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Ksivusya [100]
3 years ago
8

Let X be a normal random variable with a mean of 1.54 and a standard deviation of 2.40.

Mathematics
1 answer:
allsm [11]3 years ago
6 0

Answer:

a) z = 1.40

b) X is greater than or equal to 4.9

Step-by-step explanation:

Population mean (μ) = 1.54

Standard deviation (σ) = 2.40

The z-score for any given value X is:

z=\frac{X-\mu}{\sigma}

a) For X= 4.9:

z=\frac{4.9 -1.54}{2.40}\\z= 1.40

The corresponding z-score for x = 4.9 is z=1.40

b) Z-scores higher than 1.40 correspond to values of X higher than 4.9. Therefore, the area under the standard normal probability density function from z to infinity, P(z ≥ 1.40), is interpreted as the probability that X is greater than or equal to 4.9.

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klemol [59]

Answer:

10) 9x - 2° = 5x + 54° (corresponding angles are equal)

9x - 5x = 54 + 2

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x = 56/4

x = 14°

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10y + 6° = 9(14)° - 2°

10y + 6° = 126° - 2°

10y + 6° = 124°

10y = 124° - 6°

10y = 118

y = 118/10

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2 years ago
If the length of a leg of a triangle is 39.2 and the area of the triangle is 63 what is the length of the other leg
Nadya [2.5K]

Answer:

The length of the other leg is 3.21\ units

Step-by-step explanation:

I will assume that the triangle is a right triangle

In a right triangle the legs are perpendicular

so

The area of a right triangle is equal to

A=\frac{1}{2}(a)(b)

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a and b are the legs of the triangle

In this problem we have

a=39.2\ units

A=63\ units^{2}

substitute the values

63=\frac{1}{2}(39.2)(b)

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3 years ago
What is 99 pounds in kilograms
Bumek [7]
99 pounds in kilograms would be 44.906kg
Hope this helped you!
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3 years ago
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Find an explicit rule for the nth term of a geometric sequence where the second and fifth terms are -6 and 162, respectively. Pl
kicyunya [14]
Hello,

u_{0} =a\\
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....\\
\boxed{ u_{n} =a*r^{n}} \\

\dfrac{ 162}{-6} = \dfrac{ u_{5} }{ u_{3}} = \dfrac{ a*r^{5}}{ a*r^{2}} =r^3= -27\\
==\ \textgreater \  r=-3\\

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\boxed{ u_{n} =-\frac{2}{3}*(-3)^{n}=(-1)^{n+1}*2*3^{n-1}} \\


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3 years ago
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