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2 years ago

A line has a slope of 1/2 and run =50. What is the rise

2 answers:

nataly862011 [7]2 years ago

6 0

Since tan(theta) is slope of the line:

$\tan(theta) = \frac{rise}{run } = \frac{x}{50 } \\ \frac{1}{2 } = \frac{x}{50 } \\ x = 25$
Hence rise is 25

Svetllana [295]2 years ago

3 0

Since 1/2 is one half, the rise must be half of the run, meaning that the rise would be 25.

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VEEL

Andre45 [30]

Answer:

$a_n=-3(3)^{n-1}$ ; {-3,-9, -27,- 81, -243, ...}

$a_n=-3(-3)^{n-1}$ ; {-3, 9,-27, 81, -243, ...}

$a_n=3(\frac{1}{2})^{n-1}$ ; {3, 1.5, 0.75, 0.375, 0.1875, ...}

$a_n=243(\frac{1}{3})^{n-1}$ ; {243, 81, 27, 9, 3, ...}

Step-by-step explanation:

The first explicit equation is

$a_n=-3(3)^{n-1}$

At n=1,

$a_1=-3(3)^{1-1}=-3$

At n=2,

$a_2=-3(3)^{2-1}=-9$

At n=3,

$a_3=-3(3)^{3-1}=-27$

Therefore, the geometric sequence is {-3,-9, -27,- 81, -243, ...}.

The second explicit equation is

$a_n=-3(-3)^{n-1}$

At n=1,

$a_1=-3(-3)^{1-1}=-3$

At n=2,

$a_2=-3(-3)^{2-1}=9$

At n=3,

$a_3=-3(-3)^{3-1}=-27$

Therefore, the geometric sequence is {-3, 9,-27, 81, -243, ...}.

The third explicit equation is

$a_n=3(\frac{1}{2})^{n-1}$

At n=1,

$a_1=3(\frac{1}{2})^{1-1}=3$

At n=2,

$a_2=3(\frac{1}{2})^{2-1}=1.5$

At n=3,

$a_3=3(\frac{1}{2})^{3-1}=0.75$

Therefore, the geometric sequence is {3, 1.5, 0.75, 0.375, 0.1875, ...}.

The fourth explicit equation is

$a_n=243(\frac{1}{3})^{n-1}$

At n=1,

$a_1=243(\frac{1}{3})^{1-1}=243$

At n=2,

$a_2=243(\frac{1}{3})^{2-1}=81$

At n=3,

$a_3=243(\frac{1}{3})^{3-1}=27$

Therefore, the geometric sequence is {243, 81, 27, 9, 3, ...}.

6 0

3 years ago

A social psychologist interested in cultural differences compared women of two ethnic groups on a Role Approval Index on which h

Levart [38]

Answer:

a. 5.26

Step-by-step explanation:

As, the standard deviation of the distribution of the difference between mean is 0.76 so the t score will be simply calculated by dividing the difference of means by standard deviation of the distribution of the difference between mean as hypothesized difference is zero.

t score=Xbar A- Xbar B/Sd of difference of means

t score=55-51/0.76

t score=4/0.76=5.26

3 0

3 years ago

A town’s January high temperatures average 36 degrees (F) with a standard deviation of 10 degrees while in July the mean high

n200080 [17]

Answer:

55 degrees is more unusual in July as it is away from mean temperature as compared to January.

Step-by-step explanation:

We are given the following information in the question:

January:

Mean, μ = 36 degrees

Standard Deviation, σ = 10 degrees

July:

Mean, μ = 74 degrees

Standard Deviation, σ = 8 degrees

We calculate z scores corresponding to x = 55.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

January:

$z_{score} = \displaystyle\frac{55-36}{10} = 1.9$

July:

$z_{score} = \displaystyle\frac{55-74}{8} = -2.375$

Thus, 55 degrees is more unusual in July as it is away from mean temperature as compared to January.

3 0

3 years ago

Thinking the perimeter of the rectangle is 3x + 2 + x + 3x + 2 + X.

givi [52]

They are both right because the first expression is the perimeter but expanded in terms of length.The second expression is shortened to be a factor expression but they both represent the length.
The benefit of the first one is you can see how it’s expanded and plug the numbers in and the second one is more easy to plug in the numbers with ease.

7 0

3 years ago

X(15-x)=50 quadratic equations

meriva

Answer:

$15x - {x}^{2} =50 \\ {x }^{2} - 15x + 50 = 0 \\ {x}^{2} - 10x - 5x + 50 = 0 \\ x(x - 10) - 2(x - 10) = 0 \\ (x - 2)(x - 10) = 0 \\ x = 2 \: and \: x = 10$

3 0

2 years ago