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3 years ago

Can you please help me with this very hard math problem?

Mathematics

2 answers:

padilas [110]3 years ago

6 0

Answer:

how would I solve this monstrosity

ElenaW [278]3 years ago

5 0

Sorry man I can’t answer that the screen is too dark (I DONT MEAN TO BE RACIST)

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Ludwig borrowed $700,000 for a period of 2 years. His interest rate is 10%. How much interest will Ludwig pay in all?

harkovskaia [24]

By yearly interest or monthly? Yearly: he pays 840000 monthly: 2380000 sorry it’s so long just tryna help a bruthu or sister out I didn’t check yo username

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3 years ago

A restaurant bill is 56.39.Estimate a 15% tip

hjlf

15% would be 8.4585

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3 years ago

Who was the first president of the United States

Phoenix [80]

Answer: George Washington

Step-by-step explanation:

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3 years ago

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Danielle earns $36.80 for 4 hours of yardwork. MacKenzie earns $50.40 for 6 hours of yardwork.Who earns more money per hour? How

Fantom [35]

Answer:

Danielle earns $1.20 more than Mackenzie (per hour)

Step-by-step explanation:

To find out how much money per hour each person earns, you would divide the earinings by the number of hours;

Danielle: $36.80/4 = $9.20 per hour

MackKenzie: $50.40/6 = $8.40 per hour

To find out how much extra Danielle earns, you would subtract 8.40 from 9.20;

9.40 - 8.20 = $1.20

Thus, Danielle earns $1.20 more than Mackenzie per hour.

6 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