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AnnyKZ [126]
3 years ago
14

You are playing a game of Scrabble and have the 5 A tiles, 3 E tiles, 1 Z tile, 2 M tiles, 3 L tiles, and 1 Y tile to choose fro

m. What is the probability that you choose E or Y on your next draw?
Mathematics
1 answer:
BigorU [14]3 years ago
7 0
When calculating probability, the easiest thing to do is first count how many things there are to choose from.

5 + 3 + 1 + 2 + 3 + 1 = 15        Also, there are 3 E tiles, and 1 Y tile.  3 + 1 = 4

The probability of you drawing a E or Y tile is 4 / 15
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Y doesn't vary directly with x
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she would need 166 boxes. there is a remainder of 1. one book wont fit in a box

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15 points !!!! Diameter Radius Area Circumference of 84mm
SSSSS [86.1K]

Answer:

lets \:  \:  \: find \:  \: the \:  \: radius \\  = 2\pi \: r \\  = 2\pi \: r = 84 \\  =  \frac{2\pi \: r}{2\pi }  =  \frac{84}{2\pi}  \\  = r = 13.37mm

lets \:  \:  \: find \:  \: the \:  \:  \: area \\  = \pi {r}^{2}  \\  = \pi \times  {13.37}^{2}  \\  = 561.55 {mm}^{2}

lets \:  \:  \: find \:  \:  \: the \:  \: diameter \\  = d = 2r \\  = d = 2 \times 13.37 \\  = d = 26.74mm

6 0
3 years ago
Day care centers expose children to a wider variety of germs than the children would be exposed to if they stayed at home more o
Ostrovityanka [42]

Answer:

a) The study is an observational study

b) The proportion of children with significant social activity in children with acute lymphoblastic leukemia is approximately 0.802

The proportion of children with significant social activity in children without acute lymphoblastic leukemia is approximately 0.8565

c) The odds ratio is approximately 0.6780

d) The 95% CI is approximately 0.523 < OR < 0.833

e) i) Yes

ii) Children with more social activity have a higher occurrence of acute lymphoblastic leukemia

Step-by-step explanation:

a) An experimental study is a study in which the researcher adds inputs to the study and then monitors the outcomes

An observational study is one in which the researcher observes risk factors and does not intervene in the process

Therefore, the study is an observational study

b) The proportion of children with significant social activity in children with acute lymphoblastic leukemia is \hat p_1 = 1020/1272 = 85/106 = 0.8\overline {0188679245283} ≈ 0.802

The proportion of children with significant social activity in children without acute lymphoblastic leukemia is \hat p_2 = 5343/6238 ≈ 0.8565

c) We have the following two way table;

\begin{array}{ccc}Lymphoblastic \ Leukemia &With \ Social \ Activity & Without \ Social \ Activity\\With&1020 \ (a)&252 \ (b)\\Without&5343 \ (c)&895 \ (d) \end{array}

The \ odds \ ratio = \dfrac{1,020 \times 895}{5,343 \times 252} \approx 0.6780

The odds ratio ≈ 0.6780

d) The 95% confidence interval for the odds ratio is given as follows;

95 \% \ CI = OR \ \pm \ 1.96 \times \sqrt{\dfrac{1}{a} +\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}}

Where;

a = 1020, b = 252, c = 5343, d = 895

Therefore;

95 \% \ CI \approx  0.6780 \ \pm \ 1.96 \times \sqrt{\dfrac{1}{1,020} +\dfrac{1}{252}+\dfrac{1}{5,343}+\dfrac{1}{895}} \approx 0.678 \pm0.155

The 95% CI ≈ 0.523 < OR < 0.833

e) i) Given that the observed Odds Ratio is within the Confidence Interval, therefore, there is an indication that the amount of social activity is associated with acute lymphoblastic leukemia

ii) Based on the proportion of the study findings, children with more social activity have a higher occurrence of acute lymphoblastic leukemia

6 0
3 years ago
The percentage of people under the age of 18 was 23.5% in New York City, 25.8% in Chicago, and 26% in Los Angeles.
Hatshy [7]

Answer:

0.0158 = 1.58% probability that all of them are under 18.

Step-by-step explanation:

Probability of independent events:

If three events, A, B and C are independent, the probability of all happening is the multiplication of the probabilities of each happening, that is:

P(A \cap B \cap C) = P(A)P(B)P(C)

The percentage of people under the age of 18 was 23.5% in New York City, 25.8% in Chicago, and 26% in Los Angeles.

This means that P(A) = 0.235, P(B) = 0.258, P(C) = 0.26

If one person is selected from each city, what is the probability that all of them are under 18?

Since the three people are independent:

P(A \cap B \cap C) = 0.235*0.258*0.26 = 0.0158

0.0158 = 1.58% probability that all of them are under 18.

8 0
3 years ago
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