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mrs_skeptik [129]
2 years ago
11

A bag contains five yellow tickets numbered one to five. The bag also contains five green tickets numbered one to five. You rand

omly pick a ticket. It is green or has a number greater than four. Find the probability of this occuring.
Mathematics
1 answer:
tamaranim1 [39]2 years ago
5 0

The probability of picking a ticket that is green or has a number greater than four is 3/5

<h3>How to determine the probability?</h3>

The given parameters are:

Yellow = 1 - 5

Green = 1 - 5

Total = 10

There are 2 cards whose numbers are greater than 4 i.e. Yellow 5 and Green 5

So, we have:

P(Number greater than 4) = 2/10

There are 5 green cards.

So, we have:

P(Green) = 5/10

Also, 1 green card is numbered greater than 4

So, we have:

P(Green greater than 4) = 1/10

The required probability is:

P = P(Green) + P(Number greater than 4) - P(Green greater than 4)

This gives

P = 5/10 + 2/10 - 1/10

Evaluate

P = 6/10

Simplify

P =3/5

Hence, the probability of picking a ticket that is green or has a number greater than four is 3/5

Read more about probability at:

brainly.com/question/11234923

#SPJ1

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The population of a city is 218720.
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Correct Answer is 205,857
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The school is planting a circular garden. If the
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Answer:

<u>The area of the circular garden is 28.3 square feet</u>

Step-by-step explanation:

Let's recall that the area of a circle is π * r², therefore if the diameter of the circular garden is 6 feet, the area is:

Diameter = 6 feet ⇒ radius = (6/2) = 3 feet

Area of the circular garden = π * 3²

Area of the circular garden = 3.1416 * 9

Area of the circular garden = 28.2744

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8 0
3 years ago
Given that a 90% confidence interval for the mean height of all adult males in Idaho measured in inches was [62.532, 76.478]. Us
grigory [225]

Answer:

The confidence interval on this case is given by:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}} (1)

For this case the confidence interval is given by (62.532, 76.478)[/tex]

And we can calculate the mean with this:

\bar X = \frac{62.532+76.478}{2}= 69.505

So then the mean for this case is 69.505

Step-by-step explanation:

Previous concepts

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

\bar X represent the sample mean  

\mu population mean (variable of interest)  

\sigma represent the population standard deviation  

n represent the sample size  

Assuming the X follows a normal distribution  

X \sim N(\mu, \sigma)

The confidence interval on this case is given by:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}} (1)

For this case the confidence interval is given by (62.532, 76.478)[/tex]

And we can calculate the mean with this:

\bar X = \frac{62.532+76.478}{2}= 69.505

So then the mean for this case is 69.505

5 0
3 years ago
If LN XZ, is ALMNAXYZ?<br> Y<br> M<br> Х<br> N<br> Z<br> Yes<br> No
Nikolay [14]

Answer:

Yes

Step-by-step explanation:

Yes

because we are given that

LN ≅ XZ ( side)

the picture shows that

LM ≅XY ( side)

and

MN ≅ YZ (side)

By the <u>SSS theorem of congruency</u> we conclude that ΔLMN ≅ ΔXYZ

5 0
2 years ago
Read 2 more answers
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