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expeople1 [14]
3 years ago
13

How do I solve this operation

Mathematics
1 answer:
ivann1987 [24]3 years ago
6 0
4 / ( 3 + 5i ) = [ 4( 3 - 5i)] / [ 3^2 - (5i)^2 ] = [ 4( 3 - 5i)] / 34 =  [ 2( 3 - 5i)] / 17 =
 6 / 17 - 10i / 17;
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Which of the following Office Online apps is most effective for writing letters?
Fynjy0 [20]

Answer:

word

Step-by-step explanation:

7 0
3 years ago
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Consider the statements below. Which are not propositions? Mark all that apply.
Troyanec [42]

Answer:

In Idaho, there are 15 temples for the Church of Jesus Christ of Latter-day Saints.

A delicious chocolate cake.

The great and spacious building.

3+5=8

5+3=53

Step-by-step explanation:

To determine which statement is a propositions or not first, we need to define the concept of preposition:

<em>A statement that can be either true or false.</em>

That means it should declare unequivocal information about something.

A question is not declaring something to be true or false. Therefore, it is not a proposition.

The recommendation of "should get" is not a direct statement.

No way! is not referring to anything to be either true or false.

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
Find the sum or difference of the polynomials (x2−xy+y2) − (x2+xy+y2)
stepan [7]
The correct answer is-2xy if it’s subtract
6 0
4 years ago
I figured out part A each notebook is $2.75 and each pencil is $1.10 but I need some help on part B. I figured out that he can b
tangare [24]

Answer:

2 notebooks

Step-by-step explanation:

I understand you already figured out that it was 3 notebooks, but when I calculated it, I believe he could only buy 2...

Personally, I would show my work by first doing the calculation: 2.75[1 notebook price]+ (3[pens/pencils per notebook]*1.10[pen/pencil cost])= $9.075 per notebook and 3 pens

If you then showed that you multiplied 9.075 by 2... you would get $18.15 which is the closest you could get to his limit of $22 without going over

If we applied your answer of 3 notebooks, then the total would be $27.225 which goes over the money limit.

Hope I helped!

6 0
3 years ago
Read 2 more answers
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