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Dimas [21]
2 years ago
13

Help please!! Thank you!! :)

Mathematics
1 answer:
Naddik [55]2 years ago
5 0

Answer:

  A.  m^2+7m+10=0

Step-by-step explanation:

This is a problem in pattern matching, and in substituting a variable for a pattern.

  (x^2+3)^2 +7x^2 +21 = -10 . . . . . . given

  (x^2 +3)^2 +7(x^2 +3) = -10 . . . . . factor the last two terms

  m^2 +7m = -10 . . . . . . . . . . . . subsitute m for x^2 +3

  m^2 +7m +10 = 0 . . . . . . . . add 10 to both sides; matches A

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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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igomit [66]
Looking from left to right it is increasing
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Add x^3 - 4x^2 + 1 to 3x^2 + x <br> Show Your work!
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\qquad \qquad\huge \underline{\boxed{\sf Answer}}

Let's solve ~

\qquad \sf  \dashrightarrow \: (x {}^{3}  - 4x {}^{2}  + 1) + (3 {x}^{2}  + x)

\qquad \sf  \dashrightarrow \:  {x}^{3}  - 4 {x}^{2}  + 1 + 3 {x}^{2}  + x

\qquad \sf  \dashrightarrow \:  {x}^{3}  - 4 {x}^{2}  + 3x {}^{2}  + x + 1

\qquad \sf  \dashrightarrow \:  {x}^{3}  -  {x}^{2}  + x + 1

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2 years ago
What are the potential solutions of In(x^2-25)=0?
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Answer:

x =\pm \sqrt{26}

Step-by-step explanation:

<u>Given </u><u>:</u><u>-</u><u> </u>

  • ln ( x² - 25 ) = 0

And we need to find the potential solutions of it. The given equation is the logarithm of x² - 25 to the base e . e is Euler's Number here. So it can be written as ,

<u>Equation</u><u> </u><u>:</u><u>-</u><u> </u>

\implies log_e {(x^2-25)}= 0

<u>In </u><u>general</u><u> </u><u>:</u><u>-</u><u> </u>

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\implies log_a b = c

Then this can be written as ,

\implies a^c = b

In a similar way we can write the given equation as ,

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\implies 1 = x^2 - 25 \\\\\implies x^2 = 25 + 1 \\\\\implies x^2 = 26 \\\\\implies x =\sqrt{26} \\\\\implies x = \pm \sqrt{ 26}

<u>Hence</u><u> the</u><u> </u><u>Solution</u><u> </u><u>of </u><u>the</u><u> given</u><u> equation</u><u> is</u><u> </u><u>±</u><u>√</u><u>2</u><u>6</u><u>.</u>

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