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2 years ago

An experiment consists of tossing 4 unbiased coins simultaneously. What is the number of simple events in this experiment?

Mathematics

1 answer:

kupik [55]2 years ago

5 0

The number of simple events in this experiment according to the probability is 16.

According to the statement

we have to find that the number of simple events in this experiment.

So, For this purpose, we know that the

Simple events are the events where one experiment happens at a time and it will be having a single outcome. The probability of simple events is denoted by P(E) where E is the event.

And according to the given information is:

Total number of coins tossed is 4.

then

the simple events become

Simple events = no. of coins * total coins tossed

Simple events = 4*4

Now solve it then

Simple events = 16.

So, The number of simple events in this experiment according to the probability is 16.

Learn more about simple events here

brainly.com/question/7965468

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3 years ago

10 mL = ____________ L ( 1L = 1000 ml

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Solution:

Note that:

1 L = 1000 ml

Using the clue above, let's solve each problem.

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3 years ago

Suppose I collected a sample and calculated the sample proportion. If I construct a 90% confidence interval for the population p

Daniel [21]

Answer:

Step-by-step explanation:

If you construct a 90% confidence interval for the population proportion and a 95% confidence interval for the population proportion, the 95% confidence will have a wider interval. This is because a higher confidence interval will provide more possible values from which the true value will be determined. Therefore, If you want more confidence that an interval contains the true parameter, then the intervals will be wider.

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3 years ago

The height of a ball thrown into the air after t seconds have elapsed is h = −16t2 + 40t + 6 feet. What is the first time, t, wh

sp2606 [1]

<h3>The first time when the ball will reach a height of 20 feet is 0.42 seconds</h3>

Solution:

Given that,

The height of a ball thrown into the air after t seconds have elapsed is:

$h = -16t^2 + 40t + 6$

What is the first time, t, when the ball will reach a height of 20 feet?

Substitute h = 20

$20 = -16t^2 + 40t + 6\\\\-16t^2 + 40t + 6 -20 = 0\\\\-16t^2 + 40t -14 = 0\\\\16t^2 -40t + 14 = 0\\\\8t^2 -20t + 7=0$

Solve by quadractic formula

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=8,\:b=-20,\:c=7$

$t = \frac{-\left(-20\right)\pm \sqrt{\left(-20\right)^2-4\cdot \:8\cdot \:7}}{2\cdot \:8}\\\\t = \frac{20 \pm \sqrt{176}}{16}\\\\t = \frac{20 \pm 4\sqrt{11}}{16}\\\\t = \frac{ 5 \pm \sqrt{11}}{4}\\\\We\ have\ two\ solutions\\\\ t=2.07915, \:t=0.42084$

Rounding off we get,

t = 2.08 , t = 0.42

Thus the first time when the ball will reach a height of 20 feet is 0.42 seconds

4 0

3 years ago