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alisha [4.7K]
3 years ago
5

If correct I’ll give brainlest pls help

Mathematics
1 answer:
VladimirAG [237]3 years ago
7 0

Answer:

C

Step-by-step explanation:

I did that test before we trusty meee

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The scatter plot shows the number of passengers at a major airport over a​ 15-year period from the year 2000. About how many pas
Sliva [168]

Answer:

The answer is 65

Step-by-step explanation:

​First, determine the​ x-value that corresponds to the given year.

Since x corresponds to the number of years since​ 2000, the year 2009 corresponds to an​ x-value of 9.

​Next, find the​ y-value on the trend line that corresponds to an​ x-value of 9. The problem statement has been updated to show lines traced from the​ x-value of 9 to the trend line and then to the​ y-value.

Notice that the point​ (9,70) appears to lie on the trend line.

​Therefore, in​ 2009, approximately 70 million passengers travelled through the airport.

Question is complete.

4 0
3 years ago
A circle has a sector with area 27pie and central angle of 120
Nat2105 [25]

Answer:

Step-by-step explanation:

27π=120°/360°*πr^2

81π=Area of circle

7 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
What is 134.76821 when rounded to the nearest hundredth.
AleksAgata [21]

Answer:

134.77

Step-by-step explanation:

3 0
4 years ago
Read 2 more answers
Rewrite without rational exponents, and simplify, if possible   (x²y²)1/7    
SIZIF [17.4K]
(x^2y^2)^\frac{1}{7}=\sqrt[7]{x^2y^2}
6 0
3 years ago
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