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zavuch27 [327]
3 years ago
10

A cookie jar has 3 dozen chocolate chip, 1 dozen peanut butter, and 2 dozen oatmeal raisin cookies. If Megan chooses a cookie, d

oes not
replace it, then chooses another, what is the probability that she chooses a peanut butter then a chocolate chip cookie?
A) 6/71 to attempt to make new friends
B) 1/12 to make people jealous
C) 2/3 to win a trophy
D) 1/10 to get attention
E) 11/72 to get others to join
Mathematics
1 answer:
nika2105 [10]3 years ago
8 0

Answer:

A) 6/71 to attempt to make new friends

Step-by-step explanation:

The probability that she chooses a peanut butter then a chocolate chip cookie is about 6/71 in an attempt to make some new friends!

I hope this helps! Good luck with your assignment/test! <3

- 7v7

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A question from my math class:
valina [46]

Answer:

A) $29.68

B) $5.60

Step-by-step explanation:

<u>Part A Explanation)</u>

To find the total cost including tax, you must multiply the total cost by .06 because .06 out of 1 is 6%. <em>Quick explanation on percentages.</em>

<u><em>DO NOT FORGET TO ADD THE TOTAL COST</em></u>

28.00 + (28.00 * .06) = 29.68

<u>Part B Explanation)</u>

This is asking only for the tip, not including the total cost. However, we need the total cost to calculate the tip.

28.00 * .20 = 5.6

<em>.20 out of 1 is 20%</em>

7 0
4 years ago
Select the correct answer.
laila [671]
B but I don’t know 100%
3 0
3 years ago
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A manufacturer uses production method to produce steel rods. A random sample of 17 steel rods resulted in lengths with a standar
Arisa [49]

Answer:

\chi^2 =\frac{17-1}{12.15} 22.09 =15.87

Traditional method

We can find a critical value in the chi square distribution with df =16 who accumulates \alpha/2 =0.05 of the area on each tail and we got:

\chi^2_{crit}= 7.962

\chi^2_{crit}=26.296

Since the calculated value is between the two critical values we don't have enough evidence to conclude that the true deviation is NOT significantly different from 3.5 cm

Confidence level method

We can find the p value like this

p_v =2*P(\chi^2 >15.87)=0.92

Since the p value is higher then the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is NOT significantly different from 3.5 cm

Step-by-step explanation:

Data provided

n=17 represent the sample selected

\alpha=0.1 represent the significance

s^2 =4.7^2 =22.09 represent the sample variance

\sigma^2_0 =3.5^2 =12.15 represent the value to check

Null and alternative hypothesis

We want to verify if the new production method has lengths with a standard deviation different from 3.5 cm, so the system of hypothesis would be:

Null Hypothesis: \sigma^2 = 12.15

Alternative hypothesis: \sigma^2 \neq 12.15

The statistic for this case is given by:

\chi^2 =\frac{n-1}{\sigma^2_0} s^2

The degrees of freedom are:

df =n-1= 17-1=16

\chi^2 =\frac{17-1}{12.15} 22.09 =15.87

Traditional method

We can find a critical value in the chi square distribution with df =16 who accumulates \alpha/2 =0.05 of the area on each tail and we got:

\chi^2_{crit}= 7.962

\chi^2_{crit}=26.296

Since the calculated value is between the two critical values we don't have enough evidence to conclude that the true deviation is significantly different from 3.5 cm

Confidence level method

We can find the p value like this

p_v =2*P(\chi^2 >15.87)=0.92

Since the p value is higher then the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is significantly different from 3.5 cm

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r=4cm. So d=8cm
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What is n/4 +n/4+n/4+n/4 equal
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n/4+n/4+n/4+n/4=4n/4=n

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4 years ago
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