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Nady [450]
2 years ago
7

Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).

Mathematics
1 answer:
Tanzania [10]2 years ago
6 0

Answer:

Approximately, 159 men weighs more than 165 pounds and  159 men weighs less than 135 pounds.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 150 pounds

Standard Deviation, σ = 15

We are given that the distribution of weights of 1000 men is a bell shaped distribution that is a normal distribution.

Formula:

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

P( men weighing more than 165 pounds)

P(x > 165)

P( x > 165) = P( z > \displaystyle\frac{165 - 150}{15}) = P(z > 1)

= 1 - P(z \leq 1)

Calculation the value from standard normal z table, we have,  

P(x > 165) = 1 - 0.8413 = 0.1587 = 15.87\%

Approximately, 159 men weighs more than 165 pounds.

P(men weighing less than 135 pounds)

P(x < 135)

P( x < 135) = P( z < \displaystyle\frac{135 - 150}{15}) = P(z < -1)

Calculation the value from standard normal z table, we have,  

P(x < 135) = 0.1587 = 15.87\%

Approximately, 159 men weighs less than 135 pounds.

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If 1 japanese yen equals 0.0079 euros, then 1 euro equals approximately how many japanese yen?
Oliga [24]
We know that

<span>applying a rule of three</span>
if <span>1 japanese yen -------------> equals 0.0079 euros
  X </span> japanese yen--------------> 1 euro
x=[1 euro]*[1 japanese yen]/[0.0079 euros]------------> 126.58 japanese yen

the answer is 
126.58 japanese yen
6 0
2 years ago
Which point on the y-axis lies on the line that passes through point C and is perpendicular to line AB?
SCORPION-xisa [38]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

Which point on the y-axis lies on the line that passes through point C and is perpendicular to line AB?

A. (-6, 0)

B. (0, -6)

C. (0, 2)

D. (2, 0)

The graph of the question is attached.

Answer:

The point is (x, y) = (0, 2)

The correct option is C.

Therefore, the point (0, 2) on the y-axis lies on the line that passes through point C and is perpendicular to line AB.

Step-by-step explanation:

From the given graph, the points A and B are

(x_1, y_1) = (-2, 4) \\\\(x_2, y_2) = (2,-8) \\\\

The slope of the equation is given by

m_1 = \frac{-8 - 4 }{2 -(-2)} \\\\ m_1 = \frac{-12 }{2+2} \\\\m_1 = \frac{-12 }{4} \\\\m_1 = -3 \\\\

We know that the slopes of two perpendicular lines are negative reciprocals of each other.

m_2 = - \frac{1}{m_1}

So the slope of the other line is

m_2 = \frac{1 }{3} \\\\

Now we can find the equation of the line that is perpendicular to the line AB and passes through the point C.

From the graph, the coordinates of point C are

(x_1, y_1) = (6, 4)

The point-slope form is given by,

y - y_1 = m(x -x_1)

Substitute the value of slope and the coordinates of point C

y - 4 = \frac{1 }{3} (x - 6)\\\\

To get the y-intercept, substitute x = 0  

y - 4 = \frac{1 }{3} (0 - 6) \\\\y - 4 = \frac{-6 }{3}\\\\y - 4 = -2\\\\y = 4 -2 \\\\y = 2 \\\\

So, the point is

(x, y) = (0, 2)

The correct option is C.

Therefore, the point (0, 2) on the y-axis lies on the line that passes through point C and is perpendicular to line AB.

6 0
2 years ago
Using approximations to 1 significant figure,
malfutka [58]

Answer:

200

Step-by-step explanation:

Given:

0.482 x 61.2^2 ÷ √98.01

61.2^2 = 3745.44

√98.01 = 9.9

So,

0.482 × 3745.44 ÷ 9.9

= 1,805.30208 ÷ 9.9

= 182.35374545454

To one significant figure

= 200

One significant figure means only 1 non zero value and others are zero

5 0
2 years ago
On a coordinate plane, a line with a 90-degree angle crosses the x-axis at (negative 4, 0), turns at (negative 1, 3), crosses th
mixer [17]

Answer:

b

Step-by-step explanation:

3 0
2 years ago
A region R in the xy-plane is given. Find equations for a transformation T that maps a rectangular region S in the uv-plane onto
Gre4nikov [31]
\begin{cases}y=2x-2\\y=2x+2\end{cases}\implies\begin{cases}-2x+y=-2\\-2x+y=2\end{cases}

For these lines, let u=-2x+y.

\begin{cases}y=2-x\\y=4-x\end{cases}\implies\begin{cases}x+y=2\\x+y=4\end{cases}

And for these, let v=x+y.

Now,

\begin{cases}u=-2x+y\\v=x+y\end{cases}\implies \begin{bmatrix}u\\v\end{bmatrix}=\underbrace{\begin{bmatrix}-2&1\\1&1\end{bmatrix}}_{\mathbf T}\begin{bmatrix}x\\y\end{bmatrix}

The vertices of S in the x-y plane are (0, 2), (2/3, 10/3), (2, 2), and (4/3, 2/3). Applying \mathbf T to each of these yields, respectively, (2, 2), (2, 4), (-2, 4), and (-2, 2), which are the vertices of a rectangle whose sides are parallel to the u-v plane.
6 0
2 years ago
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