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vladimir2022 [97]
3 years ago
14

What is the solution of y-x>-3

Mathematics
1 answer:
Over [174]3 years ago
6 0

Answer:

y > x - 3

Step-by-step explanation:

y - x > -3

This is an equation of a straight line graph.

Taking x to the right side, we get:

y > -3 + x

Rearranging, we get:

y > x - 3

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Blood type AB is the rarest blood type, occurring in only 4% of the population in the United States. In Australia, only 1.5% of
Naddik [55]

Answer:

There is a 27.62% probability that exactly 2 of the U.S. residents have blood type AB.

Step-by-step explanation:

For each U.S. resident, there are only two outcomes possible. Either they have blood type AB, or they do not. This means that we can solve this problem using binomial probability distribution concepts.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And \pi is the probability of X happening.

In this problem, we have that:

50 U.S residents are sampled, so n = 50

4% of the U.S population has blood type AB, so p = 0.04.

What is the probability that exactly 2 of the U.S. residents have blood type AB?

This is P(X = 2). So:

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 2) = C_{50,2}.(0.04)^{2}.(0.96)^{48} = 0.2762

There is a 27.62% probability that exactly 2 of the U.S. residents have blood type AB.

5 0
3 years ago
Find the direction cosines and direction angles of the vector. (Give the direction angles correct to the nearest degree.) 5, 1,
Dahasolnce [82]

Answer:

The direction cosines are:

\frac{5}{\sqrt{42} }, \frac{1}{\sqrt{42} }  and  \frac{4}{\sqrt{42} }  with respect to the x, y and z axes respectively.

The direction angles are:

40°,  81° and  52° with respect to the x, y and z axes respectively.

Step-by-step explanation:

For a given vector a = ai + aj + ak, its direction cosines are the cosines of the angles which it makes with the x, y and z axes.

If a makes angles α, β, and γ (which are the direction angles) with the x, y and z axes respectively, then its direction cosines are: cos α, cos β and cos γ in the x, y and z axes respectively.

Where;

cos α = \frac{a . i}{|a| . |i|}               ---------------------(i)

cos β = \frac{a.j}{|a||j|}               ---------------------(ii)

cos γ = \frac{a.k}{|a|.|k|}             ----------------------(iii)

<em>And from these we can get the direction angles as follows;</em>

α =  cos⁻¹ ( \frac{a . i}{|a| . |i|} )

β = cos⁻¹ ( \frac{a.j}{|a||j|} )

γ = cos⁻¹ ( \frac{a.k}{|a|.|k|} )

Now to the question:

Let the given vector be

a = 5i + j + 4k

a . i =  (5i + j + 4k) . (i)

a . i = 5         [a.i <em>is just the x component of the vector</em>]

a . j = 1            [<em>the y component of the vector</em>]

a . k = 4          [<em>the z component of the vector</em>]

<em>Also</em>

|a|. |i| = |a|. |j| = |a|. |k| = |a|           [since |i| = |j| = |k| = 1]

|a| = \sqrt{5^2 + 1^2 + 4^2}

|a| = \sqrt{25 + 1 + 16}

|a| = \sqrt{42}

Now substitute these values into equations (i) - (iii) to get the direction cosines. i.e

cos α = \frac{5}{\sqrt{42} }

cos β =  \frac{1}{\sqrt{42} }              

cos γ =  \frac{4}{\sqrt{42} }

From the value, now find the direction angles as follows;

α =  cos⁻¹ ( \frac{a . i}{|a| . |i|} )

α =  cos⁻¹ ( \frac{5}{\sqrt{42} } )

α =  cos⁻¹ (\frac{5}{6.481} )

α =  cos⁻¹ (0.7715)

α = 39.51

α = 40°

β = cos⁻¹ ( \frac{a.j}{|a||j|} )

β = cos⁻¹ ( \frac{1}{\sqrt{42} } )

β = cos⁻¹ ( \frac{1}{6.481 } )

β = cos⁻¹ ( 0.1543 )

β = 81.12

β = 81°

γ = cos⁻¹ ( \frac{a.k}{|a|.|k|} )

γ = cos⁻¹ (\frac{4}{\sqrt{42} })

γ = cos⁻¹ (\frac{4}{6.481})

γ = cos⁻¹ (0.6172)

γ = 51.89

γ = 52°

<u>Conclusion:</u>

The direction cosines are:

\frac{5}{\sqrt{42} }, \frac{1}{\sqrt{42} }  and  \frac{4}{\sqrt{42} }  with respect to the x, y and z axes respectively.

The direction angles are:

40°,  81° and  52° with respect to the x, y and z axes respectively.

3 0
3 years ago
What is the max or min and range of y=2x^2+20x-12
shtirl [24]

Answer: its x = 3

Step-by-step explanation:

The maximum or minimum of a quadratic function occurs at

x

=

−

b

2

a

. If

a

is negative, the maximum value of the function is

f

(

−

b

2

a

)

. If

a

is positive, the minimum value of the function is

f

(

−

b

2

a

)

.

f

min

x

=

a

x

2

+

b

x

+

c

occurs at

x

=

−

b

2

a

Find the value of

x

equal to

−

b

2

a

.

x

=

−

b

2

a

Substitute in the values of

a

and

b

.

x

=

−

−

12

2

(

2

)

Remove parentheses.

x

=

−

−

12

2

(

2

)

Simplify

−

−

12

2

(

2

)

.

Tap for more steps...

x

=

3

The maximum or minimum of a quadratic function occurs at

x

=

−

b

2

a

. If

a

is negative, the maximum value of the function is

f

(

−

b

2

a

)

. If

a

is positive, the minimum value of the function is

f

(

−

b

2

a

)

.

f

min

x

=

a

x

2

+

b

x

+

c

occurs at

x

=

−

b

2

a

Find the value of

x

equal to

−

b

2

a

.

x

=

−

b

2

a

Substitute in the values of

a

and

b

.

x

=

−

−

12

2

(

2

)

Remove parentheses.

x

=

−

−

12

2

(

2

)

Simplify

−

−

12

2

(

2

)

.

Tap for more steps...

x

=

3

4 0
2 years ago
If θ = 330° and α = 300°, find the exact value of tan θ − sin α.
Vilka [71]
X = tan (330) - sin (330)
x = 0.133557089 - -0.132381629
x = 0.133557089 + 0.132381629
x = 0.265938718
7 0
3 years ago
Find the ending balance $35,600 at 9% for 2 years
tino4ka555 [31]

Answer:

42,008

Step-by-step explanation:

6 0
2 years ago
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