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azamat
2 years ago
8

Does anyone know the answer to this I'm k12 btw​

Mathematics
2 answers:
VMariaS [17]2 years ago
7 0
2nd and 4th are the answers
kirill115 [55]2 years ago
7 0
The Second one and the fourth one
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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

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2 years ago
Dilation ( effects of changing dimensions on perimeter and area)
Archy [21]
The answer is C) 64in.
4 0
3 years ago
PLEASE ANSWER ITS DUE SO SOON
geniusboy [140]

the line is negative because it goes down as you move right

slope = rise/run = 4/2 = 2

y = 2x+b

use a point on the line - I picked (-4, 3)

3 = 2(-4)+b

3 = -8+b

11 = b

<h2>y = 2x +11</h2>
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3 years ago
Write an expression for the calculation add 34 and 6 and then multiply 3
Ymorist [56]

We need to write an expression for "add 34 and 6 and then multiply 3"

"Add" represents + operation .

"multiply" represents × operation.

We need to add 34 and 6.

So, we can write 34+6 and then we need multiply the sum by 3.

So, we need to keep above addition of 34 and 6 in parantehsis. We get

(34+6) × 3.

So, the final expression is  (34+6) × 3.

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3 years ago
Please Help Part C Final Part Thanks
alexandr402 [8]

3 blue + 3 yellow = 6 total marbles.

The probability of picking a blue one first would be 3/6 ( 3 blue out of 6 total marbles).

3/6 reduces to 1/2.

After picking the first marble, there would be 5 marbles left in the bag.

The probability of picking a yellow one would be 3/5 ( 3 yellow and 5 total marbles left).

The probability of picking a blue then a yellow then becomes:

1/2 x 3/5 = 3/10 = 0.30 = 30%

4 0
3 years ago
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