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professor190 [17]

2 years ago

12

Jonah found that it took him 10 minutes to do 5 math problems. If he uses this information, what should be his estimation on how

long it will take him to do 32 math problems?

1 answer:

Andrei [34K]2 years ago

5 0

We can round 32 down to 30.

10 minutes for 5 math problems is the same as 2 minutes for 1 math problem. Therefore, we can estimate that it will take Jonah approximately 60 minutes to do 32 math problems.

Hope this helps!! :)

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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

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3 years ago

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diamong [38]

So those two angles are congruent to each other because they have the same size and they are the same shape

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3 years ago

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Ivanshal [37]

Answer:

B. 57

Step-by-step explanation:

Given:

A regression line for some data transformed with logarithms predicts that when

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Solution:

When $x=7$

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We can solve for $y$ by using logarithm properties.

<em>By logarithm properties.</em>

If $\log(a)=b$

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stich3 [128]

Given $sin (\theta) = \frac{1}{5}$ , and $tan \theta = \frac{\sqrt{6} }{12}$ , then:

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<h3> Computations on trigonometric identities</h3>

From the details provided:

$sin (\theta) = \frac{1}{5}\ \\\\tan \theta = \frac{\sqrt{6} }{12} \times \frac{\sqrt{6} }{\sqrt{6} }\\\\tan \theta = \frac{\sqrt{6} \times \sqrt{6} }{12 \times \sqrt{6} }\\\\tan \theta =\frac{1}{2\sqrt{6} }$

From the relationship above:

Opposite = 1

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Hypotenuse = 5

$cos\theta=\frac{Adjacent}{Hypotenuse} \\\\cos\theta=\frac{2\sqrt{6} }{5}$

Learn more on trigonometry here: brainly.com/question/20519838

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Answer:

If the number of a set of consecutive numbers is odd, the sum of the consecutive numbers is the number of consecutive numbers * the middle number. Here, the sum is 78. We can find the middle number, in this case, the 2nd, by diving 78 by 3. The second number is 26

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