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Fynjy0 [20]
3 years ago
7

1) Si se extrae una carta de un paquete de 52 cartas de las cuales 26 son negras (13 espadas A, 2, 3, … , 10, J, Q, K); 13 son t

réboles); y 26 son rojas (13 corazones y 13 diamantes), halle la probabilidad de:
a) que la carta sea una K.

b) que la carta sea roja.

c) Que la carta sea de diamante.
Mathematics
1 answer:
Daniel [21]3 years ago
5 0
1) If a card is drawn from a pack of 52 cards of which 26 are black (13 spades A, 2, 3,…, 10, J, Q, K); 13 are clovers); and 26 are red (13 hearts and 13 diamonds), find the probability of:

a) that the letter is a K.

b) that the card is red.

c) That the card is diamond.
Help them
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HELPP<br> What is the size of the label in square cm??
elena55 [62]

Answer:

226.19

Step-by-step explanation:

area of a cylinder is 2 times pi times radius time height plus 2 times pi times radius squared

radius=6/2=3

height=9

substitute in your values and get 226.19cm squared

4 0
2 years ago
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Find the area of the shape <br>NO EXPLANATION JUST ANSWER​
Katarina [22]

Answer:

64

Step-by-step explanation:

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3 years ago
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I plant grew 3 1/4 inches over 6 1/2 month period. What was the average monthly growth rate for the plant
Sholpan [36]

Answer:

0.5 inches per a month

Step-by-step explanation:

you turn them to a decimal to make them easier and then divide them to get the answer.

6 0
3 years ago
A programmer plans to develop a new software system. In planning for the operating system that he will​ use, he needs to estimat
Vedmedyk [2.9K]

Using the z-distribution, we have that:

a) A sample of 601 is needed.

b) A sample of 93 is needed.

c) A.  ​Yes, using the additional survey information from part​ (b) dramatically reduces the sample size.

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of \alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which z is the z-score that has a p-value of \frac{1+\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

95% confidence level, hence\alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so z = 1.96.

For this problem, we consider that we want it to be within 4%.

Item a:

  • The sample size is <u>n for which M = 0.04.</u>
  • There is no estimate, hence \pi = 0.5

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.04 = 1.96\sqrt{\frac{0.5(0.5)}{n}}

0.04\sqrt{n} = 1.96\sqrt{0.5(0.5)}

\sqrt{n} = \frac{1.96\sqrt{0.5(0.5)}}{0.04}

(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.5(0.5)}}{0.04}\right)^2

n = 600.25

Rounding up:

A sample of 601 is needed.

Item b:

The estimate is \pi = 0.96, hence:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.04 = 1.96\sqrt{\frac{0.96(0.04)}{n}}

0.04\sqrt{n} = 1.96\sqrt{0.96(0.04)}

\sqrt{n} = \frac{1.96\sqrt{0.96(0.04)}}{0.04}

(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.96(0.04)}}{0.04}\right)^2

n = 92.2

Rounding up:

A sample of 93 is needed.

Item c:

The closer the estimate is to \pi = 0.5, the larger the sample size needed, hence, the correct option is A.

For more on the z-distribution, you can check brainly.com/question/25404151  

8 0
2 years ago
What is the LCM of 70 and 36?​
Alik [6]

LCM = 1260

The lowest common factor is 1260

Hope this helps! :)

6 0
2 years ago
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