In an area of the Midwest, records were kept on the relationship between the rainfall (in inches) and the yield of wheat (bushels per acre). Which is the best predicted value for y given x = 7.3? Assume that the variables x and y have a significant correlation. 36.5 36.0 36.7 36.2

Answer:

76,050 ft²

Step-by-step explanation:

If the area must be rectangular, let L be the length of the side opposite to the creek, and S be the length of the remaining two sides.

The perimeter of the fencing and the area of the pasture are:

[tex]780 = L+2S\\A= LS\\\\L=780-2S\\A=-2S^2+780S[/tex]

The value of S for which the derivate of the area function is zero is the length of S that maximizes the area of pasture:

[tex]\frac{dA}{dS}=0=-4S+780\\S= 195\\L=780-(2*195)=390[/tex]

The maximum possible area is:

[tex]A_{MAX}=390*195=76,050\ ft^2[/tex]

Answer:

The answer to your question is z ≥ -159

Step-by-step explanation:

Process

1.- Write the inequality

                                   130 ≥ - 29 - z

2.- Add 29 to both sides of the inequality

                              130 + 29 ≥ - 29 + 29 - z

3.- Simplify

                                    159 ≥   - z

4.- Divide by -1 and change the inequality

                                    159/-1 ≤ -z/-1

5.- Simplify

                                   -159 ≤ z

Answer:

The complete question is stated below:

An obstetrician knew that there were more live births during the week than on weekends. She wanted to determine whether the mean number of births was the same for each of the five days of the week. She randomly selected eight dates for each of the five days of the week and obtained the following data:

Monday:      Tuesday:    Wednesday:    Thursday:        Friday:

10,456            11,621         11,084             11,171                 11,545

10,023            11,944         11,570            11,745                12,321

10,691             11,045         11,346            12,023               11,749

10,283             12,927        11,875            12,433               12,192

10,265             12,577         12,193           12,132                12,422

11,189                11,753         11,593           11,903                11,627

11,198                12,509        11,216            11,233                11,624

11,465                13,521         11,818            12,543              12,543

a. Write the null and alternative hypotheses.

b. State the requirements that must be satisfied to use the one-way ANOVA procedure.

c On which day or dates are there more births?

Answer:

a. Null Hypothesis: There is no statistically significant difference between the mean number of babies born alive on Mondays, Tuesdays, Wednesdays, Thursdays and Fridays.

Alternative Hypothesis: The mean number of babies born alive on each weekday from Monday to Friday are not the same.

b) 1. The dependent variable should be measured at ratio or interval level

2. The independent variable should contain at least two groups that are categorical and independent

3. There must independence of observations between and within the various groups

4. No significant outliers should exist

5. for each group of the independent variable, the dependent variable should be approximately normally distributed.

6. the variances should be homogeneous

c. There are more births on Tuesdays and Fridays

Step-by-step explanation:

a. A Null Hypothesis is one that states that no statistical significance exist between the two variables in the hypothesis. It is the null hypothesis that the researcher tries to disprove. While the alternative hypothesis is simply the opposite of the null hypothesis, it hypothesizes that there is indeed a statistically significant relationship between the variables in question.

b.  1. The dependent variable should be measured at ratio or interval level; among the various levels of measurements such as ordinal, nominal, ratio or interval, the dependent variable must be either on the interval or ratio level.

2. The independent variable should contain at least two groups that are categorical and independent; the independent variable, in this case, the number of live births, should be two groups or more, and they should be categorical.

3. There must independence of observations between and within the various groups; the values observed within each group should be independent of the other. For example, no one pregnant woman should be involved in more than one group.

4. No significant outliers should exist; outliers are single data points within the measured variable that do not follow the usual point pattern of the other values, either values that are too high or too low. they reduce the accuracy of the one-way ANOVA.

5. for each group of the independent variable, the dependent variable should be approximately normally distributed; violations of normality causes the result to be invalid. It can only hold little variations to produce valid results.

6. the variances should be homogeneous; the variances (distribution or spread around the mean) of the two or more test groups must be considered equal.

c. Tuesday with a mean livebirth of 12,237 births and Friday with an average livebirth of 12,002 births have the most number of births.

Answer:

1)

[tex]H_0: p_1-p_2=0\\\\H_a: p_1-p_2>0[/tex]

2) z=3.164

3) Critic value z₀=1.282.

4) P=0.00078

5) The correct answer is: "Since the p-value is less than the given value of alpha, there is sufficient evidence to reject H_0"

Step-by-step explanation:

5.1) Being:

p₁: proportion of men who keep track of the deadlines in their head

p₂: proportion of women who keep track of the deadlines in their head

If we want to test if p₁ is larger than p₂, the null hypothesis and the alternative hypothesis should be:

[tex]H_0: p_1-p_2=0\\\\H_a: p_1-p_2>0[/tex]

In this way, if we reject the null hypothesis, it can be claimed that p₁ is larger than p₂.

5.2) Compute the test statistic for the test.

First, we have to estimate a proportion as if the null hypothesis is true. This means the average of proportion of the samples taken from men and women, weighted by the sample size.

[tex]\bar p=\frac{n_1p_1+n_2p_2}{n_1+n_2}=\frac{500*0.54+500*0.44}{500+500}= 0.49[/tex]

Then, we used this average to estimate the standard error

[tex]s=\sqrt{\frac{p(1-p)}{n_1}+{\frac{p(1-p)}{n_2}}}=\sqrt{\frac{0.49(1-0.49)}{500}+{\frac{0.49(1-0.49)}{500}}}=\sqrt{0.0004998+0.0004998}\\\\s= 0.0316[/tex]

Lastly, we calculate the statistic z

[tex]z=\frac{p_1-p_2}{s}=\frac{0.54-0.44}{0.0316}=\frac{0.10}{0.0316}=3.164[/tex]

5.3) Give the rejection region for the test, using α = 0.10

For a one-tailed test with α = 0.10, the z value to limit the rejection region is z=1.282.

For every statistic larger than 1.282, the null hypothesis should be rejected.

5.4) Find the p-value for the test.

The p-value for a z=3.164 is P=0.00078 (corresponding to the area ot the standard normal distribution for a z larger than 3.164).

5.5) Choose the correct answer below.

The correct answer is: "Since the p-value is less than the given value of alpha, there is sufficient evidence to reject H_0"

The difference between the proportions is big enough to be statistically significant and enough evidence to reject the null hypothesis.

Answer:

The probability that the coins are thrown more than three times to show the same face is 0.3164.

Step-by-step explanation:

The problem is related to Geometric distribution.

The Geometric distribution defines the probability distribution of X failures before the first success.

The probability distribution function is:

[tex]P(X=k)=(1-p)^{k}p;\ k = 0, 1, 2, ...[/tex]

First compute the probability that in the [tex]i^{th}[/tex] throw all the three coins will show the same face.

P (All the 3 coins shows the same face) = P (All the three coins shows Heads) + P (All the three coins shows Tails)

                      [tex]=(\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2} )+(\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2} )\\=\frac{1}{8}+\frac{1}{8}\\ =\frac{2}{8} \\=\frac{1}{4}[/tex]

Now compute the probability that it takes more than 3 throws for the coins to show the same face.

P (X > 3) = 1 - P (X ≤ 3)

[tex]=1-[P(X=1)+P(X=2)+P(X=3)]\\=1-[[(1-\frac{1}{4} )^{0}\times\frac{1}{4}]+[(1-\frac{1}{4} )^{1}\times\frac{1}{4}]+ [(1-\frac{1}{4} )^{2}\times\frac{1}{4}]+[(1-\frac{1}{4} )^{3}\times\frac{1}{4}]]\\=1-[0.2500+0.1875+0.1406+0.1055]\\=1-0.6836\\=0.3164[/tex]

Thus, the probability that it takes more than 3 throws for the coins to show the same face is 0.3164.

The probability that it takes more than 3 throws for the coins to show the same face is 0.3164.

Probability means possibility. It deals with the occurrence of a random event. The value of probability can only be from 0 to 1. Its basic meaning is something is likely to happen. It is the ratio of the favorable event to the total number of events.

Given

Three identical fair coins are thrown simultaneously until all three show the same face.

Then the total event will be

[tex]\rm Total \ event = 2^3 = 8[/tex]

All the possibilities.

[tex]\rm (HHH),( HHT ),(HTH),( THH),(HTT),(THT),(TTH),(TTT)[/tex]

For getting the same face, then the probability will be

[tex]\rm P(same\ face) = \dfrac{2}{8} = \dfrac{1}{4}[/tex]

Now compute the probability that it takes more than 3 throws for the coins to show the same face.

[tex]\rm P (X > 3) = 1 - P (X ≤ 3)\\\\P (X > 3) = 1 -[P(x=1) +P(x=2) + P(x=3)]\\\\P (X > 3) =1 -\{[(1-\dfrac{1}{4})^0 * \dfrac{1}{4}] +[(1-\dfrac{1}{4})^1 * \dfrac{1}{4}] +[(1-\dfrac{1}{4})^2 * \dfrac{1}{4}] +[(1-\dfrac{1}{4})^3 * \dfrac{1}{4}] \}\\\\P (X > 3) =1 - 0.6836\\\\P (X > 3) =0.3164[/tex]

Thus, the probability that it takes more than 3 throws for the coins to show the same face is 0.3164.

More about the probability link is given below.

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$ 320

41 miles ............ 1 gallon

4700 miles .......x gallon

x = 4700×1/41 = 114.63 gallons

114.63 gallons×$2.79 ≈  $ 320

To find the cost of gasoline for a 4,700-mile automobile trip, divide the total distance by the car's miles per gallon, then multiply by the price of gas per gallon, finally we get a total cost of $318.65.

To find the cost of gasoline for a 4,700-mile automobile trip, we need to divide the total distance by the car's miles per gallon to get the number of gallons of gas needed.

Then, we multiply the number of gallons by the price of gas per gallon to get the total cost.

In this case, the car gets 41 miles per gallon, so the number of gallons needed is 4700 miles / 41 miles per gallon = 114.63 gallons.

Multiplying this by the average price of gas, $2.79 per gallon, we get a total cost of $318.65.

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Answer:

Try Photomath !

Step-by-step explanation:

Answer: Tara earned $775 from her summer job.

Step-by-step explanation:

Let x represent the amount of money that Tara earned from her summer job.

Lewis earned $1800 from his summer job at the grocery store. This is $250 more than twice what his friend Tara earned. The equation would be

2x + 250 = 1800

Subtracting 250 from the left hand side and the right hand side of the equation, it becomes

2x + 250 - 250 = 1800 - 250

2x = 1550

Dividing the left hand side and the right hand side of the equation by 2, it becomes

2x/2 = 1550/2

x = 775

Answer:

The probability that a person has HIV given that the test negative is 0.0033.

Step-by-step explanation:

Denote the events as follows:

X = a person in Uganda had HIV

Y = a person in Uganda was tested positive for HIV.

The information provided is:

[tex]P(Y^{c}\cap X)=\frac{4}{1517}\\[/tex]

[tex]P(Y\cap X)=\frac{166}{1517}\\[/tex]

[tex]P(Y\cap X^{c})=\frac{129}{1517}\\[/tex]

[tex]P(Y^{c}\cap X^{c})=\frac{1218}{1517}\\[/tex]

The probability that a person has HIV given that he/she was tested negative is:

[tex]P(X|Y^{c})=\frac{P(Y^{c}\cap X)}{P(Y^{c})}[/tex]

Compute the probability of a person not having HIV as follows:

[tex]P(Y^{c})=P(Y^{c}\cap X)+P(Y^{c}\cap X^{c})=\frac{4}{1517}+\frac{1218}{1517} =\frac{4+1218}{1517} =\frac{1222}{1517}[/tex]

Compute the value of [tex]P(X|Y^{c})[/tex] as follows:

[tex]P(X|Y^{c})=\frac{P(Y^{c}\cap X)}{P(Y^{c})}=\frac{4}{1517}\times\frac{1517}{1222}=0.0033[/tex]

Thus, the probability that a person has HIV given that he/she was tested negative is 0.0033.

Final answer:

The probability that a randomly selected person from this population who tests negative actually has HIV is approximately 0.327%. This accounts for the possibility of false negatives. However, confirmatory testing is necessary to accurately diagnose HIV.

Explanation:

To calculate the probability that a randomly selected person from the population who tested negative for HIV actually has the virus, we need to focus on the false negatives and true negatives provided in the figures from Uganda.

Out of 1517 tested individuals, there were 4 false negatives and 1218 true negatives.

Thus, for someone who tests negative, to find out the probability that they actually have HIV (false negative), we use the formula: Probability = Number of false negatives / (Number of false negatives + Number of true negatives).

In this case, we have Probability = 4 / (4 + 1218) = 4 / 1222. When we calculate this probability, we get approximately 0.00327, or 0.327%.

Answer:

70 combinations

Step-by-step explanation:

Given:

- The total number of designated location n = 8

- The number of location the ointment is to be applied r = 4

Find:

How many different choices were there for the four locations to apply the new ointment?

Solution:

- For this question we will apply combinations to the given problem. We have total of 8 available places out of which we have to "choose" 4 locations to apply the new ointment. We will use combinations " C " operator.

- Out of 8 "choose" 4 = 8_C _4 = 70 possible combinations.

- Since there is requirement of "order" so its only limited to the selection process and not an arrangement.

Answer:

[tex] P(D|P)= \frac{0.11}{0.59}=0.186[/tex]

Step-by-step explanation:

For this case we have defined the following events:

D= "An international flight leaving the United States is delayed in departing"

P="An international flight leaving the United States is a transpacific flight "

And we have defined the probabilities:

[tex] P(D)= 0.29 , P(P) = 0.59[/tex]

And for the event: "an international flight leaving the U.S. is a transpacific flight and is delayed in departing" [tex] D \cap P [/tex] we know the probability:

[tex] P(D \cap P) =0.11[/tex]

We want to find this probability:

What is the probability that an international flight leaving the United States is delayed in departing given that the flight is a transpacific flight

So we want this probability:

[tex] P(D|P)[/tex]

And we can use the conditional formula from the Bayes theorem given two events A and B:

[tex] P(A|B) = \frac{P(A \cap B)}{P(B)}[/tex]

And if we use this formula for our case we have:

[tex] P(D|P)= \frac{P(D \cap P)}{P(P)}[/tex]

And if we replace the values we got:

[tex] P(D|P)= \frac{0.11}{0.59}=0.186[/tex]

Answer:

u=6 1/21

make 6 5/7 a mixed number = 47/7

subtract 2/3 from 47/7

If my ans is helpful u can follow me.

The given series is an alternating harmonic series. It is convergent because the sequence of absolute values decreases to zero.

The given series is: 1/2 , 3/4 , 1/8 , 3/16 , 1/32 , 3/64 . . . As you can see, the numerator alternates between 1 and 3, while the denominator doubles each time. This type of series is known as an alternating harmonic series.

To determine if an alternating series converges or diverges, we can apply the Alternating Series Test. This test says that an alternating series converges if the sequence of absolute values decreases to zero. In our case, the absolute value of the terms based solely on the denominator is decreasing to zero, hence, the series converges.

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Answer:

a) m = 0.78

b) Average annual earnings by Males

c)  F = $ 30,548.685

Step-by-step explanation:

Given:

- The relationship between the average annual earning of Females F is expressed as a function of earnings by males M as:

                                          F = 0.78*M - 1.315

Find:

(a) Viewing F as a function of M, what is the slope of the graph of this function?

(b) what is M?

(c) When the average annual earnings for males reach $65,000, what does the equation predict for the average annual earnings for females?

Solution:

a)

- The given expression can be compared with a linear equation of the form:

                                            y = m*x + c

Where,

y is the dependent variable

m is the slope of the graph

x is the independent variable

c is the intercept when x = 0

- So for our case , y = F , m = 0.78 , x = M , c = -1.315

b)

- The variable M is the independent variable with the domain [ 0 , infinity ] denotes the average annual earnings of Males in dollars ($).

c)

- When average annual earnings by males is $ 65,000 i.e M = 65,000, We compute F by the given expression:

                                          F = 0.47 * 65,000 - 1.315

                                          F = $ 30,548.685

- The average annual earning by females is $ 30,549 when earning by males reach $65,000.

Answer:

20

Step-by-step explanation:

Answer:

It does not appear ​that, as a​ group, the students are reasonably good at estimating one minute.

Step-by-step explanation:

We are given the following data in the question:

75, 88, 51, 73, 49, 31, 69, 74, 72, 59, 72, 81, 99, 101, 73

Formula:

[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]  

where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.  

[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]

[tex]Mean =\displaystyle\frac{1067}{15} = 71.13[/tex]

Sum of squares of differences = 4739.733

[tex]S.D = \sqrt{\frac{4739.733}{14}} = 18.39[/tex]

Population mean, μ = 60 minutes

Sample mean, [tex]\bar{x}[/tex] = 71.13 minutes

Sample size, n = 15

Alpha, α = 0.10

Sample standard deviation, s = 18.39 minutes

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 60\text{ minutes}\\H_A: \mu \neq 60\text{ minutes}[/tex]

We use Two-tailed t test to perform this hypothesis.

Formula:

[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{71.13 - 60}{\frac{18.39}{\sqrt{15}} } = 2.34[/tex]

Calculating the p-value from the table, we have,

P-value = 0.034354

Since the p-value is lower than the significance level, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

Thus, we conclude that it does not appear ​that, as a​ group, the students are reasonably good at estimating one minute.

Option A is the correct set of hypotheses for testing if students can estimate one minute. The test statistic and p-value are needed to make a decision, typically using a t-test. The conclusion depends on the p-value relative to the significance level.

The null and alternative hypotheses you are working with for testing whether students are good at estimating one minute are:

To determine if students are good at estimating one minute, we have option A as correct, which is H0: \\(\mu = 60\\) and Ha: \\(\mu \not= 60\\).

The test statistic and p-value should be calculated using appropriate statistical methods (typically a t-test assuming we do not know the population standard deviation, which would require the mean, standard deviation, and sample size). After calculating the test statistic and p-value, you will compare the p-value to the significance level (alpha). If the p-value is less than alpha, you reject H0; otherwise, you fail to reject H0.

Based on the decision to reject or fail to reject H0, you can make a conclusion regarding the original claim about whether students are reasonably good at estimating one minute.

Answer: a) {x | x ∈ N , 1 ≤x≤7} b ) {10 x  | x ∈ Z ,  0≤x≤4 }

Step-by-step explanation:

A set builder form is the the mathematical way to represent aset by using its particular property of characteristic.

For example :  A = {2,4,5,6,......} is represented as

A= {2x | x ∈ N} , where N is a set of natural numbers.

a)  {1, 2, 3, 4, 5, 6, 7}

Let D = {1, 2, 3, 4, 5, 6, 7}

Then, in set builder form , we have

D = {x | x ∈ N , 1 ≤x≤7}

b) ) {1, 10, 100, 1000, 10000}

Let B = {1, 10, 100, 1000, 10000}

[tex]={10^{0} , 10^1 , 10^2 , 10^3 , 10^4}[/tex]

We can see that each element in set B is a multiple of 10.

So in set builder form , we will have

B = {10 x  | x ∈ Z ,  0≤x≤4 } , where Z is the set of integers.

Set-builder notation is used to describe sets using mathematical notation. For the given sets {1, 2, 3, 4, 5, 6, 7} and {1, 10, 100, 1000, 10000}, we can describe them using set-builder notation.

To describe the set {1, 2, 3, 4, 5, 6, 7} using set-builder notation, we can write it as:

{x | x is an element of the set of natural numbers and 1 ≤ x ≤ 7}

Similarly, to describe the set {1, 10, 100, 1000, 10000}, we can write it as:

{x | x is an element of the set of natural numbers and x is a power of 10}

Answer:

(c) Skewed to the left

Step-by-step explanation:

To describe the distribution of the data determine the mean, median and mode.

The provided data arranged in ascending order is:

{69, 92, 97, 98, 100, 104, 111, 114, 121, 135, 135, 136, 138, 189}

        [tex]Mean=\frac{Sum\ of\ observations}{Number\ of\ observations}\\ =\frac{69+92+97+ 98+ 100+ 104+ 111+ 114+ 121+ 135+ 135+ 136+ 138+ 189}{14} \\=117.07[/tex]

         [tex]Median=Mean (7^{th}, 8^{th}\ observation)\\=\frac{7^{th}\ obs.+8^{th}\ obs.}{2}\\ =\frac{111+114}{2}\\ =112.5[/tex]

       The value 135 has the highest frequency of 2.

        [tex]Mode=135[/tex]

So Mean < Mode and Median < Mode.

For a distribution that is skewed to the left the mean and median is less than the mode of the data.

Thus, the data is left-skewed.

Final answer:

The distribution of the 25-inch TV prices in Roseville, based on the provided prices, appears to be skewed to the right due to a long tail of higher values, with most other prices being lower and closer together.

Explanation:

To describe the distribution of TV prices from different stores in Roseville, it is necessary to organize the data into a histogram or look for indicators of its shape such as measures of central tendency (mean, median, mode) and measures of spread (range, interquartile range, standard deviation). In the dataset provided: 100, 98, 121, 111, 97, 135, 136, 104, 135, 138, 189, 114, 92, 69, the distribution appears to have a long tail to the right because there is a significant jump to the higher price of 189, while most other values are closer together on the lower end. This suggests that the distribution of TV prices is skewed to the right.

Answer:

242,784,360 hands.

Step-by-step explanation:

The number of hands that contain at least one hearts card is given by the total possible number of hands subtracted by the number of hands with no hearts card.

The total possible number of hands is:

[tex]N_T=\frac{52!}{(52-5)!}=\frac{52!}{(47)!} =52*51*50*49*48\\N_T=311,875,200[/tex]

Since there are 13 hearts cards in a deck, the number of possible hands with no hearts card is:

[tex]N_{H=0} =\frac{52-13!}{(52-13-5)!}=\frac{39!}{(34)!} =39*38*37*36*35\\N_{H=0}=69,090,840[/tex]

The number of hands with at least one hearts card is:

[tex]N_{H>0} = N_T-N{H=0}\\N_{H>0}=311,875,200-69,090,840\\N_{H>0}=242,784,360[/tex]

242,784,360 hands contain at least one hearts card.

Answer:

a) The the approximate number of babies between 2363 and 3234 are:

n = 0.683*621= 423.95 and that's approximately 424 babies

b) The approximate number of babies between 1492 and 4976 are:

n = 0.955*621= 592.74 and that's approximately 593 babies

Step-by-step explanation:

Assuming this problem:"A study conducted by the Center for Population Economics at the University of Chicago studied the birth weights of 621 babies born in New York. The mean weight was 3234 grams with a standard deviation of 871 grams. Assume that birth weight data are approximately bell-shaped.

Estimate the number of newborns who weighed between 2363 grams and 4105 grams. Round to the nearest whole number.

The number of newborns who weighed between 1492 grams and 4976 grams is . "

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(3234,871)[/tex]  

Where [tex]\mu=3234[/tex] and [tex]\sigma=871[/tex]

We select a sample size of n = 621. And we want to find this probability:

[tex] P(2363 < X < 4105) [/tex]

[tex]P(2363<X<4105)=P(\frac{2363-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{4105-\mu}{\sigma})=P(\frac{2363-3234}{871}<Z<\frac{4105-3234}{871})=P(-1<z<1)[/tex]

And we can find this probability taking this difference:

[tex]P(-1<z<1)=P(z<1)-P(z<-1)=0.841-0.159=0.683 [/tex]

And the the approximate number of babies between 2363 and 3234 are:

n = 0.683*621= 423.95 and that's approximately 424 babies

Part b

[tex] P(1492 < X < 4976) [/tex]

[tex]P(1492<X<4976)=P(\frac{1492-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{4976-\mu}{\sigma})=P(\frac{1492-3234}{871}<Z<\frac{4976-3234}{871})=P(-1<z<1)[/tex]

And we can find this probability taking this difference:

[tex]P(-1<z<1)=P(z<1)-P(z<-1)=0.977-0.0228=0.955 [/tex]

And the the approximate number of babies between 1492 and 4976 are:

n = 0.955*621= 592.74 and that's approximately 593 babies

Answer:

-2

1

3125

Step-by-step explanation:

Answer:

1. 12/52 or 0.2308

2. 10/13 or 0.7692

3.  3/26 or 0.1154

4. 3/26 or 0.1154

5. 9/52 or 0.1731

Step-by-step explanation:

1.

The face card are king,queen and jack.

There are 12 face card in a deck of 52 cards i.e. 4 king, 4 queen and 4 jack.

P(FC)=12/52 or 0.2308

So, the probability that the card is a face card is 0.2308.

2.

The card is not a face card.

P(FC')=1-P(FC)=1-(12/52)=(52-12)/52=40/52=10/13

P(FC')=10/13 or 0.7692

So, the probability that the card is not a face card is 0.7692.

3.

The card is a red face card

There are 6 red face card in a deck of 52 cards i.e. 2 king, 2 queen and 2 jack.

P(RFC)=6/52

P(RFC)=3/26 or 0.1154

So, the probability that the card is a red face card is 0.1154.

4.

The card is a 6 or lower.

There are 6 cards in a deck of 52 cards that are 6 or lower i.e. 6,5,4,3,2 and ace.

P(6 or lower)=6/52

P(6 or lower)=3/26 or 0.1154

So, the probability that the card is a 6 or lower is 0.1154.

5.

The card is a 9 or lower.

There are 9 cards in a deck of 52 cards that are 9 or lower i.e. 9, 8, 7, 6, 5, 4, 3, 2 and ace.

P(9 or lower)=9/52

P(9 or lower)=9/52 or 0.1731

So, the probability that the card is a 9 or lower is 0.1731.

In a pack or deck of 52 playing cards, they are divided into 4 suits of 13 cards each i.e. spades , hearts , diamonds  and clubs.  King, Queen and Jack  are face cards. So, there are 12 face cards in the deck of 52 playing cards.

In a Deck of 52 playing cards,

Number of face card = 12

number of non face card = 40

Number of red face card = 6

Number of Ace cards = 4

So, Probability of the card is face card is , = 12/52 =3/13

Probability of the card is not face card ,= 40/52 =  10/13

Probability of the card is a red face card is,=6/52 =  3/26

The probability of  card is a 6 or lower is, = 24/52 =  6/13

The probability of  card is a 9 or lower is, = 36/52 =  9/13.

Learn more:

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Answer:

1) 0.375

2) 0.5

3) 0.875

4) 0.5                  

Step-by-step explanation:

We are given the following in the question:

Sample space, S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.

[tex]\text{Probability} = \displaystyle\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}[/tex]

1. The probability of getting exactly one tail

P(Exactly one tail)

Favorable outcomes ={HHT, HTH, THH}

[tex]\text{P(Exactly one tail)} = \dfrac{3}{8} = 0.375[/tex]

2. The probability of getting a head on the first toss

P(head on the first toss)

Favorable outcomes ={HHH, HHT, HTH, HTT}

[tex]\text{P(head on the first toss)} = \dfrac{4}{8} = \dfrac{1}{2} = 0.5[/tex]

3. The probability of getting at least one head

P(at least one head)

Favorable outcomes ={HHH, HHT, HTH, HTT, THH, THT, TTH}

[tex]\text{P(at least one head)} = \dfrac{7}{8} = 0.875[/tex]

4. The probability of getting at least two heads

P(Exactly one tail)

Favorable outcomes ={HHH, HHT, HTH,THH}

[tex]\text{P(Exactly one tail)} = \dfrac{4}{8} = \dfrac{1}{2} = 0.5[/tex]

Final answer:

The question involves calculating probabilities based on the outcomes of tossing a coin three times. The probabilities for getting exactly one tail, a head on the first toss, at least one head, and at least two heads are found to be 3/8, 1/2, 7/8, and 1/2 respectively, by counting the favorable outcomes within the complete sample space.

Explanation:

The question asks for the probability of various outcomes when tossing a coin three times, given the sample space S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}. Let's solve each part step-by-step.

Probability of getting exactly one tail: There are three outcomes (HTT, THT, TTH) out of eight that satisfy this condition, so the probability is 3/8.

Probability of getting a head on the first toss: Four outcomes (HHH, HHT, HTH, HTT) satisfy this, so the probability is 4/8 or 1/2.

Probability of getting at least one head: Every outcome except TTT includes at least one head, so the probability is 7/8.

Probability of getting at least two heads: There are four outcomes (HHH, HHT, HTH, THH) that satisfy this condition, leading to a probability of 4/8 or 1/2.

Each probability is based on the fundamental principle of counting favorable outcomes over the total number of outcomes.

No long division needed here, really, since

[tex]\dfrac{x^3+8x^2+6}{x+8}=\dfrac{x^2(x+8)+6}{x+8}=x^2+\dfrac6{x+8}[/tex]

Then the integral is trivial:

[tex]\displaystyle\int\frac{x^3+8x^2+6}{x+8}\,\mathrm dx=\int x^2+\frac6{x+8}\,\mathrm dx=\frac{x^3}3+6\ln|x+8|+C[/tex]

To calculate the probabilities, we can use the binomial distribution formula. For each event, substitute the appropriate values into the formula and calculate the probabilities using the combination symbol and the probabilities of winning and losing a game. For part c, sum the probabilities of winning at least 7 games.

To find the probabilities for the given events, we can use the binomial distribution formula. Let p be the probability of winning a game for both the computer and the human, and q be the probability of losing a game. Since they are evenly matched, p = q = 0.5.

a. The probability that they each win five games is P(X = 5), where X follows a binomial distribution with n = 10 (number of games) and p = 0.5. We can calculate this probability using the formula P(X = 5) = C(10, 5) * (0.5)^5 * (0.5)^5.

b. The probability that the computer wins seven games is P(X = 7), where X follows a binomial distribution with n = 10 and p = 0.5. We can calculate this probability using the formula P(X = 7) = C(10, 7) * (0.5)^7 * (0.5)^3.

c. The probability that the human chess champion wins at least seven games can be calculated by summing the probabilities of winning exactly 7, 8, 9, and 10 games: P(X >= 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10).

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The student has 210 ways to schedule the remaining six hours of classes across the week, after meeting specific daily minimum class hour requirements.

The student wants to schedule twelve hours of classes across a week while meeting minimum class hour requirements on specific days.

Monday requires at least three hours, Tuesday at least two, and Friday at least one hour of classes.

Considering the constraints, we must calculate the number of ways she can distribute her remaining class hours across the week.

Steps to Calculate the Distribution:

Therefore, the student has 210 ways to schedule the remaining six hours of classes across the week.

Here is the complete question.

1). What are the odds for rolling a sum of 5 in a single roll of two fair dice?

2). If you bet $1 that a sum of 5 will turn up, what should the house pay (plus returning your $1 bet) if a sum of 5 turns up in order for the game to be fair?

Answer:

1).  [tex]\frac{1}{8}[/tex]

2).  $8.

Step-by-step explanation:

1).

The sample space (S) for rolling two fair dice is given as the following parameters illustrated below:

[tex]\left[\begin{array}{cccccc}(1,2)&(1,2)&(1,3)&(1,4)&(1,5)&(1,6)\\(2,1)&(2,2)&(2,3)&(2,4)&(2,5)&(2,6)\\(3,1)&(3,2)&(3,3)&(3,4)&(3,5)&(3,6)\end{array}\right] \left[\begin{array}{cccccc}(4,1)&(4,2)&(4,3)&(4,4)&(4,5)&(4,6)\\(5,1)&(5,2)&(5,3)&(5,4)&(5,5)&(5,6)\\(6,1)&(6,2)&(6,3)&(6,4)&(6,5)&(6,6)\end{array}\right][/tex]

Let F represent the event that the sum of both dice turns up is 5.

F = [(1,4),(2,3),(3,2),(4,1)}

∴ The probability for an event F  is illustrated below as:

[tex]\frac{P(F)}{P(F')}[/tex]  [tex]=\frac{\frac{4}{36} }{1-\frac{4}{36} }[/tex]

[tex]=\frac{\frac{4}{36} }{\frac{32}{36} }[/tex]

[tex]={\frac{4}{36}}*\frac{36}{32}[/tex]

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

2). From above, we can see that the probability for rolling a sum of 5 are 1 to 8. Therefore, if you roll a sum of 5, in order for the game to be fair, the house is required to pay $8.

Full Question

Eighteen telephones have just been received at an authorized service center. Six of these telephones are cellular, six are cordless, and the other six are corded phones. Suppose that these components are randomly allocated the numbers 1, 2, . . . , 18 to establish the order in which they will be serviced.

What is the probability that after servicing twelve of these phones, phones of only two of the three types remain to be serviced?

What is the probability that two phones of each type are among the first six serviced?

Answer:

a. 0.149

b. 0.182

Step-by-step explanation:

Given

Number of telephone= 18

Number of cellular= 6

Number of cordless = 6

Number of corded = 6

a.

There are 18C6 ways of choosing 6 phones

18C6 = 18564

From the Question, there are 3 types of telephone (cordless, Corded and cellular)

There are 3C2 ways of choosing 2 out of 3 types of television

3C2 = 3

There are 12C6 ways of choosing last 6 phones from just 2 types (2 types = 6 + 6 = 12)

12C6 = 924

There are 2 * 6C6 * 6C0 ways of choosing none from any of these two types of phones

2 * 6C6 * 6C0 = 2 * 1 * 1 = 2.

So, the probability that after servicing twelve of these phones, phones of only two of the three types remain to be serviced is

3 * (924 - 2) / 18564

= 3 * 922/18564

= 2766/18564

= 0.149

b)

There are 6C2 * 6C2 * 6C2 ways of choosing 2 cellular, 2 cordless, 2 corded phones

= (6C2)³

= 3375

So, the probability that two phones of each type are among the first six serviced is

= 3375/18564

= 0.182

Answer:

You will need to buy 5 1.50lb bags of sugar to make your recipe.

Step-by-step explanation:

This problem can be solved by consecutive rules of three.

A large recipe calls for 3.18kg of sugar. Your bags of sugar are measured in pounds (lb). Remembering that 1kg=2.20462lb

How many lbs of sugar you will need to buy?

1 kg - 2.20462lb

3.18 kg - x lb

[tex]x = 3.18*2.20462[/tex]

[tex]x = 7 lb[/tex]

How many 1.50lb bags of sugar will you need to buy to make your recipe?

1 bag - 1.50 lb

x bags - 7 lb

[tex]1.5x = 7[/tex]

[tex]x = \frac{7}{1.5}[/tex]

[tex]x = 4.67[/tex]

You will need to buy 5 1.50lb bags of sugar to make your recipe.

Final answer:

To make a recipe requiring 3.18kg of sugar, you need to buy 5 bags of 1.50lb sugar, after converting kilograms to pounds and rounding up to the nearest whole number.

Explanation:

To find out how many 1.50lb bags of sugar you need for 3.18kg of sugar, first convert kilograms to pounds using the conversion factor 1kg = 2.20462lb. Multiply 3.18kg by 2.20462lb/kg:

3.18kg * 2.20462lb/kg = 7.01068lb

Next, divide the total pounds of sugar by the weight of each bag to determine how many bags you need:

7.01068lb ÷ 1.50lb/bag = 4.67379 bags

Since you can't buy a fraction of a bag, round up to the nearest whole number:

5 bags of sugar (rounded up from 4.67379)

You'll need to purchase 5 bags of 1.50lb sugar to have enough for your recipe.

Answer:

546.94 meters is the height of the mountain.

Step-by-step explanation:

Pressure at the top of mountain = [tex]P_1=93.8 kPa[/tex]

Pressure at the bottom of the mountain = [tex]P_2=100.5 kPa[/tex]

Pressure difference  =[tex]P_2-P_1=100.5kPa-93.8kPa=6.7 kPa[/tex]

6.7 kPa = 6.7 × 1000 Pa (1 kPa= 1000 pa)

Density of the air = d = [tex]1.25 kg/m^3[/tex]

Acceleration due to gravity = g = [tex]9.8 m/s^2[/tex]

Height of the mountain = h

[tex]P=h\times d\times g[/tex]

[tex]h=\frac{P}{d\times g}=\frac{6.7\times 1000 Pa}{1.25 kg/m^3\times 9.8 m/s^2}[/tex]

[tex]h=546.94 m[/tex]

546.94 meters is the height of the mountain.

Answer:

From a statistical point of view: A correct decision was made.

so that the extent of the problem may be ascertained.

Step-by-step explanation:

"In statistical hypothesis testing, a type I error is the rejection of a true null hypothesis (also known as a "false positive" finding or conclusion), while a type II error is the non-rejection of a false null hypothesis (also known as a "false negative" finding or conclusion)."

Final answer:

The scenario described indicates that a Type I error was made during the hypothesis test, as the machine was working correctly but was stopped due to the rejection of the null hypothesis.

Explanation:

The question asks about the consequences of a hypothesis test related to the functioning of a machine filling quart size bottles of olive oil. When the hypothesis is rejected and the machine is shut down, but no issue is found upon inspection, this scenario describes a Type I error. This error occurs because the null hypothesis, which would be the assumption that the machine is functioning correctly, is rejected even though it is actually true.