You are mixing solution to a gallon of water it is one part solution to 3 parts water how much solution would you use

The area of a circle is πr2. To find the area of a sector with a central angle of Θ, measured in radians, by what should you mul

Ipatiy [6.2K]

There are 2pi radians in a circle

so the fraciton is Θ/(2pi)

so
pir^2 times Θ/(2pi)=pir^2Θ/2pi=Θr^2/2
multiply it by Θ/(2pi)

6 0

3 years ago

Read 2 more answers

Classify the triangle by its sides, and then by its angles.

Gekata [30.6K]

Answer:

Equilateral, acute

Step-by-step explanation:

Equilateral triangles have all sides the same length (all sides are 4 ft in this triangle, so it is equilateral).

Acute triangles have no angles that are greater than or equal to 90 degrees (all angles are 60, which is less than 90, so it is acute).

3 0

3 years ago

Brian scored 92, 87 and 88 on three science tests. Write and solve an equation to find the score Brian will need on the fourth s

Marta_Voda [28]

Answer:

You would need to score a 93% to have an average of 90.

Step-by-step explanation:

Well, to check my work you would need to find the average.

So, 93 + 92 + 88 + 87 = 360. Then you divide by four; 360 / 4 = 90. ^^

3 0

2 years ago

Inequalities part 2<br> I need help pls i dont understand

Bad White [126]

Step-by-step explanation:

For y/x to be as high as possible, y must have the highest possible value and x must have the lowest possible value. (12 and 6)

Hence, y/x < 12/6, which is 2.

For y/x to be as low as possible, y must have the lowest possible value and x must have the highest possible value (10 and 7)

Hence, y/x > 10/7.

Combining the 2 inequalities, we have 10/7 < y/x < 2.

8 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