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katrin2010 [14]
3 years ago
10

Out of 200 students in a school, 30 have brown sandals. What is the probability that a student chosen at random would have brown

sandals?
Mathematics
2 answers:
diamong [38]3 years ago
8 0

Answer:

There is a 15% chance a student chosen at random would have brown sandals.

Step-by-step explanation:

Mariana [72]3 years ago
6 0

(200-30):170

Step-by-step explanation:

170:200. 17;20

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A 41 gram sample of a substance that's used to detect explosives has a k-value of 0.1392. Find the substance's half life, in day
-BARSIC- [3]

Answer:

The substance half-life is of 4.98 days.

Step-by-step explanation:

Equation for an amount of a decaying substance:

The equation for the amount of a substance that decay exponentially has the following format:

A(t) = A(0)e^{-kt}

In which k is the decay rate, as a decimal.

k-value of 0.1392.

This means that:

A(t) = A(0)e^{-0.1392t}

Find the substance's half life, in days.

This is t for which A(t) = 0.5A(0). So

A(t) = A(0)e^{-0.1392t}

0.5A(0) = A(0)e^{-0.1392t}

e^{-0.1392t} = 0.5

\ln{e^{-0.1392t}} = \ln{0.5}

-0.1392t = \ln{0.5}

t = -\frac{\ln{0.5}}{0.1392}

t = 4.98

The substance half-life is of 4.98 days.

5 0
3 years ago
A researcher used a sample of n = 60 individuals to determine whether there are any preferences among six brands of pizza. Each
Blizzard [7]

Answer:

1) χ² ≥ 11.07

2) Goodness of fit test, df: χ²_{3}

Independence test, df: χ²_{1}

The goodness of fit test has more degrees of freedom than the independence test.

3) e_{females.} = 80

4) H₀: P_{ij}= P_{i.} * P_{.j} ∀ i= 1, 2, ..., r and j= 1, 2, ..., c

5) χ²_{6}

Step-by-step explanation:

Hello!

1)

The researcher took a sample of n=60 people and made them taste proof samples of six different brands of pizza and choose their favorite brand, their choose was recorded. So the study variable is the following:

X: favorite pizza brand, categorized in brand 1, brand 2, brand 3, brand 4, brand 5 and brand 6.

The Chi-square goodness of fit test is done with the following statistic:

χ²= ∑\frac{(O_i-E_i)^2}{E_i} ≈χ²_{k-1}

Where k represents the number of categories of the study variable. In this example k= 6.

Remember, the rejection region for the Chi-square tests of "goodnedd of fit", "independence", and "homogeneity" is allways one-tailed to the right. So you will only have one critical value.

χ²_{k-1; 1 - \alpha }

χ²_{6-1; 1 - 0.05 }

χ²_{5; 0.95 } = 11.070

This means thar the rejection region is χ² ≥ 11.07

If the Chi-Square statistic is equal or greather than 11.07, then you reject the null hypothesis.

2)

The statistic for the goodness of fit is:

χ²= ∑\frac{(O_i-E_i)^2}{E_i} ≈χ²_{k-1}

Degrees of freedom: χ²_{k-1}

In the example: k= 4 (the variable has 4 categories)

χ²_{4-1} = χ²_{3}

The statistic for the independence test is:

χ²= ∑∑\frac{(O_ij-E_ij)^2}{E_ij} ≈χ²_{(r-1)(c-1)} ∀ i= 1, 2, ..., r & j= 1, 2, ..., c

If the information is in a contingency table

r= represents the total of rows

c= represents the total of columns

In the example: c= 2 and r= 2

Degrees of freedom: χ²_{(r-1)(c-1)}

χ²_{(2-1)(2-1)} = χ²_{1}

The goodness of fit test has more degrees of freedom than the independence test.

3)

To calculate the expected frecuencies for the independence test you have to use the following formula.

e_{ij} = n * P_i. * P_.j = n * \frac{o_i.}{n} * \frac{o_.j}{n}

Where o_i. represents the total observations of the i-row, o_.j represents the total of observations of the j-column and n is the sample size.

Now, this is for the expected frequencies in the body of the contingency table, this means the observed and expected frequencies for each crossing of categories is not the same.

On the other hand, you would have the totals of each category and population in the margins of the table (subtotals), this is the same when looking at the observed frequencies and the expected frequencies. Wich means that the expected frequency for the total of a population is the same as the observed frequency of said population. A quick method to check if your calculations of the expected frequencies for one category/population are correct is to add them, if the sum results in the subtotal of that category/population, it means that you have calculated the expected frequencies correctly.

The expected frequency for the total of females is 80

Using the formula:

(If the females are in a row) e_{females.} = 100 * \frac{80}{100} * \frac{0}{100}

e_{females.} = 80

4)

There are two ways of writing down a null hypothesis for the independence test:

Way 1: using colloquial language

H₀: The variables X and Y are independent

Way 2: Symbolically

H₀: P_{ij}= P_{i.} * P_{.j} ∀ i= 1, 2, ..., r and j= 1, 2, ..., c

This type of hypothesis follows from the definition of independent events, where if we have events A and B independent of each other, the probability of A and B is equal to the product of the probability of A and the probability of B, symbolically: P(A∩B) = P(A) * P(B)

5)

In this example, you have an independence test for two variables.

Variable 1, has 3 categories

Variable 2, has 4 categories

To follow the notation, let's say that variable 1 is in the rows and variable 2 is in the columns of the contingency table.

The statistic for this test is:

χ²= ∑∑\frac{(O_ij-E_ij)^2}{E_ij} ≈χ²_{(r-1)(c-1)} ∀ i= 1, 2, ..., r & j= 1, 2, ..., c

In the example: c= 3 and r= 4

Degrees of freedom: χ²_{(r-1)(c-1)}

χ²_{(3-1)(4-1)} = χ²_{6}

I hope you have a SUPER day!

4 0
3 years ago
Which angle is congruent to BGC?
lawyer [7]

Answer:

BFC makes this angle add up to 90 degree at BFD

Therefore it becomes complementary to angle CFD

4 0
3 years ago
Read 2 more answers
5. Thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately norma
Triss [41]

Answer:

percent (%) of ancient prehistoric thinner than 3 is 0.0013

Step-by-step explanation:

Given data

mean =  5.1 millimeters (mm)

standard deviation = 0.7 mm

to find out

percent (%) of ancient prehistoric thinner than 3

solution

we have given mean and SD

so Z will be = x -mean  / SD

Z = ( x - 5.1 ) / 0.7

so P( x < 3 ) = P ( (x - 5.1 ) / 0.7  <  ( 3 - 5.1 ) / 0.7 )

P( Z  = ( 3 - 5.1 ) / 1.5 )

P( Z  =  -3 ) = 0.0013

percent (%) of ancient prehistoric thinner than 3 is 0.0013

4 0
3 years ago
Determine whether the graph shows a positive correlation, a negative correlation, or no correlation. If there is a positive or n
RUDIKE [14]
If the points are scattered it’s no correlation
7 0
3 years ago
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