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3 years ago

A or B Choose 1 answer:

Mathematics

2 answers:

viva [34]3 years ago

8 0

Answer:

Step-by-step explanation:

since you are dividing by four, the graph must show 1/4 in for divisions.

Serggg [28]3 years ago

7 0

Answer:

HOPE THIS HELPS

-Todo <3

Step-by-step explanation:

4 parts broken into 4 parts

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Question is stated in picture. The figure is a triangular piece of cloth

Anastasy [175]

Answer:

Alternative D - 8 sin(35°)

Step-by-step explanation:

Sin(x) is defined as:

$\begin{gathered} \sin (x)=\frac{\text{Opposite side}}{Hypotenuse\text{ }} \\ \end{gathered}$

In this exercise,

BC is the opposite side to 35°

AC is the hypotenuse and measures 8 in

Then:

$\begin{gathered} \sin (35\degree)=\frac{BC}{8} \\ \sin (35\degree)\cdot8=BC \\ BC=8\sin (35\degree) \end{gathered}$

4 0

2 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

Suppose that the population P(t) of a country satisfies the differential equation dP/dt = kP (600 - P) with k constant. Its pop

jeka94

Answer:

The country's population for the year 2030 is 368.8 million.

Step-by-step explanation:

The differential equation is:

$\frac{dP}{dt}=kP(600 - P)\\\frac{dP}{P(600 - P)} =kdt$

Integrate the differential equation to determine the equation of P in terms of t as follows:

$\int\limits {\frac{1}{P(600-P)} } \, dP =k\int\limits {1} \, dt \\(\frac{1}{600} )[(\int\limits {\frac{1}{P} } \, dP) - (\int\limits {\frac{}{600-P} } \, dP)]=k\int\limits {1} \, dt\\\ln P-\ln (600-P)=600kt+C\\\ln (\frac{P}{600-P} )=600kt+C\\\frac{P}{600-P} = Ce^{600kt}$

At t = 0 the value of P is 300 million.

Determine the value of constant C as follows:

$\frac{P}{600-P} = Ce^{600kt}\\\frac{300}{600-300}=Ce^{600\times0\times k}\\\frac{1}{300} =C\times1\\C=\frac{1}{300}$

It is provided that the population growth rate is 1 million per year.

Then for the year 1961, the population is: P (1) = 301

Then $\frac{dP}{dt}=1$ .

Determine k as follows:

$\frac{dP}{dt}=kP(600 - P)\\1=k\times300(600-300)\\k=\frac{1}{90000}$

For the year 2030, P (2030) = P (70).

Determine the value of P (70) as follows:

$\frac{P(70)}{600-P(70)} = \frac{1}{300} e^{\frac{600\times 70}{90000}}\\\frac{P(70)}{600-P(70)} =1.595\\P(70)=957-1.595P(70)\\2.595P(70)=957\\P(70)=368.786$

Thus, the country's population for the year 2030 is 368.8 million.

3 0

3 years ago

What is equivalent to 4/16 as a whole number?

il63 [147K]

I think just divide 4 by 16 and that will be your answer? I'm not sure but the answer that I'm giving you is 4 your welcome if it helped in anyway

8 0

4 years ago

Someone answer please :)

vlabodo [156]

Answer:

1 : 11

Step-by-step explanation:

9 : 99 = divide both numbers by 9
1 : 11

5 0

3 years ago