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olga2289 [7]
3 years ago
8

Chris has 150 cookies to sell each cookie is covered with one topping 1 /5 of cookie are covered with coconut 1/3 of the cookie

are covered with chocolate chips 3/10 of the cookies are covered with yellow icing the rest is sprinkles how many cookies are covered with sprinkles ?
Mathematics
1 answer:
Elan Coil [88]3 years ago
3 0

Answer:

1/6 or 25

Step-by-step explanation:

1/3=10/30

3/10=9/30

1/5=6/30

together is 25/30, which leaves the rest, 5/30 or 1/6 or 25 cookies.

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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

7 0
3 years ago
Simplify ...can I get help
Vinvika [58]

Answer:

6^21 = B

Step-by-step explanation:

[6^{7} ]^{3} \\(6)^7^*^3\\6 ^{21} \\

4 0
3 years ago
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What is the lenght in miles of the field
maw [93]
(Perimeter/6)2= Length
8 0
3 years ago
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The following histogram shows the number of items sold at a grocery store at various prices:
monitta

Answer:

<h2><u><em>its B</em></u></h2>

Step-by-step explanation:

8 0
3 years ago
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The length of a rectangle is 5 in. more than 3 times its width. the area of the rectangle is 50in.2. a quadratic function repres
Goshia [24]
The domain is actually the x value of the function, so we need to find the value of x
suppose the width is x, the length is then 3x+5
the area is 50^2 inches, so x(3x+5)=50 => 3x^2+5x-50=0
Factor this quadratic equation: (x-5)(3x+10)=0 =>x=5 or x=-10/3
width can not be negative, so the width is 5
the domain is x=5
6 0
3 years ago
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