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2 years ago

Solve y/2 + 22 =38 for y

Mathematics

2 answers:

KATRIN_1 [288]2 years ago

4 0

Answer:

y = 32

Step-by-step explanation:

$\frac{y}{2}+22=38\\\frac{y}{2}=16\\y=32$

stellarik [79]2 years ago

4 0

Answer:

y=32

Step-by-step explanation:

y/2 + 22 =38

Subtract 22 from each side

y/2 + 22-22 =38-22

y/2 = 16

Multiply each side by 2

y/2 *2 =16*2

y = 32

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The graph illustrates a normal distribution for the prices paid for a particular model of HD television. The mean price paid is

marysya [2.9K]

Answer:

(a) 0.14%

(b) 2.28%

(d) 68%

(e) 34%

(f) 50%

Step-by-step explanation:

Let X be a random variable representing the prices paid for a particular model of HD television.

It is provided that X follows a normal distribution with mean, μ = $1600 and standard deviation, σ = $100.

(a)

Compute the probability of buyers who paid more than $1900 as follows:

$P(X>1900)=P(\frac{X-\mu}{\sigma}>\frac{1900-1600}{100})$

$=P(Z>3)\\=1-P(Z$

*Use a z-table.

Thus, the approximate percentage of buyers who paid more than $1900 is 0.14%.

(b)

Compute the probability of buyers who paid less than $1400 as follows:

$P(X$

$=P(Z$

*Use a z-table.

Thus, the approximate percentage of buyers who paid less than $1400 is 2.28%.

(c)

Compute the probability of buyers who paid between $1400 and $1600 as follows:

$P(1400$

$=P(-2$

*Use a z-table.

Thus, the approximate percentage of buyers who paid between $1400 and $1600 is 48%.

(d)

Compute the probability of buyers who paid between $1500 and $1700 as follows:

$P(1500$

$=P(-1$

*Use a z-table.

Thus, the approximate percentage of buyers who paid between $1500 and $1700 is 68%.

(e)

Compute the probability of buyers who paid between $1600 and $1700 as follows:

$P(1600$

$=P(0$

*Use a z-table.

Thus, the approximate percentage of buyers who paid between $1600 and $1700 is 34%.

(f)

Compute the probability of buyers who paid between $1600 and $1900 as follows:

$P(1600$

$=P(0$

*Use a z-table.

Thus, the approximate percentage of buyers who paid between $1600 and $1900 is 50%.

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3 years ago

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rusak2 [61]

Answer:

Step-by-step explanation:

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7 0

3 years ago

Simplify the expression 5^4

lana66690 [7]

Raise 5 to the power of 4 and get 625

7 0

3 years ago

Read 2 more answers

Determine whether the set of all linear combinations of the following set of vector in R^3 is a line or a plane or all of R^3.a.

Temka [501]

Answer:

a. Line

b. Plane

c. All of R^3

Step-by-step explanation:

In order to answer this question, we need to study the linear independence between the vectors :

1 - A set of three linearly independent vectors in R^3 generates R^3.

2 - A set of two linearly independent vectors in R^3 generates a plane.

3 - A set of one vector in R^3 generates a line.

The next step to answer this question is to analyze the independence between the vectors of each set. We can do this by putting the vectors into the row of a R^(3x3) matrix. Then, by working out with the matrix we will find how many linearly independent vectors the set has :

a. Let's put the vectors into the rows of a matrix :

$\left[\begin{array}{ccc}-2&5&-3\\6&-15&9\\-10&25&-15\end{array}\right]$ ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒

$\left[\begin{array}{ccc}-2&5&-3\\0&0&0\\0&0&0\end{array}\right]$

We find that the second vector is a linear combination from the first and the third one (in fact, the second vector is the first vector multiply by -3).

We also find that the third vector is a linear combination from the first and the second one (in fact, the third vector is the first vector multiply by 5).

At the end, we only have one vector in R^3 ⇒ The set of all linear combinations of the set a. is a line in R^3.

b. Again, let's put the vectors into the rows of a matrix :

$\left[\begin{array}{ccc}1&2&0\\1&1&1\\4&5&3\end{array}\right]$ ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒

$\left[\begin{array}{ccc}1&1&1\\0&1&-1\\0&0&0\end{array}\right]$

We find that there are only two linearly independent vectors in the set so the set of all linear combinations of the set b. is a plane (in fact, the third vector is equivalent to the first vector plus three times the second vector).

c. Finally :

$\left[\begin{array}{ccc}0&0&3\\0&1&2\\1&1&0\end{array}\right]$ ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒

$\left[\begin{array}{ccc}1&1&0\\0&1&2\\0&0&3\end{array}\right]$

The set is linearly independent so the set of all linear combination of the set c. is all of R^3.

4 0

2 years ago

Find all solutions to 2sin(θ)=√3 on the interval 0≤θ<2π

makkiz [27]

Answer:

2sin(3θ) - √3 = 0

Step-by-step explanation:

sin(3θ) = √3/2

3θ = π/3 + 2kπ or 2π/3 + 2kπ, k = 0, ±1, ±2, ±3,...

θ = π/9 + 2kπ/3, 2π/9 + 2kπ/3

If k = 0, we get θ = π/9, 2π/9

If k = 1, we get θ = 7π/9, 8π/9

If k = 2, we get θ = 13π/9, 14π/9

Other values of k give values of θ lying outside of the interval [0, 2π).

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2 years ago