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3 years ago

Pls help asap due tomorrow

Mathematics

1 answer:

Svet_ta [14]3 years ago

4 0

Based on the information given, it should be noted that the least number of marbles that can be in the bag will be 10 marbles.

<h2>Solution to the probability.</h2>

If the bag contains 60 marbles, the number of red marbles will be:

= 1/5 × 60

= 12 marbles.

The number of white marbles will be:

= 3/10 × 60

= 18 marbles.

The number of blue marbles will be:

= 60 - (12 + 18)

= 60 - 30

= 30 marbles.

Furthermore, when the bag has 4 red marbles and 6 white marbles, there'll be 10 blue marbles.

Lastly, to find the probability of choosing a blue marble, we've to multiply 1/2 by the number given.

Learn more about probability on:

brainly.com/question/25870256

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Can y’all help me please

likoan [24]

Answer:

y = -4x + 60

Step-by-step explanation:

(0, 60) and (4,44)

Slope:

m=(y2-y1)/(x2-x1)

m=(44 - 60)/(4-0)

m= (-16)/4

m= -4

Slope-intercept:

y - y1 = m(x - x1)

y - 60 = -4(x - 0)

y - 60 = -4x

y = -4x + 60

8 0

3 years ago

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Which has a positive value in Quadrant IV?

denis23 [38]

Cosine is positive in Quadrant IV.

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3 years ago

An object falls from a position of rest and

kumpel [21]

Answer:

About 4.1323 meters.

Step-by-step explanation:

We can use the following kinematic equations:

$\displaystyle v_f = v_0 + at \text{ and } \Delta d = v_0 t + \frac{1}{2} at^2$

We want to determine the distance the object has dropped after falling from rest and reaching an instantenous speed of 9 m/s.

We are given that the acceleration due to gravity is 9.8 m/s².

Using this fact and the first equation, find the time for which it took the object to reach 9 m/s. Note that the initial velocity is 0 m/s since the object started from rest.

$\displaystyle \begin{aligned} v_f = v_0 + at \\ \\ (9\text{ m/s}) & = (0\text{ m/s}) + (9.8\text{ m/s$^2$})t \\ \\ 9\text{ m/s} & = (9.8 \text{ m/s$^2$})t \\ \\ t &\approx0.9184 \text{ s} \end{aligned}$

To find how far the object dropped, we can use the second equation:

$\displaystyle\begin{aligned} \Delta d & = v_0 t + \frac{1}{2} at^2 \\ \\ & = (0\text{ m/s})(0.9184 \text{ s}) + \frac{1}{2}(9.8 \text{ m/s$^2$})(0.9184 \text{ s})^2 \\ \\ & = 4.1323 \text{ m} \end{aligned}$