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Llana [10]
3 years ago
5

There are 35 students in the drama club. 2 out of every 5 students in the club are boys.

Mathematics
2 answers:
mrs_skeptik [129]3 years ago
8 0

Answer: In the drama club there’s 14 boys

Step-by-step explanation:

35/5=7

7*2=14 boys

ankoles [38]3 years ago
3 0

Answer:

14

Step-by-step explanation:

35 divided by 5=7

7times 2 =14

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Decide which chi-square test (goodness-of fit, homogeneity, or independence) would be most appropriate for the given situation.
sasho [114]

Answer:

C. Test for Goodness-of-fit.

Step-by-step explanation:

C. Test for Goodness-of-fit would be most appropriate for the given situation.

A. Test Of  Homogeneity.

The value of q is large when the sample variances differ greatly and is zero when all variances are zero . Sample variances do not differ greatly in the given question.

B. Test for Independence.

The chi square  is used to test the hypothesis about the independence of two variables each of which is classified into number of attributes. They are not classified into attributes.

C. Test for Goodness-of-fit.

The chi square test is applicable when the cell probabilities depend upon unknown parameters provided that the unknown parameters are replaced with  their estimates and provided that one degree of freedom is deducted for each parameter estimated.

4 0
2 years ago
If the circumference of a circle measures 3/2TT cm, what is the area of the circle in terms of TT?
diamong [38]

Answer:

A = pi 9/16  cm^2

Step-by-step explanation:

The circumference is 3/2 pi

The circumference is given by

C = 2 * pi *r

3/2 pi = 2 * pi *r

Divide each side by 2 *pi

3/2 pi / 2 pi  = 2 pi r /2pi

3/4 cm = r

Now we can find the area

A = pi r^2

A = pi ( 3/4) ^2

A = pi 9/16  cm^2

7 0
3 years ago
Read 2 more answers
The probability that a professor arrives on time is 0.8 and the probability that a student arrives on time is 0.6. Assuming thes
saul85 [17]

Answer:

a)0.08  , b)0.4  , C) i)0.84  , ii)0.56

Step-by-step explanation:

Given data

P(A) =  professor arrives on time

P(A) = 0.8

P(B) =  Student aarive on time

P(B) = 0.6

According to the question A & B are Independent  

P(A∩B) = P(A) . P(B)

Therefore  

{A}' & {B}' is also independent

{A}' = 1-0.8 = 0.2

{B}' = 1-0.6 = 0.4

part a)

Probability of both student and the professor are late

P(A'∩B') = P(A') . P(B')  (only for independent cases)

= 0.2 x 0.4

= 0.08

Part b)

The probability that the student is late given that the professor is on time

P(\frac{B'}{A}) = \frac{P(B'\cap A)}{P(A)} = \frac{0.4\times 0.8}{0.8} = 0.4

Part c)

Assume the events are not independent

Given Data

P(\frac{{A}'}{{B}'}) = 0.4

=\frac{P({A}'\cap {B}')}{P({B}')} = 0.4

P({A}'\cap {B}') = 0.4 x P({B}')

= 0.4 x 0.4 = 0.16

P({A}'\cap {B}') = 0.16

i)

The probability that at least one of them is on time

P(A\cup B) = 1- P({A}'\cap {B}')  

=  1 - 0.16 = 0.84

ii)The probability that they are both on time

P(A\cap  B) = 1 - P({A}'\cup {B}') = 1 - [P({A}')+P({B}') - P({A}'\cap {B}')]

= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56

6 0
3 years ago
In an effort to estimate the mean of amount spent per customer for dinner at a major Lawrence restaurant, data were collected fo
larisa86 [58]

Answer:

The 99% confidence interval for the population mean is 22.96 to 26.64

Step-by-step explanation:

Consider the provided information,

A sample of 49 customers. Assume a population standard deviation of $5. If the sample mean is $24.80,

The confidence interval if 99%.

Thus, 1-α=0.99

α=0.01

Now we need to determine z_{\frac{\alpha}{2}}=z_{0.005}

Now by using z score table we find that  z_{\frac{\alpha}{2}}=2.58

The boundaries of the confidence interval are:

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Hence, the 99% confidence interval for the population mean is 22.96 to 26.64

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pav-90 [236]
Why....are you asking..$50 is least expensive. Because you use about 36 gigabytes at least so 5•x=
Or $50+12
5 0
3 years ago
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