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umka2103 [35]
3 years ago
15

Can 4:10 be switched to 10:4

Mathematics
2 answers:
ValentinkaMS [17]3 years ago
5 0
It depends on the question . What is the question asking about?
VLD [36.1K]3 years ago
3 0
No, that would be saying that 0.4 (4/10) and 2.5 (10/4) are equal, which they aren't.
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Hope this helps!
==jding713==
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What is the solution to this equation 3(80 - 9x) = x - 96
andreev551 [17]

240-27x = x-96

240 - 28x = -96

28x = 336

x = 12

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3 years ago
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in the given kite SZ= 10yards , WZ = 10 yards TZ=12 yards , and RZ=32 yards Determine the area of the kite
Fynjy0 [20]

If we assume the given segments are those from the vertices to the point of intersection of the diagonals, it seems one diagonal (SW) is 20 yards long and the other (TR) is 44 yards long. The area (A) of the kite is half the product of the diagonals:

... A = (1/2)·SW·TR = (1/2)·(20 yd)·(44 yd)

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3 years ago
The weekly amount spent by a small company for in-state travel has approximately a normal distribution with mean $1450 and stand
Llana [10]

Answer:

0.0903

Step-by-step explanation:

Given that :

The mean = 1450

The standard deviation = 220

sample mean = 1560

P(X > 1560) = P( Z > \dfrac{x - \mu}{\sigma})

P(X > 1560) = P(Z > \dfrac{1560 - 1450}{220})

P(X > 1560) = P(Z > \dfrac{110}{220})

P(X> 1560) = P(Z > 0.5)

P(X> 1560) = 1 - P(Z < 0.5)

From the z tables;

P(X> 1560) = 1 - 0.6915

P(X> 1560) = 0.3085

Let consider the given number of weeks = 52

Mean \mu_x = np = 52 × 0.3085 = 16.042

The standard deviation =  \sqrt {n \time p (1-p)}

The standard deviation = \sqrt {52 \times 0.3085 (1-0.3085)}

The standard deviation = 3.3306

Let Y be a random variable that proceeds in a binomial distribution, which denotes the number of weeks in a year that exceeds $1560.

Then;

Pr ( Y > 20) = P( z > 20)

Pr ( Y > 20) = P(Z > \dfrac{20.5 - 16.042}{3.3306})

Pr ( Y > 20) = P(Z >1 .338)

From z tables

P(Y > 20) \simeq 0.0903

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3 years ago
50 PoInts for math homework Waxton
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Check out the attached photo

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9 months ago
Y=2(x+1)^2 has how many real roots
oee [108]

The equation y= 2(x+1)^2 has one real root and that is x=-1.

What is real roots of the equation?

    We are aware that when we resolve a linear or quadratic equation, we always arrive at the value variable of the equation, or, to put it another way, we always locate the equation's solution. This "solution" is what we refer to as the real roots. For instance, when the equation X^2-7x+12=0 is solved, the actual roots are 3 and 4.

Here given,

=> y = 2(x+1)^2

Take y=0 then,

=> 2(x+1)^2=0

=> (x+1)^2=0

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Hence the given equation has one real root and that is x=-1.

To learn more about real roots refer the below link

brainly.com/question/24147137

#SPJ1

8 0
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