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xxTIMURxx [149]
3 years ago
5

5 plz help me out I really need help

Mathematics
2 answers:
frez [133]3 years ago
6 0
Perimeter 32. area 63
jeka57 [31]3 years ago
3 0
32cm because the perimeter was originally 28 but theyre adding 1 cm to each side, and the are 4 sides so 4 extra cm in total
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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

6 0
2 years ago
Please help❤️❤️ It’s for 20 pts!!
solong [7]
If you need part A, draw a person and label them 6 feet tall and a shadow coming off of him that is five feet long and draw a basketball hoop that and label it X feet tall and draw a shadow coming from it that is 8 feet long. To solve, you use proportions 6/5=X/8 solving this you get X=9.6 feet which is the height of the rim. Therefore, the rim does not meet regulation height, since the required height (10 feet) is greater than the actual height (9.6 feet).   
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PLEASE SOMEONE HELP ME. I WILL BRAINLIEST YOU AND GIVE YOU POINTS.
almond37 [142]

Answer:

C).

Step-by-step explanation:

-3x+2 and[(-)(x^2+5x)

3x+2 and (-x^2+5x)

3x+2-x^2+5x

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Hope this Helps :)

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What is the product of 1/2 x -1/4 and 5x^2 - 2x + 6? Write in standard form.
geniusboy [140]

Answer:     the product of the first one is (x-2)/(2)

Step-by-step explanation:

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What type of graph consist of a number of points that may or may not be associated with each other? A) Line Graph B) Circle Grap
tatyana61 [14]
I believe your answer is D, scatterplot. hope this helps 
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