If that dot in the middle is the middle of the circle, then that should mean the center is 5 away from both of those lines. Therefore, I believe x would 9/2 or 4.5

4 0

2 years ago

Please give me the answer<br> I will give you 29 points

yanalaym [24]

Sammy made 6 baskets so whever mad 6*3 baskets or 18 is Jefferson . mult statement =6*3

Cassidy made x times as mitchel or 24=3x so the answer is 8
24/3=8

Jefferson made ____ times as many baskets as ____
you could use sammy or mitchell so if we use sammy
jeff made 3 times as many as sammy
18/6=3

sammy made twice the baskets as ____
sammy divided by 2 = person=6/2=3 and the thre person is mitchell so 6/2=3

8 0

3 years ago

Read 2 more answers

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

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.

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.

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

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

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

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

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

I hope this helps!

5 0

3 years ago