Answer:
(a) The expected value of the prize for one play of Instant Lotto is $3.50.
(b) The probability that the visitor wins some prize at least twice in the 20 free plays is 0.2641.
(c) The probability that a randomly selected day has at least 1000 people play Instant Lotto is 0.2579.
Step-by-step explanation:
(a)
The probability distribution of the monetary prizes that can be won at the game called Instant Lotto is:
X P (X = x)
$10 0.05
$15 0.04
$30 0.03
$50 0.01
$1000 0.001
$0 0.869
___________
Total = 1.000
Compute the expected value of the prize for one play of Instant Lotto as follows:
[tex]E(X)=\sum x\cdot P (X=x)[/tex]
[tex]=(10\times 0.05)+(15\times 0.04)+(30\times 0.03) \\+ (50\times 0.01)+(1000\times 0.001)+(0\times 0.869)\\=0.5+0.6+0.9+0.5+1+0\\=3.5[/tex]
Thus, the expected value of the prize for one play of Instant Lotto is $3.50.
(b)
Let X = number of times a visitor wins some prize.
A visitor to the casino is given n = 20 free plays of Instant Lotto.
The probability that a visitor wins at any of the 20 free plays is, p = 1/20 = 0.05.
The event of a visitor winning at a random free play is independent of the others.
The random variable X follows Binomial distribution with parameters n = 20 and p = 0.05.
Compute the probability that the visitor wins some prize at least twice in the 20 free plays as follows:
P (X ≥ 2) = 1 - P (X < 2)
= 1 - P (X = 0) - P (X = 1)
[tex]=1-[{20\choose 0}0.05^{0}(1-0.05)^{20-0}]-[{20\choose 1}0.05^{1}(1-0.05)^{20-1}]\\=1-0.3585-0.3774\\=0.2641[/tex]
Thus, the probability that the visitor wins some prize at least twice in the 20 free plays is 0.2641.
(c)
Let X = number of people who play Instant Lotto each day.
The random variable X is normally distributed with a mean, μ = 800 people and a standard deviation, μ = 310 people.
Compute the probability that a randomly selected day has at least 1000 people play Instant Lotto as follows:
Apply continuity correction:
P (X ≥ 1000) = P (X > 1000 + 0.50)
= P (X > 1000.50)
[tex]=P(\frac{X-\mu}{\sigma}>\frac{1000.50-800}{310})[/tex]
[tex]=P(Z>0.65)\\=1-P(Z<0.65)\\=1-0.74215\\=0.25785\\\approx0.2579[/tex]
Thus, the probability that a randomly selected day has at least 1000 people play Instant Lotto is 0.2579.
What is y={-\dfrac{1}{3}}x-9y=− 3 1 x−9y, equals, minus, start fraction, 1, divided by, 3, end fraction, x, minus, 9 written in standard form?
Answer:
[tex]x+3y=-27[/tex]
Step-by-step explanation:
We are given that
[tex]y=(-\frac{1}{3})x-9[/tex]
We have to find the standard form of given equation
[tex]y=\frac{-x-27}{3}[/tex]
[tex]3y=-x-27[/tex]
By using multiplication property of equality
[tex]x+3y=-27[/tex]
We know that
Standard form of equation
[tex]ax+by=c[/tex]
Therefore, the standard form of given equation is given by
[tex]x+3y=-27[/tex]
The equation y = -1/3x - 9 can be converted to standard form by eliminating fractions and rearranging. The final equation in standard form is x - 3y = 27.
Explanation:The equation you provided is already in slope-intercept form, which is y = mx + b, where m is the slope and b is the y-intercept. However, you're asked to convert this equation into standard form, which is Ax + By = C, with A, B, and C being integers, and A and B not both equal to zero.
To convert the given equation y = -1/3x - 9 into standard form, we first need to eliminate the fractions. We can achieve this by multiplying every term by -3, giving us 3y = x + 27. To make it fit the standard form, we can rearrange as -x + 3y = -27.
Remember, standard form shouldn't have any negatives in front of the x term, so we multiply everything by -1. The final equation in standard form is x - 3y = 27.
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Her fee is 5 hours of work is $350
Convert 2.575 into a mixed number in simplest form
Step-by-step explanation:
(2575÷25)(1000÷25) = 10340 when reduced to the simplest form. As the numerator is greater than the denominator, we have an IMPROPER fraction, so we can also express it as a MIXED NUMBER, thus 25751000 is also equal to 22340 when expressed as a mixed number.
The amounts (in ounces) of juice in eight randomly selected juice bottles are: 15.3 15.3 15.7 15.7 15.3 15.9 15.3 15.9 Construct a 98% confidence interval for the mean amount of juice in all such bottles
Answer:
[tex]15.55-2.997\frac{0.278}{\sqrt{8}}=15.26[/tex]
[tex]15.55+2.997\frac{0.278}{\sqrt{8}}=15.84[/tex]
The 98% confidence interval would be given by (15.26;15.84)
Step-by-step explanation:
Notation
[tex]\bar X[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s represent the sample standard deviation
n represent the sample size
Solution to the problem
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
We can calculate the mean and the sample deviation we can use the following formulas:
[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)
[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)
The mean calculated for this case is [tex]\bar X=15.55[/tex]
The sample deviation calculated [tex]s=0.278[/tex]
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=8-1=7[/tex]
Since the Confidence is 0.98 or 98%, the value of [tex]\alpha=0.02[/tex] and [tex]\alpha/2 =0.01[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.01,7)".And we see that [tex]t_{\alpha/2}=2.997[/tex]
And the confidence interval is given by:
[tex]15.55-2.997\frac{0.278}{\sqrt{8}}=15.26[/tex]
[tex]15.55+2.997\frac{0.278}{\sqrt{8}}=15.84[/tex]
The 98% confidence interval would be given by (15.26;15.84)
A study is run to estimate the mean total cholesterol level in children 2 to 6 years of age. A sample of 9 participants is selected and their total cholesterol levels are measured as follows. 180 220 240 196 175 183 195 140 200 What is the sample mean?A)196B)200C)192.11D)180
Answer:
C) 192.11
Step-by-step explanation:
To find the mean of any group of numbers, you simply add up all the numbers and then divide by how many numbers there are.
Using a calculator or a pencil, add the following:
180 + 220 + 240 + 196 + 175 + 183 + 195 + 140 + 200
= 1,729
The last step is to divide 1,729 by 9:
1,729 ÷ 9 = 192.11
A study dealing with health care issues plans to take a sample survey of 1500 Americans to estimate the proportion who have health insurance and the mean dollar amount that Americans spent on health care this past year. a. Identify the population parameters that this study will estimate. b. Identify the statistics that can be used to estimate these parameters.
Answer:
Step-by-step explanation:
(a) A parameter of a population measures the characteristics of the population, In the question the proportion of all the persons who have health insurance and the mean of the entire dollar amount that Americans spent on health care in the past year measure the population.
Invariably, the proportion of persons having health insurance, and the mean dollars spent on health care for all Americans are the population parameter
(b) A statistics measure the characteristics of the sample.
In the question, the sample of 1500 Americans are considered to estimate the proportion of all Americans, proportion of all the persons who have health insurance among 1500 and the sample mean of all the dollar amounts that the selected Americans spent on healthy care in the past year describe the sample.
Invariably, the sample proportion of persons having health insurance, and the mean dollars spent on health care for 1500 selected Americans are sample statistics
A sample of n = 4 scores is obtained from a population with a mean of 70 and a standard deviation of 8. If the sample mean corresponds to a z score of 2.00, then what is the value of the sample mean?
Answer:
The value of the sample mean is 78.
Step-by-step explanation:
We are given that a sample of n = 4 scores is obtained from a population with a mean of 70 and a standard deviation of 8.
Also, the sample mean corresponds to a z score of 2.00.
Let [tex]\bar X[/tex] = sample mean
The z-score probability distribution for a sample mean is given by;
Z = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean = 70
[tex]\sigma[/tex] = standard deviation = 8
n = sample size = 4
The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.
Now, we are given that the sample mean corresponds to a z score of 2.00 for which we have to find the value of sample mean;
So, z-score formula is given by ;
z-score = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] = [tex]\frac{\bar X-70}{\frac{8}{\sqrt{4} } }[/tex]
2.00 = [tex]\frac{\bar X-70}{\frac{8}{\sqrt{4} } }[/tex]
2.00 = [tex]\frac{\bar X-70}{4 } }[/tex]
[tex]\bar X = 70+(2 \times 4)[/tex]
[tex]\bar X[/tex] = 70 + 8 = 78
Therefore, the value of the sample mean is 78.
The value of the sample mean is 78.
Explanation:To find the value of the sample mean, we can use the formula:
sample mean = population mean + (z score * (standard deviation / square root of sample size))
In this case, the population mean is 70, the z score is 2.00, and the standard deviation is 8. Since the sample size is 4, we calculate the square root of 4, which is 2. Plugging these values into the formula gives us:
sample mean = 70 + (2.00 * (8 / 2)) = 70 + (2.00 * 4) = 70 + 8 = 78...
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A triangle has an area of 54 square inches the height of the triangle is 12 inches what is the length of the base of the triangle?
Answer:
The base of the triangle = 9inches
Step-by-step explanation:
Explanation:-
The area of the triangle = [tex]\frac{1}{2} bh[/tex] square units
Given area of the triangle(A) = 54 square inches
Given the height of the triangle (h) = 12 inches
now equating [tex]\frac{1}{2} b h = 54[/tex]
[tex]\frac{1}{2} b (12) = 54[/tex]
now simplification, we get
6 b = 54
Dividing '6' on both sides, we get
b = 9
The base of the triangle = 9 inches
Conclusion:-
The base of the triangle = 9 inches
Help meeeeeewwww :):):)):
Answer: Top Right, A rectangle has all the properties of a square.
Step-by-step explanation: Nobody how hard you try to make a rectangle a square, it won't work. A rectangle cannot have 4 equal slides.
There is a spinner with 15 equal areas, numbered 1 through 15. If the spinner is spun one time, what is the probability that the result is a multiple of 5 and a multiple of 2?
Answer:
6.67% probability that the result is a multiple of 5 and a multiple of 2
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes.
Desired outcomes:
Multiples of 2 AND 5
Between 1 and 15, the multiples of 2 are: 2,4,6,8,10,12,14
Between 1 and 15, the multiples of 5 are: 5,10,15
So only 10 is a multiply of both 2 and 5, so only one desired outcome, which means that [tex]D = 1[/tex]
Total outcomes:
Any number between 1 and 15, there are 15, so [tex]T = 15[/tex]
Probability:
[tex]p = \frac{D}{T} = \frac{1}{15} = 0.0667[/tex]
6.67% probability that the result is a multiple of 5 and a multiple of 2
Answer:
Step-by-step explanation:
3/5 or 9/15
5 of 5
It is worked out that if 5 ladles full of soup are given to
each person,
140 people can be fed.
The customers have complained in the past that the
portions are too small.
The cook decides to give 7 ladles full of soup to each
person.
How many people can now be fed soup?
Answer:100 people
Step-by-step explanation:
140 people x 5 ladle/1 person = 700 ladles full
700 ladles x 1 person/7 ladles = 100 people fed
Let z denote a random variable that has a standard normal distribution. Determine each of the probabilities below. (Round all answers to four decimal places.) (a) P(z < 2.36) = (b) P(z 2.36) = (c) P(z < -1.22) = (d) P(1.13 < z < 3.35) = (e) P(-0.77 z -0.55) = (f) P(z > 3) = (g) P(z -3.28) = (h) P(z < 4.98) =
Answer:
(a) P (Z < 2.36) = 0.9909 (b) P (Z > 2.36) = 0.0091
(c) P (Z < -1.22) = 0.1112 (d) P (1.13 < Z > 3.35) = 0.1288
(e) P (-0.77< Z > -0.55) = 0.0705 (f) P (Z > 3) = 0.0014
(g) P (Z > -3.28) = 0.9995 (h) P (Z < 4.98) = 0.9999.
Step-by-step explanation:
Let us consider a random variable, [tex]X \sim N (\mu, \sigma^{2})[/tex], then [tex]Z=\frac{X-\mu}{\sigma}[/tex], is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, [tex]Z \sim N (0, 1)[/tex].
In statistics, a standardized score is the number of standard deviations an observation or data point is above the mean. The z-scores are standardized scores.
The distribution of these z-scores is known as the standard normal distribution.
(a)
Compute the value of P (Z < 2.36) as follows:
P (Z < 2.36) = 0.99086
≈ 0.9909
Thus, the value of P (Z < 2.36) is 0.9909.
(b)
Compute the value of P (Z > 2.36) as follows:
P (Z > 2.36) = 1 - P (Z < 2.36)
= 1 - 0.99086
= 0.00914
≈ 0.0091
Thus, the value of P (Z > 2.36) is 0.0091.
(c)
Compute the value of P (Z < -1.22) as follows:
P (Z < -1.22) = 0.11123
≈ 0.1112
Thus, the value of P (Z < -1.22) is 0.1112.
(d)
Compute the value of P (1.13 < Z > 3.35) as follows:
P (1.13 < Z > 3.35) = P (Z < 3.35) - P (Z < 1.13)
= 0.99960 - 0.87076
= 0.12884
≈ 0.1288
Thus, the value of P (1.13 < Z > 3.35) is 0.1288.
(e)
Compute the value of P (-0.77< Z > -0.55) as follows:
P (-0.77< Z > -0.55) = P (Z < -0.55) - P (Z < -0.77)
= 0.29116 - 0.22065
= 0.07051
≈ 0.0705
Thus, the value of P (-0.77< Z > -0.55) is 0.0705.
(f)
Compute the value of P (Z > 3) as follows:
P (Z > 3) = 1 - P (Z < 3)
= 1 - 0.99865
= 0.00135
≈ 0.0014
Thus, the value of P (Z > 3) is 0.0014.
(g)
Compute the value of P (Z > -3.28) as follows:
P (Z > -3.28) = P (Z < 3.28)
= 0.99948
≈ 0.9995
Thus, the value of P (Z > -3.28) is 0.9995.
(h)
Compute the value of P (Z < 4.98) as follows:
P (Z < 4.98) = 0.99999
≈ 0.9999
Thus, the value of P (Z < 4.98) is 0.9999.
**Use the z-table for the probabilities.
To find the probabilities, we use the standard normal distribution table or a calculator to calculate probabilities for a standard normal distribution. We calculate each probability step by step and round the answers to four decimal places.
Explanation:To determine the probabilities, we will use the standard normal distribution table or a calculator that can calculate probabilities for a standard normal distribution.
(a) P(z < 2.36) = 0.9900
(b) P(z > 2.36) = 1 - P(z < 2.36) = 1 - 0.9900 = 0.0100
(c) P(z < -1.22) = 0.1103
(d) P(1.13 < z < 3.35) = P(z < 3.35) - P(z < 1.13) = 0.9993 - 0.8708 = 0.1285
(e) P(-0.77 < z < -0.55) = P(z < -0.55) - P(z < -0.77) = 0.2896 - 0.2823 = 0.0073
(f) P(z > 3) = 1 - P(z < 3) = 1 - 0.9987 = 0.0013
(g) P(z > -3.28) = 1 - P(z < -3.28) = 1 - 0.0005 = 0.9995
(h) P(z < 4.98) = 1 - P(z > 4.98) = 1 - 0.0000 = 1.0000
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Make COS the subject of the formula
c^2 = a^2 + b^2 - 2(ab)(cos C)
c^2 + 2(ab)(cos C) = a^2 + b^2
2(ab)(cos C) = a^2 + b^2 - c^2
cos C = (a^2 + b^2 - c^2) / 2ab - Answer choice E
Hope this helps! :)
To make COS the subject of a formula, the equation is rearranged such that COS is isolated. An example can be in the equation a = b cos(x) which can be rearranged as cos(x) = a/b. A complete formula is necessary for an accurate step-by-step guide.
Explanation:To make COS the subject of a formula, it typically involves other known quantities represented by variables and constants. For example, in the equation a = b cos(x), we can make cos(x) the subject of the equation by rearranging it to: cos(x) = a/b. However, to provide a more accurate step-by-step guide, the complete formula is necessary. This principle can be applied to various trigonometric formulas so that COS becomes the main focus.
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Proof: Suppose A, B, and C are any sets. [To show that A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C), we must show that A ∩ (B ∪ C) ⊆ (A ∩ B) ∪ (A ∩ C) and that (A ∩ B) ∪ (A ∩ C) ⊆ A ∩ (B ∪ C).] g
I suppose you're supposed to prove that set intersection is distributive across a union,
[tex]A\cap(B\cup C)=(A\cap B)\cup(A\cap C)[/tex]
Two sets are equal if they are subsets of one another. To prove a set [tex]X[/tex] is a subset of another set [tex]Y[/tex], you have to show that any element [tex]x\in X[/tex] also belongs to [tex]Y[/tex].
Let [tex]x\in A\cap(B\cup C)[/tex]. By definition of intersection, both [tex]x\in A[/tex] and [tex]x\in B\cup C[/tex]. By definition of union, either [tex]x\in B[/tex] or [tex]x\in C[/tex]. If [tex]x\in B[/tex], then clearly [tex]x\in A\cap B[/tex]; if [tex]x\in C[/tex], then [tex]x\in A\cap C[/tex]. Either way, [tex]x\in(A\cap B)\cup(A\cap C)[/tex]. Hence [tex]A\cap(B\cup C)\subseteq(A\cap B)\cup(B\cap C)[/tex].
The proof in the other direction uses the same sort of reasoning. Let [tex]x\in(A\cap B)\cup(A\cap C)[/tex]. Then either [tex]x\in A\cap B[/tex] or [tex]x\in A\cap C[/tex]. If [tex]x\in A\cap B[/tex], then both [tex]x\in A[/tex] and [tex]x\in B[/tex]; if [tex]x\in A\cap C[/tex], then both [tex]x\in A[/tex] and [tex]x\in C[/tex]. So certainly [tex]x\in A[/tex], and either [tex]x\in B[/tex] or [tex]x\in C[/tex] so that [tex]x\in B\cup C[/tex]. Hence [tex](A\cap B)\cup(A\cap C)\subseteq A\cap(B\cup C)[/tex].
Both sets are subsets of one another, so they are equal.
Given that €1 =£0.72 a) how much is €410
Answer: £569.40p
Step-by-step explanation:
410÷£0.72=£569.40p
Answer:
£295.20
Step-by-step explanation:
€410 x 0.72 = £295.20
A business journal investigation of the performance and timing of corporate acquisitions discovered that in a random sample of 2,684 firms, 715 announced one or more acquisitions during the year 2000. Does the sample provide sufficient evidence to indicate that the true percentage of all firms that announced one or more acquisitions during the year 2000 is less than 29%? Use alpha equals0.05 to make your decision.
Calculate the value of the z-statistic for this test.
Answer:
z = 1.960
Step-by-step explanation:
The sample proportion is:
p = 715 / 2684 = 0.2664
The standard error is:
σ = √(pq/n)
σ = √(0.266 × 0.734 / 2684)
σ = 0.0085
For α = 0.05, the confidence level is 95%. The z-statistic at 95% confidence is 1.960.
The margin of error is 1.960 × 0.0085 = 0.0167.
The confidence interval is 0.2664 ± 0.0167 = (0.2497, 0.2831).
The upper limit is 28.3%, so the journal can conclude with 95% confidence that the true percentage is less than 29%.
Yes, the considered sample provides sufficient evidence to indicate that the true percentage of all firms that announced one or more acquisitions during the year 2000 is less than 29%.
The z-test statistic came out to be -2.69What is the z test statistic for one sample proportion?Suppose that we have:
n = sample size[tex]\hat{p}[/tex] = sample proportionp = population proportion (hypothesised)Then, the z test statistic for one sample proportion is:
[tex]Z = \dfrac{\hat{p} - p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]
For this case, we're provided that:
Size of sample = n = 2684Sample proportion of those who announced one or more acquisitions during year 2000 is: [tex]\hat{p}[/tex] = 715/n = 715/2684 Level of significance = 0.05We want to determine if true percentage of all firms that announced one or more acquisitions during the year 2000 is less than 29% = 0.29 (converted percent to decimal).
Hypotheses:
Null hypothesis: [tex]H_0: p \geq p_0 = 0.29\\[/tex] (it nullifies what we want to test for, thus assumes that true mean of the population proportion is not less than 0.29).Alternate hypothesis: [tex]H_A: p < p_0 = 0.29[/tex] Assumes that the true population mean is less than 0.29Thus, the test is left tailed test.
where [tex]p_0[/tex] = 29% = 0.29 is the hypothesized mean value of population proportion.
The test statistic is:
[tex]Z = \dfrac{\hat{p} - p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\\\ Z = \dfrac{715/2684 - 0.29}{\sqrt{\dfrac{0.29(1-0.29)}{2684}}} \approx -2.69[/tex]
The critical value of Z at level of significance 0.05 is -1.6449
Since the test statistic = -2.69 < critical value = -1.6449, so the test statistic lies in the rejection region (the rejection region for the left tailed test is all the values below critical value).
Thus, we reject the null hypothesis and accept the alternative hypothesis that the true population mean is less than 0.29.
Thus, the considered sample provides sufficient evidence to indicate that the true percentage of all firms that announced one or more acquisitions during the year 2000 is less than 29%.
The z-test statistic came out to be -2.69
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6. The Brinson family is going camping. Their tent is shaped like a rectangular pyramid. The volume
of the tent is 6,000 cubic inches. The area of the base of the tent is 1,200 cubic inches. What is
the height of the tent? *
(1 Point)
Given:
Given that Brinson's family is going camping. Their tent is shaped like a rectangular pyramid.
The volume of the tent is 6000 cubic inches.
The area of the base of the tent is 1200 square inches.
We need to determine the height of the tent.
Height of the tent:
The height of the tent can be determined using the formula,
[tex]V=\frac{1}{3}Bh[/tex]
where B is the area of the base and h is the height of the pyramid.
Substituting V = 6000 and B = 1200, we get;
[tex]6000=\frac{1}{3}(1200)h[/tex]
[tex]6000=400h[/tex]
[tex]15=h[/tex]
Thus, the height of the tent is 15 inches.
Consider the reduction of the rectangle.
2.3 ft
16.8 ft
Not drawn to scale
Rounded to the nearest tenth, what is the value of x?
Answers-
0.1 feet
0.6 feet
1.6 feet
2.0 feet
Answer:
1.6
Step-by-step explanation:
if you divide 16.8 by 4.5 you get 3.7 now 3.7 divided 2.3 = 1.6
At Denver International Airport, 86% of recent flights have arrived on time. A sample of 11 flights is studied. What is the variance in a sample of 11 flights? Round your answer to 2 decimal places if needed.
Answer:
[tex]s^2=1.32[/tex]
Step-by-step explanation:
-This is a binomial probability distribution problem.
-Given that p=0.86 and n=11, the sample variance can be calculated using the formula:
[tex]\sigma^2=np(1-p)[/tex]
#We substitute the given parameters in the formula to solve for variance:
[tex]\sigma^2=np(1-p)\\\\\\=11\times 0.86(1-0.86)\\\\=1.324\approx1.32[/tex]
Hence, the sample variance is 1.32
Final answer:
The variance in a sample of 11 flights that have arrived on time at Denver International Airport, where 86% of flights are on time, is calculated using a binomial distribution with the formula variance = np(1-p), resulting in a variance of 1.33 after rounding to two decimal places.
Explanation:
The subject of this question is about finding the variance in a sample of flights that have arrived on time at Denver International Airport, given that 86% of recent flights have arrived on time and that a sample of 11 flights is studied. To calculate the variance for a binomial distribution, which is applicable in this context because each flight can either be on time or not, we use the formula:
Variance = np(1-p).
Where 'n' is the number of trials (or flights, in this case), which is 11, and 'p' is the probability of success on each trial (a flight arriving on time), which is 0.86. Thus, the variance for the 11 flights can be calculated as follows:
Variance = 11 × 0.86 × (1 - 0.86) = 11 × 0.86 × 0.14 = 1.3284.
After rounding to two decimal places, the variance for a sample of 11 flights is 1.33.
When a certain prescription drug is taken orally by an adult, the amount of the drug (in mg/L) in the bloodstream at t hours is given by the function y=f(t)
1. What is the concentration of the drug in the bloodstream at t= 8 hours?
2. During what time interval is the drug level in the bloodstream increasing? Decreasing?
Answer:
1. y = f(8)
2. So for t which f'(t) > 0, the drug level in the bloodstream is increasing. And for t which f'(t) < 0, it is decreasing.
Step-by-step explanation:
The concentration of the drug in the bloodstream at t hours is:
y = f(t)
1. What is the concentration of the drug in the bloodstream at t= 8 hours?
At t hours, y = f(t)
So at 8 hours, y = f(8)
2. During what time interval is the drug level in the bloodstream increasing? Decreasing?
A function f(t) is increasing when
f'(t) > 0
And is decreasing when
f'(t) < 0
So for t which f'(t) > 0, the drug level in the bloodstream is increasing. And for t which f'(t) < 0, it is decreasing.
(1) The concentration of drug in the bloodstream in 8 hours is given by
[tex]\rm \bold{y = f (8)}[/tex]
(2) The time interval for which [tex]\rm y'=f'(t)>0[/tex] the drug level of the bloodstream is increasing.
The time interval for which [tex]\rm y' = f'(t) <0[/tex] the drug level of the bloodstream is decreasing.
When a certain prescription drug is taken orally by an adult.
the amount of the drug (in mg/L) in the bloodstream at t hours is given by the function y=f(t)
To be determined
(1) The concentration of the drug in the bloodstream at t= 8 hours
(2) During what time interval is the drug level in the bloodstream increasing or deceasing
The amount of the drug (in mg/L) in the bloodstream at t hours is given by the function
y=f(t).......(1)
(1) The concentration of drug in the bloodstream in 8 hours is given by putting t= 8 in the equation (1) which can be formulated as below
[tex]\rm y = f (8)[/tex]
(2) From the definition of increasing and decreasing function we can write that
[tex]\rm y = f(x) \; is \; increasing \; when \; f' (x)>0 \\and\; y = f(x) ; is \; decreasing \; when \; f' (x)<0 \\\\\\\\[/tex]
By the definition of increasing and decreasing function we can say that
The time interval for which [tex]\rm y'=f'(t)>0[/tex] the drug level of the bloodstream is increasing.
The time interval for which [tex]\rm y'=f'(t) <0[/tex] the drug level of the blood stream is decreasing.
For more information please refer to the link below
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Simplify to create equivalent to 19-6(-k+4)
Answer: don’t use k for the other numbers
Step-by-step explanation:
Answer:
6k -5
Step-by-step explanation:
19-6(-k+4)
Distribute
19 -6*-k -6*4
19 +6k -24
Combine like terms
6k +19-24
6k -5
A consumer group selected 100 different airplanes at random from each of two large airlines. The mean seat width for the 100 airplanes was calculated for each airline, and the difference in the sample mean widths was calculated. The group used the sample results to construct a 95 percent confidence interval for the difference in population mean widths of seats between the two airlines. Suppose the consumer group used a sample size of 50 instead of 100 for each airline. When all other things remain the same, what effect would the decrease in sample size have on the interval
Answer: B The width of the interval would increase
Step-by-step explanation:
i just took this and go the answers back
Decreasing the sample size from 100 to 50 would widen the 95 percent confidence interval for the difference in mean seat widths between two airlines. This is because smaller samples have more variability, requiring a larger interval to capture the population mean with the same level of certainty.
Explanation:If the consumer group used a sample size of 50 instead of 100 for each airline, that would increase the width of the 95 percent confidence interval for the difference in population mean widths of seats between the two airlines. This is because smaller sample sizes result in more variability, requiring a wider interval to capture the true population mean with the same level of certainty.
As with the unoccupied seats example, where the sample mean of 11.6 and standard deviation of 4.1 were used to form a confidence interval, the size of the interval is dependent on the variability within the sample, and smaller samples generally have higher variability. Similarly in the case of exam scores, with a lower confidence level of 90 percent, a narrower interval is needed compared to a higher confidence level of 95 percent.
Therefore, in effect, decreasing the sample size from 100 to 50 would make the confidence interval wider, as more variability is expected and a larger interval is needed to capture the true population mean with a 95 percent confidence level.
Learn more about Confidence Intervals here:https://brainly.com/question/34700241
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Mismatch went to a store spent half of her money and then $10 more she went to the second store spend half the money she has left and then $10 more then she had no money left how much money did mismatch have when she started our
Answer: I had a stroke trying to read and understand this sorry
Step-by-step explanation:
Answer:
$60.
Step-by-step explanation:
x / 2 - 10 = 0
x / 2 = 10
x = 20
Before she spent her money:
x / 2 - 10 = 20
x / 2 = 30
x = 60
She had $60 before she went shopping.
Feel free to let me know if you need more help. :)
What is the median of this set of data?
1,2,5, 6, 9
Answer:
5
Step-by-step explanation:
It be 5 because its in the middle
3,4,7,8
Step-by-step explanation:
1 2 3 4 5 6 7 8 9
3 4 7 8
Bob's z-score for the last exam was 1.52 in Prof. Axolotl's class BIO 417, "Life Cycle of the Ornithorhynchus." Bob said, "Oh, good, my score is in the top 10 percent." Assuming a normal distribution of scores, is Bob right?
Answer:
yes
Step-by-step explanation:
The 90-th percentile of a normal distribution corresponds to a z-score of 1.282. Bob's z-score is above that, so he is definitely in the top 10%.
_____
If you haven't memorized the percentiles associated with the normal distribution, it is convenient to use a calculator or table of values.
In this case, ob is correct in saying that his score is in the top 10 percent.
To determine if Bob's z-score corresponds to a score in the top 10 percent of the distribution, we need to find the percentile associated with his z-score.
We can then compare this percentile to 90%, as the top 10% corresponds to the highest scores.
Using a standard normal distribution table or a calculator, we find that a z-score of 1.52 corresponds approximately to the 93rd percentile.
This means that 93% of the scores are below Bob's score, indicating that he is indeed in the top 10% of the class.
So, Bob is correct in saying that his score is in the top 10 percent.
Find the value of g(-1)
Answer:
g(-1) = -3
Step-by-step explanation:
g(-1) is the y value of the graph when x = -1
when x=-1 y=-3
g(-1) = -3
The surface area of the cube is 150in squared. What is the length of the side of the cube?
Answer:
Cube surface area = 6 * side^2
150 square inches = 6 * side^2
sq root (side) = 25
side = 5 inches
Step-by-step explanation:
In a study of the nicotine patch, 21% of those who used the patch for 2 months reported no smoking incidents in the following year. The 95% confidence interval is (17.4%, 24.8%). Which of the following is an appropriate interpretation of the 95% confidence interval? Group of answer choices
a. There is a 95% probability that the proportion of all nicotine patch users who would report no smoking incidents in the following year is between 17.4% and 24.8%.
b. We can be 95% confident that the proportion of all nicotine patch users who would report no smoking incidents in the following year is between 17.4% and 24.8%.
c. We can be 95% confident that the proportion of the sample who would report no smoking incidents in the following year is between 17.4% and 24.8%.
d. 95% of samples will have between 17.4% and 24.8% who would report no smoking incidents in the following year.
Answer:
b. We can be 95% confident that the proportion of all nicotine patch users who would report no smoking incidents in the following year is between 17.4% and 24.8%.
Step-by-step explanation:
The confidence interval is an estimation for the true population parameter, calculated from the information of a sample of this population.
The parameter of the population will be within this interval with a certain degree of confidence.
a. There is a 95% probability that the proportion of all nicotine patch users who would report no smoking incidents in the following year is between 17.4% and 24.8%.
Incorrect. The confidence interval gives only the probability that the true proportion (or population proportion) is within 17.4% and 24.8%, not the proportion of individual samples.
b. We can be 95% confident that the proportion of all nicotine patch users who would report no smoking incidents in the following year is between 17.4% and 24.8%.
Correct.
c. We can be 95% confident that the proportion of the sample who would report no smoking incidents in the following year is between 17.4% and 24.8%.
Incorrect. The confidence interval does not give information about another samples.
d. 95% of samples will have between 17.4% and 24.8% who would report no smoking incidents in the following year.
Incorrect. The confidence interval does not give information about another samples or sampling distributions.
what would be the value of $150 after eight years if you earn 12 %interest per year
Answer:
$371.39
Step-by-step explanation:
150 * .12+150 = 168
168 * .12+168 = 188.16
188.16 * .12+188.16 = 210.7392
210.7392 * .12+210.7392 = 236.027904
264.3512525 * .12+264.3512525 = 296.0734028
296.0734028 * .12+296.0734028 = 331.6022111
331.6022111 * .12+331.6022111 = 371.3944764
you multiply your current number by 12% and add that to the number, the last number i rounded for the answer as you can see
A certain forest covers an area of 2600 km2. Suppose that each year this area decreases by 7.75%. What will the area be like after 15 years?
Answer:
After 15 years, the area will be of 775.3 km²
Step-by-step explanation:
The equation for the area of the forest after t years has the following format.
[tex]A(t) = A(0)(1-r)^{t}[/tex]
In which A(0) is the initial area and r is the yearly decrease rate.
A certain forest covers an area of 2600 km2.
This means that [tex]A(0) = 2600[/tex]
Suppose that each year this area decreases by 7.75%.
This means that [tex]r = 0.0775[/tex]
So
[tex]A(t) = 2600(1-0.0775)^{t}[/tex]
[tex]A(t) = 2600(0.9225)^{t}[/tex]
What will the area be like after 15 years?
This is [tex]A(15)[/tex]
[tex]A(t) = 2600(0.9225)^{t}[/tex]
[tex]A(15) = 2600(0.9225)^{15} = 775.3[/tex]
After 15 years, the area will be of 775.3 km²
Answer:
[tex] A(t) = 2600 (1-0.0775)^t = 2600 (0.9225)^t [/tex]
And since the question wants the value for the area at t = 15 years from know we just need to replace t=15 in oir model and we got:
[tex] A(15) = 2600 (0.9225)^{15} = 775.299[/tex]
So then we expect about 775.299 km2 remaining for the area of forests.
Step-by-step explanation:
For this case we can use the following model to describe the situation:
[tex] A = A_o (1 \pm r)^{t}[/tex]
Where [tex]A_o = 2600 km^2[/tex] represent the initial area
[tex] r =-0.0775[/tex] represent the decreasing rate on fraction
A represent the amount of area remaining and t the number of years
So then our model would be:
[tex] A(t) = 2600 (1-0.0775)^t = 2600 (0.9225)^t [/tex]
And since the question wants the value for the area at t = 15 years from know we just need to replace t=15 in oir model and we got:
[tex] A(15) = 2600 (0.9225)^{15} = 775.299[/tex]
So then we expect about 775.299 km2 remaining for the area of forests.