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Fudgin [204]
3 years ago
15

Find, as a ratio, the scale of a map given as 2cm to 1km. If two towns are 23.4cm apart on the map, what is the actual distance

between them?
Mathematics
1 answer:
Rina8888 [55]3 years ago
8 0

Answer:

The actual distance is 11.7km

Step-by-step explanation:

2cm = 1km

1cm = 0.5km

23.4cm = ?

23.4 × 0.5 = 11.7km

(Correct me if i am wrong)

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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
Find the perimeter of the figure?
zhannawk [14.2K]

Answer:

36

Step-by-step explanation:

9+9+9+9=36

8 0
3 years ago
Read 2 more answers
Drag the slider to 2 inches. How many centimeters are in 2 inches? There are centimeters in 2 inches.
choli [55]

Answer: 5


Step-by-step explanation: There are 2.5 cm per in, so its just 2.5*2


4 0
3 years ago
Read 2 more answers
how many solutions does the equation 2(2x- 10)-8 =-2(14-3x) have?A.exactly one solution B. exactly two solutions C. no solution
Zarrin [17]

Answer: it has no solutions

Step-by-step explanation:

Factor then solve to find the complex solutions.

5 0
2 years ago
The Academic Computing Center has five trainers available in its computer labs to provide training sessions to students. Assume
kondor19780726 [428]

Answer:

87.72%

Step-by-step explanation:

Data provided in the question:

Design capacity of the system = 1900 students per semester

Effective capacity = 90% of design capacity

Actual number of students = 1500

Now,

Efficiency = [ [ Actual capacity ] ÷ [ Effective capacity ] ] × 100%

also,

Effective capacity = 90% of 1900

= 0.90 × 1900

= 1710

Efficiency = [ 1500 ÷ 1710 ] × 100%

= 0.8772 × 100%

= 87.72%

7 0
3 years ago
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