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makkiz [27]
3 years ago
7

And one season the top four players on one major league team had batting averages of 0.332, 0.361, 0.343 and 0.360 which is the

highest average
Mathematics
1 answer:
DanielleElmas [232]3 years ago
8 0
ANSWER: 0.361

This is as:

0.361 > 0.332

0.361 > 0.343

0.361 > 0.36

Therefore, as 0.361 is greater than all the other averages, it must be the highest average.

Hope this helps! :)
Have a lovely day! <3
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Solve the equation. 8x2 – 4 = 28
Flura [38]
Hello,

Your answer would be 2, -2.

8x^2 - 4 = 28

8x^2-4+4=28+4

8x^2 = 32

\frac{8x^2}{8}=  \frac{32}{8}  &#10;

x^2 = 4

x = +/4 

x = 2 or x = -2

2 , -2 would be the answer! :)


Mark Brainliest if this helped you :)
4 0
3 years ago
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Solve: -5√x= -15<br><br> a: x=-9<br> b: x= -7<br> c: x=9<br> d: x=7
babymother [125]
Divide both sides by -5
Sqrt(x) = 3
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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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3 years ago
Which pairs of angles in the figure below are vertical angles?<br><br> Check all that apply.
erica [24]
I think it’s all of them.
7 0
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Simplify.
Hoochie [10]
C might be the answer
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