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Neko [114]
3 years ago
5

105 push ups in 3 minutes . find the unit rate and how many in five minutes

Mathematics
2 answers:
stich3 [128]3 years ago
6 0

Answer:

<u>Unit Rate:</u> 35    <u>Number of Pushups in five minutes:</u> 175

Step-by-step explanation:

In order to find the unit rate, divide 105 by 3. You'll get 35. So the <u>Unit Rate</u> is 35.

Now, in order to find out how many pushups in five minutes, simply multiply 35 by 5. You'll get 175 pushups in five minutes. So the <u>number of pushups</u> in <u>five minutes</u> is 175.

umka2103 [35]3 years ago
4 0

Step-by-step explanation:

105/3 = 35 push ups per minute

35 x 5 = 175 push ups in 5 minutes

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A researcher planned a study in which a crucial step was offering participants a food reward. It was important that three food r
lbvjy [14]

Answer:

Step-by-step explanation:

Hello!

A pilot study was conducted to test if three food rewards are equally appealing to the participants.

Of 60 participants surveyed:

16 preferred cupcakes (CC)

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18 preferred dried apricots (DA)

If the three types of food are equally appealing for the participants, you'd expect that their proportions will be equal: P(CC)=P(CB)=P(DA)= 1/3

1.

The objective of this pilot study is to test if the observed frequencies follow a theoretical model/ distribution. To analyze this, you have to apply a Chi Square Goodness to Fit test. X^2=sum \frac{(O_i-E_i)^2}{E_i} ~~X^2_{k-1} Where k= number of categories of the variable.

For this example the statistical hypotheses are:

H₀: P(CC)=P(CB)=P(DA)= 1/3

H₁: At least one of the proportions isn't equal to the others.

2.

To calculate the expected frequencies for each category you have to use the formula: E_i= n* P_i where Pi represents the theoretical proportion for the i category, stated in the null hypothesis.

E_{CC}= n* P(CC)= 60*1/3= 20

E_{CB}= n*P(CB)= 60*1/3= 20

E_{DA}= n* P(DA)= 60* 1/3= 20

3.

The cutoff or critical value indicates the beginning of the rejection region for the hypothesis test. For the Chi-Square tests, the rejection region is always one-tailed to the right, meaning that you'll reject the null hypothesis if the value of the statistic is big. For the goodness to fit test you have k-1 degrees of freedom, so the critical value will be:

Assuming α: 0.05

X^2_{k-1;1-\alpha /2}= X^2_{2;0.975}= 7.378

The rejection region is then X²₂ ≥ 7.378

4.

X^2_{H_0}= \frac{(O_{CC}-E_{CC})^2}{E_{CC}} + \frac{(O_{CB}-E_{CB})^2}{E_{CB}}  + \frac{(O_{DA}-E_{DA})^2}{E_{DA}} = \frac{(16-20)^2}{20} +\frac{(26-20)^2}{20} +\frac{(18-20)^2}{20}=  \frac{14}{5}= 2.8

I hope this helps!

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