Mathematics: Three forces of sizes 15n,7n and 19n pulling at 120 degrees to each other are acting on a particle. What is the size of the resultant force and the direction

Three forces of sizes 15n,7n and 19n pulling at 120 - id1571912011860789, jellybellyje

bankskry

10.04.2021

10.58 N

Step-by-step explanation:

Let us find component of these forces in x and y direction.

The solution along with diagram can be found in attachment.

We have found resultant in X- direction = 6√3 N

resultant in Y- direction = 2 N

therefore net force resultant force = $\sqrt{(6\sqrt3)^2+2^2}$

= 10.58 N

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kim643

02.08.2020

The coach should start recruiting players with weight 269.55 pounds or more.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 225 pounds

Standard Deviation, σ = 43 pounds

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

Formula:

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

We have to find the value of x such that the probability is 0.15

$P( X x) = P( z \displaystyle\frac{x - 225}{43})=0.15$

$= 1 -P( z \leq \displaystyle\frac{x - 225}{43})=0.15$

$=P( z \leq \displaystyle\frac{x - 225}{43})=0.85$

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

P(z < 1.036) = 0.85

$\displaystyle\frac{x - 225}{43} = 1.036\\\\x = 269.548 \approx 269.55$

Thus, the coach should start recruiting players with weight 269.55 pounds or more.

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