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A factor of 30 is chosen out random.what is the probability as a decimal,that is a two digit number.

goblinko [34]

Answer:

A factor of 30 is chosen at random. What is the probability, as a decimal, that it is a 2-digit number?

The positive whole-number factors of 30 are:

1, 2, 3, 5, 6, 10, 15 and 30.

So, there are 8 of them. Of these, 3 have two digits. Writing each factor on a slip of paper, then putting the slips into a hat, and finally choosing one without looking, get that

P(factor of 30 chosen is a 2-digit number) = number of two-digit factors ÷ number of factors

=38=3×.125=.375 as a decimal number.

8 0

4 years ago

How can you solve an equation with the variable on both sides?

masya89 [10]

Add the like terms (varibles) on both sides if their the same

7 0

3 years ago

How to determine if set of vectors is linearly independent?

zimovet [89]

The vectors $\{v_1,v_2,...,v_p\}\in \mathbb{R}^n$ are linearly independent if the ONLY solution to the following equation:

$c_1v_1+c_2v_2+...c_pv_p=0$

is $c_1=c_2=...=c_p=0$

7 0

3 years ago

Two forces with magnitudes of 100 and 50 pounds act on an object at angles of 50° and 160°, respectively. Find the direction and

MatroZZZ [7]

<h2>Hello!</h2>

Since there is no additional information, we can assume that the given angles are respect to the horizontal axis (x), so

The answer is:

Magnitude: 95.14 lbs

Direction: 80° respect to the x-axis

Since there is no additional information about the given angles, we can assume that both angles are respect to the horizontal axis (x), positive direction to x-axis positive directions (right) and above x-axis.

<h2>Calculations:</h2>

100 pounds (lb) force, 50° angle:

$100lbs*cos(50)=64.28lbs\\100lbs*sin(50)=76.60lbs$

50 pounds (lb) force, 160° angle:

$50lbs*cos(160)=-46.98lbs\\50lbs*sin(160)=17.10lbs$

So,

For x-axis forces we have:

64.28lbs (positive direction)

-46.98lbs (negative direction)

Then,

$64.28lbs-46.98lbs=17.3lbs$ (positive direction)

Fox y-axis forces we have:

76.60lbs (positive direction)

17.10 lbs (positive direction)

Then,

$76.60lbs + 17.10 lbs =93.7lbs$ (positive direction)

So, we have the two components values of the acting force:

17.3 lbs for the x-axis and 93.7 lbs for the y-axis.

Therefore,

We have to calculate the angle between both resultant force components:

Let's calll it α

So,

$\alpha=tan^{-1}(\frac{y-axisValue}{x-axisValue})=tan^{-1}(\frac{93.7}{17.3})\\\alpha=79.5=80$

Hence, the angle between both resultant force components is 80 °, meaning that the resultant force direction is positive, above the x-axis.

Now we have to calculate the magnitude of the resultant force since the sum of both resultant force components is equal to the hypotenuse of the formed triangle, we have that:

$sin(\alpha)=\frac{y-axisValue}{ResultantForceMagnitude}=\frac{93.7lbs}{ResultantForceMagnitude}\\\\ResultantForceMagnitude=\frac{93.7lbs}{sin(80)}=95.14lbs$