frac{5^{4}*e^{-5}  }{4!}=\frac{625*0.006737947}{24} =\frac{4.211}{24}=0.1754

The probability of having exactly four arrivals during a particular hour is 0.1754.

(b) The probability that at least 3 people arriving during a particular hour can be written as

P(X>3)=1-P(X\leq3)=1- (P(0)+P(1)+P(2)+P(3))

Using

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

We get

P(X>3)=1-P(X\leq3)=1- (P(0)+P(1)+P(2)+P(3))\\P(X>3)=1-(0.0067+ 0.0337+ 0.0842 + 0.1404 )\\P(X>3)=1-0.2650=0.7350\\

The probability that at least 3 people arriving during a particular hour is 0.7350.

(c) The expected arrivals in a 45 minute period (0.75 hours) is

EV=\lambda*t=5*0.75=3.75

8 0
3 years ago
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