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viktelen [127]
3 years ago
6

Keisha can jog 2 miles in 24 minutes. At the same rate , how many miles can she jog in 60 minute

Mathematics
1 answer:
asambeis [7]3 years ago
5 0
In 24 minutes Keisha jogs 2 miles . So in 60 minutes she will jog  (2/24) x 60 =5 miles . In 12 minutes she jogs 1 mile . so in 60 minutes it will be 5 miles.
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Y=2x+3<br> solve for y<br> help me plz
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Answer:

this is already solved.

Step-by-step explanation:

the only thing left would be graphing it.

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There’s a 5% chance
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A recipe for punch requires 20 grams of drink mix for 5 liters.
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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)

26 preferred candy bars (CB)

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!

5 0
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Below is a piece of a local pizza place menu where Jacob works.
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Read 2 more answers
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