The mean weight of newborn infants at a community hospital is 6.6 pounds. A sample of seven infants is randomly selected and their weights at birth are recorded as 9.0, 7.3, 6.0, 8.8, 6.8, 8.4, and 6.6 pounds. Does the sample data show a significant increase in the average birthrate at a 5% level of significance?
A. Fail to reject the null hypothesis and conclude the mean is 6.6 lb.
B. Reject the null hypothesis and conclude the mean is lower than 6.6 lb.
C. Reject the null hypothesis and conclude the mean is greater than 6.6 lb.
D. Cannot calculate because the population standard deviation is unknown

Answer:

Step-by-step explanation:

This is an example of a frequency distribution for a class interval. In order to complete the frequency distribution, we will count the number of data occurring in each group, and write that number as the frequency for that group. This is done as shown below:

 Age                Frequency          ages in class

20-29                       4                  22, 26, 27, 27                

30-39                       3                  31, 34, 36

40-49                       5                  42, 43, 44, 48, 49

50-59                       5                  52, 53, 55, 56, 57

60-69                       6                  60, 56, 65, 66, 67, 69

70-79                        6                  72, 75, 77, 78, 78, 79

80-89                       1                   87

Total                        30

Answer:

45

------

75

Step-by-step explanation:

Let x be the value of the numerator and x+30 be the value of the denominator

This is equal to 3/5

x             3

-------- = -------

x+30      5

Using cross products

5x = 3(x+30)

Distribute

5x = 3x+90

Subtract 3x from each side

2x = 90

Divide by 2

x = 45

The fraction is

45

-----

30+45

45

------

75

[tex]\dfrac{x}{x+30}=\dfrac{3}{5}\\\\5x=3(x+30)\\5x=3x+90\\2x=90\\x=45\\\\\dfrac{x}{x+30}=\dfrac{45}{45+30}=\dfrac{45}{75}[/tex]

Answer:

27/35

Step-by-step explanation:

We use combination to solve for this

C(n, r), =nCr = n!/r!(n - r)!

From the question, we are told that:

Four couples are at a party. Four of the eight people are randomly selected to win a prize.

Four couples = 8 people.

= 8C4 = 8!/4! (8 - 4)!

= 70

No person can win more than one prize. ( No person can win more than one prize of the 4 people selected)

This can happen in 4 ways

[4C1 × 3C2 ] × 4=

[4!/1! ×( 4 - 1)!] × [3!/2! ×(3-2)!] × 4 ways

= 4 × 3 × 4 ways

= 48

The probability that both of the members of at least one couple win prizes

48 + 4C2/ 8C4

4C2 = 4!/2!(4 - 2) !

= 6

8C4 = 8C4 = 8!/4! (8 - 4)!

= 70

48 + 6/ 70

= 54/70

= 27/35

Therefore, the probability that both of the members of at least one couple win prizes is 27/35.

The probability that both of the members of at least one couple win prizes is 27/35 and this can be determined by using the given data.

Given :

The following steps can be used in order to determine the probability that both of the members of at least one couple win prizes:

Step 1 - The concept of probability is used in order to determine the probability that both of the members of at least one couple win prizes.

Step 2 - According to the given data, the total number of people is 8.

Step 3 - So, the probability that both of the members of at least one couple win prizes is:

[tex]\rm P =\dfrac{ \;^4C_1\times \;^3C_2\times 4 + \;^4C_2}{\;^8C_4}[/tex]

Step 4 - Simplify the above expression.

[tex]\rm P =\dfrac{48+ 6}{70}[/tex]

[tex]\rm P = \dfrac{27}{35}[/tex]

So, the probability that both of the members of at least one couple win prizes is 27/35.

For more information, refer to the link given below:

https://brainly.com/question/795909

Answer:

The mean, variance, and standard deviation of the binomial distribution are 10, 8, and 2.83 respectively.

Step-by-step explanation:

We have to find the mean, variance, and standard deviation of the binomial distribution with the given values of n and p, i.e; n = 50 p = 0.2.

Let X = binomial random variable

So, X ~ Binom(n = 50, p = 0.2)

Now, the mean of the binomial distribution is given by;

         Mean of X, E(X) = n [tex]\times[/tex] p

                                    = 50 [tex]\times[/tex] 0.2 = 10

Now, the variance of the binomial distribution is given by;

        Variance of X, V(X) = n [tex]\times[/tex] p [tex]\times[/tex] (1 - p)

                                         = 50 [tex]\times[/tex] 0.2 [tex]\times[/tex] (1 - 0.2)

                                         = 10 [tex]\times[/tex] 0.8 = 8

Also, the standard deviation of the binomial distribution is given by;

        Standard deviation of X, S.D.(X) = [tex]\sqrt{\text{n} \times \text{p} \times (1 - \text{p})}[/tex]

                                                              = [tex]\sqrt{\text{50} \times \text{0.2} \times (1 - \text{0.2})}[/tex]

                                                              = [tex]\sqrt{8}[/tex] = 2.83

Answer:

  see the attachment

Step-by-step explanation:

When you have a number of function evaluations to do, it is convenient to let a graphing calculator or spreadsheet do them. That avoids the tedium and the mistakes in arithmetic.

Here's your completed table.

Answer:

[tex]\large \boxed{\mathrm{34.2}}[/tex]

Step-by-step explanation:

[tex]\sf B= arcsin (\frac{b \times sin(A)}{a} )[/tex]

[tex]\sf B= arcsin (\frac{7 \times sin(40\°)}{8} )[/tex]

[tex]\sf B = 0.59733 \ rad = 34.225\°[/tex]

Answer:

40

Step-by-step explanation:

In the question only lies the answer:

"and a total of forty students plan to participate in cross country."

Answer:

2

Step-by-step explanation:

2

Answer:

Step-by-step explanation:

Lets, turn this into words and use order of operations, First, we look for multiplication and division.

the sum of one fourth of 5 times of 8 and 10 gets you 1/4(5*8) + 10 = 20

what is the number if 4 is subtracted from the sum

20 - 4 = 16

Answer:

154.2

Step-by-step explanation:

143 plus

156 plus

172 plus

133 plus

167 = 771

divide by 5 equals 154.2

Answer:

7

Step-by-step explanation:

10+4 = 14

14/2  = 7

x = 7

Answer:

x = -5/2

Step-by-step explanation:

1+2x=6x+11

Subtract 2x from each side

1+2x-2x=6x-2x+11

1 = 4x+11

Subtract 11 from each side

1-11 = 4x

-10 =4x

Divide by 4

-10/4 = 4x/4

-5/2 =x

Answer:

[tex]\boxed{x=-\frac{5}{2}}\\[/tex]

Step-by-step explanation:

To begin, get the variable on one side of the equation - preferably the left for standard solution notation (for this equation, it is easier to place it on the right side to avoid negative values). Do this by subtracting 2x from both sides of the equation. Then, subtract 11. Finally, divide by 4 and get the answer in terms of x.

1 + 2x = 6x + 11

1 = 4x + 11

-10 = 4x

[tex]\boxed{x=-\frac{5}{2}}[/tex]

Answer:

51

Step-by-step explanation:

Answer:

D. the coefficient of determination is .81.

Step-by-step explanation:

SST = SSE + SSR

where

SST is the summation of square total

SSE is the summation of squared error estimate = 608

SSR is the summation of square of residual = 2593

with these in mind we put the values into the formula

= 2592 + 608

=3200

Coefficient of determination = SSR/SST

= 2592/3200

= 0.81

Therefore option D is the correct answer to the question.

Answer:

Average rate of change of the function will be = (-1.5)

Step-by-step explanation:

Average rate of change of a function f(x) is determined by the formula,

Average rate of change = [tex]\frac{f(b)-f(a)}{b-a}[/tex] If a < x < b

We have to find the average rate of change of a function g(t) between the interval [-3, 1]

From the given table,

For t = -3,

g(-3) = 6

For t = 1,

g(1) = 0

Therefore, average rate of change of the function in the given interval

= [tex]\frac{g(1)-g(-3)}{1-(-3)}[/tex]

= [tex]\frac{0-6}{1+3}[/tex]

= [tex]-\frac{3}{2}[/tex]

= - 1.5

Answer: The point estimate of the population proportion is . 485.​

The margin of error is 0.273.​

Step-by-step explanation:

Confidence interval for population proportion(p):

sample proportion ± Margin of error

Given:  Lower bound of confidence interval  = 0.212

Upper bound = 0.758

⇒sample proportion - Margin of error=0.212  (i)

sample proportion + Margin of error= 0.758  (ii)

Adding (i) and (ii) , we get

2(sample proportion) =0.970

⇒ sample proportion = 0.970÷2= 0.485

Since sample proportion is the point estimate of the population proportion.

So, the point estimate of the population proportion=  0.485

Now put sample proportion =0.485 in (ii), we get

0.485+ Margin of error= 0.758

⇒  Margin of error= 0.758 - 0.485 =0.273

i.e. The margin of error is 0.273.​

*a clearer picture containing the graph is shown in the attachment

Answer:

20% of the class earned a D

Step-by-step Explanation:

Step 1: Determine the total number of students represented on the graph:

9 students => D

5 students => C

14 students => B

17 students => A

Total number of students = 45

Step 2: Express each category of students who scored a particular grade as a fraction and as percentage.

9 students => D => [tex] \frac{9}{45} = \frac{1}{5} [/tex] => as percentage, we have [tex] \frac{1}{5} * 100 = 20 percent [/tex]

5 students => C => [tex] \frac{5}{45} = \frac{1}{9} [/tex] => as percentage, we have [tex] \frac{1}{9} * 100 = 11.1 percent [/tex]

14 students => B => [tex] \frac{14}{45} [/tex] => as percentage, we have [tex] \frac{14}{45} * 100 = 31.1 percent [/tex]

17 students => A => [tex] \frac{17}{45} [/tex] => as percentage, we have [tex] \frac{17}{45} * 100 = 37.8 percent [/tex]

Step 3: Check each statement to see if they are true or not based on the calculations above.

Statement 1: "⅕ of the students earned a C."

This is NOT TRUE From our calculation, ⅑ of the students earned a C.

Statement 2: "3% more students earned an A than a B." This is also NOT TRUE.

37.8% earned A, while 31.1% earned a B. Thus, about 6.7% more students earned an A than a B.

Statement 3: "20% of the class earned a D".  This is TRUE.

Check calculation in step 2.

Statement 4: "¼ of the class earned a B". This is NOT TRUE.

¼ is 25% of the class. Those who earned a B account for 31.1% not 25% (¼ of the class).

The correct statement is: "20% of the class earned a D"

Answer:

Following are the correct code to this question:

import re#import package for regular expression

def check_time(text):#defining a method check_time that accepts string value  

   p = r'(1[012]|[1-9]):[0-5][0-9][ ]{0,1}?(am|pm|AM|PM)'#defining string variable p that stores values

   val = re.search(p, text)#defining val variable that check serachs p and text variable values

   return val!= None#use return keyword to return val value

print(check_time("12:45pm"))#defining print method that calls method by input value  

print(check_time("9:59 AM")) #defining print method that calls method by input value

print(check_time("6:60 am")) #defining print method that calls method by input value

print(check_time("five o'clock"))#defining print method that calls method by input value

Output:

True

True

False

False

Step-by-step explanation:

In the above-given program, some data is missing that is code file so, the correct code can be defined as follows:

In the above-given method, that is "check time" it uses 12-hour time format validation, that is tested by coding the regex and  all the value validates in the "val" variables, that can be defined as  follows:  

Answer:

The number of ways is  [tex]\left n} \atop {}} \right. P_r = 336[/tex]

Step-by-step explanation:

From the question we are told that

   The number of tickets are   [tex]n = 8[/tex]

    The number of finalist are [tex]r =3[/tex]

Generally the number of way by which this winners can be drawn and arrange in the order of   [tex]1^{st} , \ 2nd , \ 3rd[/tex]    is mathematically represented as

             [tex]\left n} \atop {}} \right. P_r = \frac{n\ !}{(n-r) !}[/tex]

substituting values

             [tex]\left n} \atop {}} \right. P_r = \frac{ 8!}{(8-3) !}[/tex]

           [tex]\left n} \atop {}} \right. P_r = \frac{ 8* 7*6*5*4*3*2*1}{ 5*4*3*2*1}[/tex]

           [tex]\left n} \atop {}} \right. P_r = 336[/tex]

if they're in a bundle of 10 then theres 10 pencils

Answer:

Student t-distribution.

Step-by-step explanation:

In this scenario, a test is being conducted to test the difference between two population "means" using data that are gathered from a matched pairs experiment. If the paired differences are normal, then the distribution used for testing is the student t-distribution.

In Statistics and probability, a student t-distribution can be defined as the probability distribution which can be used to estimate population parameters when the population variance is not known (unknown) and the sample population is relatively small. The student t-distribution is a statistical distribution which was published in 1908 by William Sealy Gosset.

A student t-distribution has a similar curve with the normal distribution curve, except that it is fatter and a little bit shorter.

Hi there! :)

Answer:

[tex]P(heads) = \frac{1}{8}[/tex]

Step-by-step explanation:

Probability of a coin landing on heads:

[tex]P(heads) = \frac{1}{2}[/tex]

Find the probability of getting heads 3 times:

[tex]\frac{1}{2} * \frac{1}{2} * \frac{1}{2} = \frac{1}{8}[/tex]

Therefore, the probability of the coin showing heads for 3 tosses is:

[tex]P(heads) = \frac{1}{8}[/tex]

Answer:

The probability is  [tex]P(X < 12) = 0.99286[/tex]

Step-by-step explanation:

From the question we are told that

        The population mean is [tex]\mu = 12 \ oz[/tex]

         The  standard deviation is  [tex]\sigma = 0.1 \ oz[/tex]

          The sample mean is  [tex]\= x = 12.1 \ oz[/tex]

          The sample size is  n = 6 packs

The standard error of the mean is mathematically represented as

              [tex]\sigma_{\= x } = \frac{\sigma}{\sqrt{n} }[/tex]

substituting values

            [tex]\sigma_{\= x } = \frac{0.1}{\sqrt{6} }[/tex]

            [tex]\sigma_{\= x } = 0.0408[/tex]

Given that the can’s contents follow a Normal distribution then then  the probability that the mean contents of a six-pack are less than 12 oz is mathematically represented as

         [tex]P(X < 12) = P ( \frac{X - \mu }{ \sigma_{\= x }} < \frac{\= x - \mu }{ \sigma_{\= x }} )[/tex]

Generally  [tex]\frac{X - \mu }{ \sigma_{ \= x }} = Z (The \ standardized \ value \ of \ X )[/tex]

So

         [tex]P(X < 12) = P ( Z < \frac{\= x - \mu }{ \sigma_{\= x }} )[/tex]

substituting values

       [tex]P(X < 12) = P ( Z < \frac{12.2 -12 }{0.0408} )[/tex]

      [tex]P(X < 12) = P ( Z < 2.45 )[/tex]

From the normal distribution table the value of [tex]P ( Z < 2.45 )[/tex] is  

           [tex]P (Z < 2.45)0.99286[/tex]

=>   [tex]P(X < 12) = 0.99286[/tex]

Answer:

  linear with a common first difference of 2

Step-by-step explanation:

On the face of it, you can reject answers that ascribe a common ratio to a linear or quadratic function. (A common ratio is characteristic of an exponential function.)

You can also reject the answer that ascribes a common first difference to a quadratic function. (A quadratic function has a common second difference.)

After you reject the nonsense answers, there is only one remaining choice. It is also the correct one:

  linear with a common first difference of 2

_____

The ratio of change in y to change in x is ...

  (0 -(-2))/(-2 -(-3)) = 2

  (4 -0)/(0 -(-2)) = 2

  (12 -4)/(4 -0) = 2

That is, y increases by 2 when x increases by 1. The common first difference is 2.

Answer:

x = 6

Step-by-step explanation:

In the third line of the solution on right side of the equal sign, middle term should be 8x instead of 4x.

The final value of x will be 6.

[tex] PQ^2 + QO^2 = PO^2 \\

x^2 + 8^2 = (4+x)^2 \\

x^2 + 64 = 16 + 8x + x^2 \\

64 = 16 + 8x \\

64 - 16 = 8x \\

48 = 8x \\

6 = x\\[/tex]

Answer:

the log to the base 3 of 14 is 2.402

Step-by-step explanation:

You must find a way to indicate that 3 is the base; you cannot run this '3' together with 2, 7 or 14.

Example:  

log to the base 3 of 2 = 0.631

log to the base 3 of 7 = 1.771

Note that 2 times 7 is 14.  Thus, to obtain the log to the base 3 of 14, we must ADD the two logs shown above:

0.631

+1.771

----------

2.402

Thus, the log to the base 3 of 14 is 2.402.

Check:  Does 3^2.402 = 14?  YES

Answer:

Solution : 5.244 - 16.140i

Step-by-step explanation:

If we want to express the two as a product, we would have the following expression.

[tex]-6\left[\cos \left(\frac{2\pi }{5}\right)+i\sin \left(\frac{2\pi }{5}\right)\right]\cdot 2\sqrt{2}\left[\cos \left(\frac{-\pi }{2}\right)+i\sin \left(\frac{-\pi \:}{2}\right)\right][/tex]

Now we have two trivial identities that we can apply here,

( 1 ) cos(- π / 2) = 0,

( 2 ) sin(- π / 2) = - 1

Substituting them,

= [tex]-6\cdot \:2\sqrt{2}\left(0-i\right)\left(\cos \left(\frac{2\pi }{5}\right)+i\sin \left(\frac{2\pi }{5}\right)\right)[/tex]

= [tex]-12\sqrt{2}\sin \left(\frac{2\pi }{5}\right)+12\sqrt{2}\cos \left(\frac{2\pi }{5}\right)i[/tex]

Again we have another two identities we can apply,

( 1 ) sin(x) = cos(π / 2 - x )

( 2 ) cos(x) = sin(π / 2 - x )

[tex]\sin \left(\frac{2\pi }{5}\right)=\cos \left(\frac{\pi }{2}-\frac{2\pi }{5}\right) = \frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}[/tex]

[tex]\cos \left(\frac{2\pi }{5}\right)=\sin \left(\frac{\pi }{2}-\frac{2\pi }{5}\right) = \frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}[/tex]

Substitute,

[tex]-12\sqrt{2}(\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}) + 12\sqrt{2}(\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4})[/tex]

= [tex]-6\sqrt{5+\sqrt{5}}+6\sqrt{3-\sqrt{5}} i[/tex]

= [tex]-16.13996 + 5.24419i[/tex]

= [tex]5.24419i - 16.13996[/tex]

As you can see option d is the correct answer. 5.24419 is rounded to 5.244, and 16.13996 is rounded to 16.14.

Answer:

38.911≤p≤41.089

Step-by-step explanation:

The formula for calculating confidence interval for a population mean us as shown below;

CI = xbar ± Z×S/√N where;

xbar is the sample mean = 40

Z is the z score at 95% confidence interval = 1.96

S is the standard deviation = 5

N is the sample size = 81

Substituting this parameters in the formula we have;

CI = 40±1.96×5/√81

CI = 40±(1.96×5/9)

CI = 40±(1.96×0.556)

CI = 40±1.089

CI = (40-1.089, 40+1.089)

CI = (38.911, 41.089)

The 95% confidence interval for the population mean is 38.911≤p≤41.089

Answer:

38.9 ≤ U ≤ 41.1

Step-by-step explanation:

Mean, m = 40; standard deviation, α = 5; Confidence limit, U = 95% or 0.95

N = 81

The standard error, α(m) = α/√(N) = 5/√81 =5/9

Using table: 0.95 = 0.0379

Z(0.95) = 2 - 0.0379 = 1.9621 or 1.96

Hence, confidence interval = { m - 1.96(α/√N) ≤ U ≤ m +1.96(α/√N)}

But, 1.96(α/√N) = 1.96 X 5/9 = 1.96 X 0.56 = 1.1

(40 - 1.1 ≤ U ≤ 40 + 1.1)

∴ the confidence interval = 38.9 ≤ U ≤ 41.1

Answer:

Nine thousandths = 0.009

Step-by-step explanation:

thousandths =  1/1000 = 0.001

nine thousandths = 9/1000 = 0.009

Answer:

.009

Step-by-step explanation:

9 thousandths as a decimal is 9/1000.  Which is the same 0.009

Answer:

One - way ANOVA, the smaller the between treatment

The smaller the value of F statistic will give us significant evidence.

Step-by-step explanation:

ANOVA is a statistical technique designed to test mean of one or more quantitative populations.  In two-way ANOVA it equals the block mean. Column block means square is three-way ANOVA. It is a statistical technique designed to test mean of one or more quantitative populations. In two-way ANOVA it equals the block mean. Column block means square is three-way ANOVA.

One-way ANOVA, the smaller the between treatment

The smaller the value of F statistic will give us significant evidence.

It should be the statistical technique that are made for testing the mean for one or more than one quantitative population. In two-way ANOVA it should be equivalent to the block mean. Here the column block represent the square be the three-way ANOVA.

Therefore, One-way ANOVA, the smaller the between treatment

The smaller the value of F statistic will give us significant evidence.

Learn more about evidence here: https://brainly.com/question/6764645?referrer=searchResults