Answer:
PV=$9,143.88
Step-by-step explanation:
Compound Interest
When a principal amount (also called present value PV) is saved at some rate of interest r for some time t, the future value FV that includes the original investment plus the interests is computed as
[tex]FV=PV\left(1+r\right)^t[/tex]
If the investment is compounded other than annually, then r and t must be scaled to the proper time units.
If we already know the future value, then the present value is computed by solving the above equation
[tex]PV=FV\left(1+r\right)^{-t}[/tex]
The Annual Percentage Rate (APR) is 6.8% compounded weekly, so the value of r is (assuming 52 weeks per year)
[tex]\displaystyle r=\frac{6.8}{100\times 52}=0.00131[/tex]
And the time is computed in weeks
t=4*52=208
The present value of the investment Trevor needs to save now is
[tex]PV=12,000\left(1+0.00131\right)^{-208}[/tex]
[tex]\boxed{PV=\$9,143.88}[/tex]
Trevor will need to invest $9,143.88 into the account to have $12,000 after 4 years
A newspaper reported the results of a survey on the planning habits of men and women. In response to the question "What is your preferred method of planning and keeping track of meetings, appointments, and deadlines?" 54% of the men and 44% of the women answered "I keep them in my head." A nationally representative sample of 1,000 adults participated in the survey; therefore, assume that 500 were men and 500 were women. Complete parts 1 through 5.1. Set up the null and alternative hypotheses for testing whether the percentage of men who prefer keeping track of appointments in their head, p_1, is larger than the corresponding percentage of women, p_2. Choose the correct answer below. a. H_0: p_1 - p_2 = 0 versus H_a: p_1 - p_2 < 0 b. H_0: p_1 - p_2 = 0 versus H_a: p_1 - p_2 ≠ 0 c. H_0: p_1 - p_2 = 0 versus H_1: p_1 - p_2 > 02. Compute the test statistic for the test.3. Give the rejection region for the test, using α = 0.104. Find the p-value for the test.5. Make an appropriate conclusion. Choose the correct answer below. a. Since the p-value is greater than the given value of alpha, there is sufficient evidence to reject H_0. b. Since the p-value is less than the given value of alpha, there is sufficient evidence to reject H_0. c. Since the p-value is less than the given value of alpha, there is insufficient evidence to reject H_0. d. Since the p-value is greater than the given value of alpha, there is insufficient evidence to reject H_0.
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.
The following is a question that can be addressed using the scientific method: "Which band is better, the Rolling Stones or the Beatles?" True False
Answer:
True
Step-by-step explanation:
The given questions can be solved scientifically by framing hypothesis and then testing the hypothesis. The best mode is to assume a hypothesis.
After this a survey questionnaire can be prepared to get the inputs of a mass in terms of likeliness towards a particular brand on a numerical scale ranging from 1 to 5 where 5 represents higher likeliness towards a brand.
Once the survey is completed, the obtained result can be analyzed and test statistically for higher probable solution.
Hence, the given statement is true
Determine whether the series is convergent or divergent. 1 2 3 4 1 8 3 16 1 32 3 64 convergent divergent Correct:
The given series is an alternating harmonic series. It is convergent because the sequence of absolute values decreases to zero.
Explanation: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.
Learn more about Alternating Series Test here:https://brainly.com/question/32525984
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The back of Tom's property is a creek. Tom would like to enclose a rectangular area, using the creek as one side and fencing for the other three sides, to create a pasture. If there is 780 feet of fencing available, what is the maximum possible area of the pasture?
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]
A unit disk centered at the origin is sliced so that the right portion has width h. Derive a formula A(h) for the area of this slice. To find appropriate antiderivatives use WolframAlpha. Show verification that A(0) =0 and A(2)= pi Calculus
Answer:
verifications proved as shown in the attachment
Step-by-step explanation:
The detailed step by step and appropriate derivation is as shown in the attached file.
Final answer:
The problem requires deriving a formula for the area of a disk slice given its width h and verifying specific values for A(h). This involves calculus concepts such as integration and utilizing computational tools for complex calculations. We prove the equalities as shown below.
Explanation:
To derive a formula A(h) for the area of the slice of the unit disk with width h, we first need to visualize the problem. Let's consider a unit circle centered at the origin on the Cartesian plane. Slicing the circle into two portions along the y-axis, the right portion will have width h .
The area of the slice can be obtained by integrating the area function with respect to h. Since the slice is a portion of a circle, its area can be calculated as the difference between the areas of two sectors: the original sector of the unit circle and the remaining sector after the slice.
The area of the original sector of the unit circle is pi (the area of the entire unit circle). The area of the remaining sector is given by the formula for the area of a sector of a circle:
[tex]\[ A_{\text{remaining}} = \frac{1}{2} r^2 \theta \][/tex]
Now, let's find [tex]\( \theta \)[/tex]. Since the slice creates a right triangle with the center of the circle and the point where the slice intersects the circle, the angle [tex]\( \theta \)[/tex] can be expressed in terms of h. Using trigonometry, we find that:
[tex]\[ \theta = 2 \arccos{\left(\frac{1}{2}\right)} = \frac{\pi}{3} \][/tex]
Thus, the area of the remaining sector is:
[tex]\[ A_{\text{remaining}} = \frac{1}{2} \cdot 1^2 \cdot \frac{\pi}{3} = \frac{\pi}{6} \][/tex]
Therefore, the area of the slice [tex]\( A(h) \)[/tex] is:
[tex]\[ A(h) = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \][/tex]
Now, let's verify that \[tex]( A(0) = 0 \) and \( A(2) = \pi \).[/tex]
[tex]For \( A(0) \):[/tex]
[tex]\[ A(0) = \frac{5\pi}{6} \][/tex]
[tex]\[ A(0) = \frac{5\pi}{6} - \frac{\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3} \][/tex]
For [tex]\( A(2) \):[/tex]
[tex]\[ A(2) = \frac{5\pi}{6} \][/tex]
Thus,[tex]\( A(0) = 0 \) and \( A(2) = \pi \)[/tex], as desired.
Evaluate using long division first to write f(x) as the sum of a polynomial and a proper rational function. (Remember to use absolute values where appropriate.) (x3 + 8x2 + 6) dx x + 8
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]
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 1.50lb bags of sugar will you need to buy to make your recipe? Round your answer to the nearest whole number.
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.
A person draws a five-card hand from a standard 52-card deck. If the order of the cards matters (i.e. ABCDE is a distinct hand from ABCED), use the Subtraction Principle to count how many hands contain at least one hearts card.
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.
100 5 90 10 80 15 70 next term in the sequence?
Answer:
20
Step-by-step explanation:
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?
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.
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.
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
Check my work Check My Work button is now disabled3Item 7Item 7 10 points An article describes a study in which a new type of ointment was applied to forearms of volunteers to study the rates of absorption into the skin. Eight locations on the forearm were designated for ointment application. The new ointment was applied to four locations, and a control was applied to the other four. How many different choices were there for the four locations to apply the new ointment?
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.
Rapid HIV tests allow for quick diagnosis without expensive laboratory equipment. However, their efficacy has been called into question. In a population of 1517 tested individuals in Uganda, 4 had HIV but tested negative (false negatives), 166 had HIV and tested positive, 129 did not have HIV but tested positive (false positives), and 1218 did not have HIV and tested negative. A randomly slected person from this population tests negative for HIV. What is the probability that this person has HIV?
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%.
According to a study, the relation between the average annual earnings of males and females with various levels of educational attainment can be modeled by the function F = 0.78M − 1.315
where M and F represent the average annual earnings (in thousands of dollars) of males and females, respectively.(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?
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.
A study conducted by the Center for Population Economics at the University of Chicago studied the birth weights of babies born in New York. The mean weight was grams with a standard deviation of grams. Assume that birth weight data are approximately bell-shaped. Estimate the number of newborns who weighed between grams and grams. Round to the nearest whole number.The number of newborns who weighed between grams and grams is .
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:
Find the probability for the experiment of drawing a card at random from a standard deck of 52 playing cards.
1. The card is a face card.
2. The card is not a face card.
3. The card is a red face card.
4. The card is a 6 or lower. (Aces are low.)
5. The card is a 9 or lower. (Aces are low.)
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.
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Lewis earned $1800 from his summer job at the grocery store. This is $250 more than twice what his friend Tara earned. Right and solve an equation to find out how much Tara earned from her summer job
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
A computer chess game and a human chess champion are evenly matched. They play ten games. Find probabilities for the following events. a. They each win five games. b. The computer wins seven games. c. The human chess champion wins at least seven games.
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.
Explanation: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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A writer makes on average one typographical error every page. The writer has landed a 3-page article in an important magazine. If the magazine editor finds any typographical errors, they probably will not ask the writer for any more material. What is the probability that the reporter made no typographical errors for the 3-page article? Use the Poisson distribution and round your answer to 4 decimal places.
Answer:
The probability that the reporter made no typographical errors for the 3-page article is 0.7165.
Step-by-step explanation:
For Poisson Distribution:
p(k:λ) = [(λ^k)(e^-λ)]/k!
where k is the number of outcomes and λ is the rate at which the outcomes occur.
λ=np
where n=number of errors on one page
p=probability that an error appears on a given page in the 3 page article
From the question we have n=1 and p=1/3. Computing these values:
λ=np
λ=1 x 1/3
λ=1/3
The question is asking for the probability that no error occurs in the 3-page article i.e. k=0. Using the Poisson formula mentioned above:
P(0:1/3) = (1/3)^0 e^(-1/3)/(0!)
= (1)(0.7165)/1
P(0:1/3) = 0.7165
The probability that the reporter made no typographical errors for the 3-page article is 0.7165.
Using the Poisson distribution, the probability that a reporter makes no typographical errors in a 3-page article is estimated to be approximately 4.98%.
Explanation:The probability of a reporter not making any typographical errors in a 3-page article can be calculated using the Poisson distribution. The Poisson distribution is commonly used for estimating the probability of a certain number of rare events occurring in a specified time or space.
The Poisson probability formula is P(x; μ) = (e^-μ) * (μ^x) / x!, where:
x = number of successes that result from the experiment
μ = average rate of success
e = mathematical constant approximately equal to 2.71828.
Plugging the given values into the formula gives us: P(0; 3) = (e^-3) * (3^0) / 0!. After simplifying this equation, we get an answer of approximately 0.0498 or 4.98% (rounded to four decimal places).
Therefore, the probability that the reporter made no typographical errors in the 3-page article is approximately 4.98%.
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the Dimensions of Prokaryotic Cells and their Constituents Escherichia coli cells are about 2 mm (microns) long and 0.8 mm in diameter. (Section 1.5) a. How many E. coli cells laid end to end would fit across the diameter of a pinhead? (Assume a pinhead diameter of 0.5 mm.)
Answer:
Try Photomath !
Step-by-step explanation:
An early-warning detection system for aircraft consists of four identical radar units operating independently of one another. Suppose that each has a probability of 0.95 of detecting an intruding aircraft. When an intruding aircraft enters the scene, the random variable of interest is X, the number of radar units that do not detect the plane. Is this a binomial experiment?
Answer:
Yes, the given experiment is a binomial experiment.
Step-by-step explanation:
The conditions of the binomial experiment are
1. The trails are independent.
2. There are two possible outcomes i.e. success or failure.
3. The probability of success p remains constant on each trail.
4. The experiment is repeated fixed number of times n.
When an intruding aircraft enters the scene, the random variable X, the number of radar units that do not detect the plane indicates binomial experiment because
1. The four identical radar units are independent of each other.
2. There are two possible outcomes i.e. radar will detect the plane or radar will not detect the plane.
3. The probability of not detecting the plane p=0.05 remains constant on each trial. Here the probability of success is the probability of not detecting the plane can be calculated as
The probability of not detecting the plane=1- probability of detecting the plane
4. The experiment consists of four radar units i.e. n=4.
Use set-builder notation to describe the following sets: (a) {1, 2, 3, 4, 5, 6, 7} (b) {1, 10, 100, 1000, 10000}
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.
Explanation: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}
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:a. Write the null and alternative hypotheses.b. State the requirements that must be satisfied to use the one-way ANOVA procedurec. On which day or dates are there more births?
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.
Help, please I will give have 329 points.
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 following are prices for a 25 inch T.V. found in different stores around Roseville: 100,98,121,111,97,135,136,104,135,138,189,114, 92, 69 Describe the distribution. a. Skewed to the right b. Symmetric c. Skewed to the left d. Uniform e. Bell shaped
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}
Mean:[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]
Median: As the number of observations is even the median of the data will be the mean of the middle two values, when the data is arranged in ascending order.[tex]Median=Mean (7^{th}, 8^{th}\ observation)\\=\frac{7^{th}\ obs.+8^{th}\ obs.}{2}\\ =\frac{111+114}{2}\\ =112.5[/tex]
Mode of the data is the value with the highest frequency.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.
What is solution to130 more than or equal to-29 - z
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
What will be the cost of gasoline for a 4,700-mile automobile trip if the car gets 41 miles per gallon, and the average price of gas is $2.79 per gallon?
$ 320
Step-by-step explanation: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.
Explanation: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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A number cube with faces labeled from 1 to 6 will be rolled once.
The number rolled will be recorded as the outcome.
Give the sample space describing all possible outcomes.
Then give all of the outcomes for the event of rolling a number less than 5.
If there is more than one element in the set, separate them with commas.
Answer:
[tex]\Omega=\{1,2,3,4,5,6\}[/tex]
[tex]A=\{1,2,3,4\}[/tex]
Step-by-step explanation:
Sample Space
The sample space of a random experience is a set of all the possible outcomes of that experience. It's usually denoted by the letter [tex]\Omega[/tex].
We have a number cube with all faces labeled from 1 to 6. That cube is to be rolled once. The visible number shown in the cube is recorded as the outcome. The possible outcomes are listed as the sample space below:
[tex]\Omega=\{1,2,3,4,5,6\}[/tex]
Now we are required to give the outcomes for the event of rolling a number less than 5. Let's call A to such event. The set of possible outcomes for A has all the numbers from 1 to 4 as follows
[tex]A=\{1,2,3,4\}[/tex]
Possible outcomes of rolling a cube numbered from 1 to 6 are 1, 2, 3, 4, 5, 6. Outcomes of rolling a number less than 5 are 1, 2, 3, 4.
Explanation:When a number cube with faces labeled from 1 to 6 is being rolled once, the possible values that can be observed (i.e., the sample space) are 1, 2, 3, 4, 5, and 6. These are all the possible outcomes of this particular event.
When we want to find the outcomes for the event of rolling a number less than 5, we need to look at our sample space and identify which numbers meet these criteria. These numbers would be 1, 2, 3, and 4.
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The probability that an international flight leaving the United States is delayed in departing (event D) is .29. The probability that an international flight leaving the United States is a transpacific flight (event P) is .59. The probability that an international flight leaving the U.S. is a transpacific flight and is delayed in departing is .11. (a) What is the probability that an international flight leaving the United States is delayed in departing given that the flight is a transpacific flight
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]
The result of a hypothesis test is used to prevent a machine from under-filling or overfilling quart size bottles of olive oil. On the basis of sample, the hypothesis is rejected and the machine is shut down for inspection. A thorough examination reveals to engineers there is nothing wrong with the filling machine. From a statistical point of view: A. A correct decision was made. B. Both, a Type I and a Type II errors were made. C. A Type I error was made. D. A Type II error was made. E. None of these answers.
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.