Two-fifths of one less than a number is less than three-fifths of one more than that number. What numbers are in the solution se

Response:

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Detailed explanation of the process:

Answer:

6.79 miles

Step-by-step explanation:

Consider triangle ABC where A marks the starting point and C indicates the end point.

|AB|=5.2 miles

|BC|=3.0 miles

∠ABC = 90 + 75 = 165°

We're given two side lengths and an angle that is not opposite to either side.

The approach to resolve this scenario is known as the Cosine Rule.

Each side opposite an angle is designated by the corresponding lowercase letter.

The Cosine Rule asserts that:

b² = a² + c² - 2acCosB

|AC|²=3² + 5.2² - (2 × 5.2 × Cos 165°)

|AC|²=9 + 27.04 + 10.05

|AC|²=46.09

|AC|=√46.09=6.79 miles

Thus, the straight-line distance from A to C measures 6.79 miles

Answer:

link:

Step-by-step explanation:

I will designate the hourly rate for weekdays as x and for weekends as y. The equations are arranged as follows:

13x + 14y = $250.90
15x + 8y = $204.70

This gives us a system of equations which can be solved by multiplying the first equation by 4 and the second by -7. This leads to:

52x + 56y = $1003.60
-105x - 56y = -$1432.90

By summing these two equations, we arrive at:

-53x = -$429.30 --> 53x = $429.30 --> (dividing both sides by 53) x = 8.10. This represents her hourly wage on weekdays.

Substituting our value for x allows us to determine y. I will utilize the first equation, but either could work.

$105.30 + 14y = $250.90. To isolate y, subtract $105.30 from both sides --> 14y = $145.60 divide by 14 --> y = $10.40

Thus, we find that her earnings are $8.10 per hour on weekdays and $10.40 per hour on weekends. The difference shows she earns $2.30 more on weekends than on weekdays.

c) Step-by-step breakdown: The collision rate is 1.2 incidents per 4 months, which can be expressed as 0.3 incidents monthly. Therefore, the Poisson distribution for the variable X representing monthly collisions is defined as P(X = x) =... for x ∈ N ∪ {0} = 0 otherwise. (1) Where X = 0 denotes no collisions during a 4-month timeframe, substituting gives P(X = 0) =... (2). For a 4-month period, P(No collision in 4 month period) =... (3). Two collisions in a 2-month span translate to 1 per month, thus P(X =1) =... (4). Over 2 months, P(2 collisions in a 2 month period) =... (5). One collision over a 6-month period equates to P(1 collision in 6 months period) =... (6). Consequently, P(1 collision in 6 month period) results in... (7). For no collisions in a 6-month period, P(No collision in 6 months period) =... (8). Finally, the probability of 1 or fewer collisions over six months is P(1 or fewer collision in 6 months period) = (8) + (7) = 0.0785 + 0.1653.