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3 years ago

A sailboat leaves port and sails 12 kilometers west and then 9 kilometers north. The sailboat is now kilometers from port

Mathematics

1 answer:

Luden [163]3 years ago

5 0

Answer: The sailboat is at a distance of 15 km from the port.

Step-by-step explanation: Given that a sail boat leaves port and sails 12 kilometers west and then 9 kilometers north.

We are to find the distance between the sailboat from the port in kilometers.

Since the directions west and north are at right-angles, we can visualize the movement of the sailboat in the form of a right-angled triangle as shown in the attached figure.

The sailboat moves leaves the port at P and reach O after sailing 12 km west. From point O, again it moves towards north 9 km and reach the point S.

PS = ?

Using the Pythagoras theorem, we have from right-angled triangle SOP,

$PS^2=OS^2+OP^2=9^2+12^2=81+144=225\\\\\Rightarrow PS=15~\textup{km}.$

Thus, the sailboat is at a distance of 15 km from the port.

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Samples of emissions from three suppliers are classified for conformance to air-quality specifications. The results from 100 sam

tigry1 [53]

Answer:

$P(A) = \frac{30}{100}$

$P(B) = \frac{77}{100}$

$P(A\ n\ B) = \frac{22}{100}$

$P(A\ u\ B) = \frac{85}{100}$

Step-by-step explanation:

Given

See attachment for proper format of table

$n = 100$ --- Sample

A = Supplier 1

B = Conforms to specification

Solving (a): P(A)

Here, we only consider data in sample 1 row.

In this row:

$Yes = 22$ and $No = 8$

So, we have:

$n(A) = Yes + No$

$n(A) = 22 + 8$

$n(A) = 30$

P(A) is then calculated as:

$P(A) = \frac{n(A)}{Sample}$

$P(A) = \frac{30}{100}$

Solving (b): P(B)

Here, we only consider data in the Yes column.

In this column:

$(1) = 22$ $(2) = 25$ and $(3) = 30$

So, we have:

$n(B) = (1) + (2) + (3)$

$n(B) = 22 + 25 + 30$

$n(B) = 77$

P(B) is then calculated as:

$P(B) = \frac{n(B)}{Sample}$

$P(B) = \frac{77}{100}$

Solving (c): P(A n B)

Here, we only consider the similar cell in the yes column and sample 1 row.

This cell is: [Supplier 1][Yes]

And it is represented with; n(A n B)

So, we have:

$n(A\ n\ B) = 22$

The probability is then calculated as:

$P(A\ n\ B) = \frac{n(A\ n\ B)}{Sample}$

$P(A\ n\ B) = \frac{22}{100}$

Solving (d): P(A u B)

This is calculated as:

$P(A\ u\ B) = P(A) + P(B) - P(A\ n\ B)$

This gives:

$P(A\ u\ B) = \frac{30}{100} + \frac{77}{100} - \frac{22}{100}$

Take LCM

$P(A\ u\ B) = \frac{30+77-22}{100}$

$P(A\ u\ B) = \frac{85}{100}$

8 0

3 years ago

Given the slope of the line containing the points (7 , b) and (-2 , 5) is 4/9, find the value of b

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This would give us 4/9 which is the slope we already had.

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saul85 [17]

Answer:

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Step-by-step explanation:

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