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3 years ago

Solve for t. 4=12gt^2

Mathematics

1 answer:

ivann1987 [24]3 years ago

6 0

Answer:

$t = \sqrt{ \frac{8}{g} }$

Step-by-step explanation:

We are solving for t.

$4 = \frac{1}{2} g {t}^{2}$

Multiply both sides by 2.

$8 = gt {}^{2}$

Divide both sides by

$\frac{8}{g} = {t}^{2}$

Take square root of both sides.

$t = \sqrt{ \frac{8}{g} }$

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How do you find the area of this triangle ?

marin [14]

Multiply the base which is 18.65 and multiply it by the height which is 9 3/10 and then multiply by half and that estimates to B. 90 cm

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3 years ago

② The complement of an<br> angle. is 10 more<br> than 3 times the<br> angle. Find both

Mrac [35]

Let x be the angle. The complement is 3x + 10. The sum of two complementary angles is 90. So,
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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

2 years ago

Hcf of 14 154 and 28

anyanavicka [17]

Answer:

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Question 13

ArbitrLikvidat [17]

Answer:

Step-by-step explanation:

Here in this question , we are interested in calculating the total distance covered by the train in the 10 minutes that it ran.

From the first part of the question, we already know the distances in the first 5 minutes.

Now, to calculate the total distance in the second 5 minutes, we use the distance formula since we have the average speed and the time;

Mathematically; Total distance = average speed * time

From the question, average speed is 33km/h, while time is 5 minutes. To achieve a consistent unit, we convert 5 minutes to hours.

That would be 5/60 = 1/12 hours

So the total distance in the second 5 minutes is;

33 * 1/12 = 2.75 km

Now, to calculate the total distance traveled, let’s add up the distances in the first 5 minutes and convert to kilometers;

That would be;

68 + 127 + 208 + 312 + 535 = 1,250 m

Let’s convert this to km.

We simply divide by 1000 = 1250/1000 = 1.25 km

The total distance is thus ;

1.25 + 2.75 = 4 km

4 0

3 years ago