Answer:
[tex] 7p + 5 \le 50 [/tex]
Step-by-step explanation:
The number of pizzas is p.
One pizza costs $7.
p number of pizzas cost 7p.
The total cost is the cost of the pizzas plus the $5 tip.
The total cost is 7p + 5
He can spend $50 or less, so the total cost must be less than or equal to $50.
Answer: [tex] 7p + 5 \le 50 [/tex]
What is the value of the interquartile range of the data below?
+
10
20
30
40
50
A(6)
B(12)
C(14)
D(24)
What is the percent of 18% of 87
The answer is 15.66
Answer:
15.66
Step-by-step explanation:
[tex]18\% \: of \: 87 \\ \\ = \frac{18}{100} \times 87 \\ \\ = 0.18 \times 87 \\ \\ = 15.66 \\ [/tex]
A random sample of n = 83 measurements is drawn from a binomial population with probability of success 0.4. Complete parts a through d below.
a. Give the mean and standard deviation of the sampling distribution of the sample proportion, p. The mean of the sampling distribution of p is The standard deviation of the sampling distribution of p is (Round to four decimal places as needed.)
b. Describe the shape of the sampling distribution of p. 0 A The shape of the sampling distribution of p is approximately normal because the sample size is small. The shape of the sampling distribution of p is approximately normal because the sample size is large The shape of the sampling distribution of p is approximately uniform because the sample size is smal ○ C. O D. The shape of the sampling distribution of p is approximately uniform because the sample size is large.
c. Calculate the standard normal z-score corresponding to a value of p=0.41. The standard normal z-score corresponding to a value of p: 041 is . Round to two decimal places as needed.) Finn)
Answer:
a) The mean of the sampling distribution of p is 0.4.
The standard deviation of the sampling distribution of p is 0.0538.
b) The shape of the sampling distribution of p is approximately normal because the sample size is large.
c) z=0.19
Step-by-step explanation:
We have a random sample of size n=83, drawn from a binomial population with proabiliity p=0.4. We have to compute the characteristic of the sampling distribution of the sample proportion.
a) The mean of the sampling distribution is equal to the mean of the distribution p:
[tex]\mu_p=p=0.4[/tex]
The standard deviation of the sampling distribution is:
[tex]\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.4*0.6}{83}}=\sqrt{0.0029}=0.0538[/tex]
b) The shape of sampling distribution with enough sample size tend to be approximately normal. In this case, n=83 is big enough for a binomial distribution.
c) The z-score fof p-0.41 can be calculated as:
[tex]z=\dfrac{p-\mu_p}{\sigma_p}=\dfrac{0.41-0.4}{0.0538}=\dfrac{0.01}{0.0538}=0.19[/tex]
The mean of the sampling distribution is 0.4 and the standard deviation is 0.0538.
How to solve the sampling distribution?From the information given, the sample is drawn from a binomial population with probability of success 0.4. Therefore, the mean is 0.4.
The standard deviation will be:
= [(✓0.4 × ✓0.6) / ✓83)]
= 0.0538
Furthermore, the shape of the sampling distribution of p is approximately normal because the sample size is large.
The standard normal z-score will be:
= (0.41 - 0.4) / 0.0538
= 0.10/0.0538
= 0.19.
Therefore, the standard normal z-score is 0.19.
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In a study comparing various methods of gold plating, 7 printed circuit edge connectors were gold-plated with control-immersion tip plating. The average gold thickness was 1.5 μm, with a standard deviation of 0.25 μm. Five connectors were masked and then plated with total immersion plating. The av- erage gold thickness was 1.0 μm, with a standard deviation of 0.15 μm. Find a 99% confidence interval for the difference between the mean thicknesses produced by the two methods.
Answer:
99% confidence interval for the difference between the mean thicknesses produced by the two methods is [0.099 μm , 0.901 μm].
Step-by-step explanation:
We are given that in a study comparing various methods of gold plating, 7 printed circuit edge connectors were gold-plated with control-immersion tip plating. The average gold thickness was 1.5 μm, with a standard deviation of 0.25 μm.
Five connectors were masked and then plated with total immersion plating. The average gold thickness was 1.0 μm, with a standard deviation of 0.15 μm.
Firstly, the pivotal quantity for 99% confidence interval for the difference between the population mean is given by;
P.Q. = [tex]\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] ~ [tex]t__n__1+_n__2-2[/tex]
where, [tex]\bar X_1[/tex] = average gold thickness of control-immersion tip plating = 1.5 μm
[tex]\bar X_2[/tex] = average gold thickness of total immersion plating = 1.0 μm
[tex]s_1[/tex] = sample standard deviation of control-immersion tip plating = 0.25 μm
[tex]s_2[/tex] = sample standard deviation of total immersion plating = 0.15 μm
[tex]n_1[/tex] = sample of printed circuit edge connectors plated with control-immersion tip plating = 7
[tex]n_2[/tex] = sample of connectors plated with total immersion plating = 5
Also, [tex]s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }[/tex] = [tex]\sqrt{\frac{(7-1)\times 0.25^{2}+(5-1)\times 0.15^{2} }{7+5-2} }[/tex] = 0.216
Here for constructing 99% confidence interval we have used Two-sample t test statistics as we don't know about population standard deviations.
So, 99% confidence interval for the difference between the mean population mean, ([tex]\mu_1-\mu_2[/tex]) is ;
P(-3.169 < [tex]t_1_0[/tex] < 3.169) = 0.99 {As the critical value of t at 10 degree of
freedom are -3.169 & 3.169 with P = 0.5%}
P(-3.169 < [tex]\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] < 3.169) = 0.99
P( [tex]-3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] < [tex]{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}[/tex] < [tex]3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] ) = 0.99
P( [tex](\bar X_1-\bar X_2)-3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] < ([tex]\mu_1-\mu_2[/tex]) < [tex](\bar X_1-\bar X_2)+3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] ) = 0.99
99% confidence interval for ([tex]\mu_1-\mu_2[/tex]) =
[ [tex](\bar X_1-\bar X_2)-3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] , [tex](\bar X_1-\bar X_2)+3.169 \times {s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] ]
= [ [tex](1.5-1.0)-3.169 \times {0.216\sqrt{\frac{1}{7}+\frac{1}{5} } }[/tex] , [tex](1.5-1.0)+3.169 \times {0.216\sqrt{\frac{1}{7}+\frac{1}{5} } }[/tex] ]
= [0.099 μm , 0.901 μm]
Therefore, 99% confidence interval for the difference between the mean thicknesses produced by the two methods is [0.099 μm , 0.901 μm].
Answer:
The 99% of confidence intervals for difference between the mean thicknesses produced by the two methods.
( 0.17971 , 0.82028)
Step-by-step explanation:
Step:-(i)
Given data the average gold thickness was 1.5 μm, with a standard deviation of 0.25 μ m.
Given sample size n₁ = 7
mean of first sample x₁⁻ =1.5 μ m.
Standard deviation of first sample S₁ = 0.25 μ m
Given data Five connectors were masked and then plated with total immersion plating. The average gold thickness was 1.0 μ m, with a standard deviation of 0.15 μ m.
Given second sample size n₂ = 5
The mean of second sample x⁻₂ = 1.0 μ m
Standard deviation of first sample S₂ = 0.15 μ m
Level of significance ∝ =0.01 or 99%
Degrees of freedom γ = n₁+ n₂ -2 = 7+5-2=10
tabulated value t = 2.764
Step(ii):-
The 99% of confidence intervals for μ₁- μ₂ is determined by
(x₁⁻ - x⁻₂) - z₀.₉₉ Se((x₁⁻ - x⁻₂) , (x₁⁻ - x⁻₂) + z₀.₉₉ Se((x₁⁻ - x⁻₂)
where [tex]se(x^{-} _{1}-x^{-} _{2} ) = \sqrt{\frac{s^2_{1} }{n_{1} } +\frac{s^2_{2} }{n_{2} } }[/tex]
[tex]se(x^{-} _{1}-x^{-} _{2} ) = \sqrt{\frac{0.25^2 }{7 } +\frac{0.15^2 }{5 } } = 0.115879[/tex]
[1.5-1.0 - 2.764 (0.115879) , (1.5+1.0) + 2.764(0.115879 ]
(0.5-0.32029,0.5+0.32029
( 0.17971 , 0.82028)
Conclusion:-
The 99% of confidence intervals for μ₁- μ₂ is determined by
( 0.17971 , 0.82028)
In order to estimate the proportion of students that text while driving, a school administrator selects a simple random sample of students from a list of all students at the school who have parking permits. The students are called to the office, one at a time. The administrator asks each student, “Do you text while driving, even though you are not supposed to?” Based on the survey, the administrator estimates that only 2% of students with parking passes text while driving. What potential bias is present in the design of this survey and what is the likely direction of the bias?
Answer:the answer is (A)
Step-by-step explanation:
Because the survey face-to-face, one at a time, with an administrator of the school the students would likely reply that they do not text and drive even if they actually do. This response bias would cause the sample proportion to underestimate the value of the true proportion of all students at this school that text and drive.
The design of the survey contains response bias, which is likely to underestimate the true proportion of students who text while driving due to students' reluctance to admit to prohibited and risky behaviors to an authority figure.
Explanation:The potential bias present in the design of this survey is response bias.
This type of bias occurs when the way a question is asked leads the respondents to give a certain answer
. In this case, by directly asking the students if they engage in the prohibited behavior of texting while driving, there is a likelihood that students may not answer truthfully due to fear of repercussions, the desire to give socially acceptable answers, or reluctance to admit to risky behaviors in a face-to-face setting with a school authority figure.
The likely direction of the bias is that it would underestimate the true proportion of students who text while driving since students might not want to admit to a behavior that is both dangerous and frowned upon while talking to a school administrator.
Dakota earned $4.50 in interest in Account A and $15 in interest in Account B after 18 months. If the simple interest rate is 3% for Account A and 4% for Account B, which account has the greater principal? Explain.
this is probably easy for most people but i need help ASAP
Answer:
Account B
Step-by-step explanation:
Pls Help if u Know thx
Answer:
x = 42
Step-by-step explanation:
The two angles are complementary so the add to 90 degrees.
x+48 = 90
Subtract 48 from each side
x+48-48=90-48
x = 42
Answer:
The angle x°=42.
Step-by-step explanation:
∠PQS equals 90° because it's a right angle (denoted with the square on the bottom).
∠PQS = ∠PQR + ∠RQS
So, ∠RQS = ∠PQS - ∠PQR where ∠PQR = 48°
Plug in the Values:
∠RQS = 90° - 48° = 42°
What is the range of the data below?
50
60
70
80
90
100
A(22)
B(28)
C(41)
D(42)
Answer:
it d (42)
Step-by-step explanation:
Answer:
d: 42
Step-by-step explanation:
The product of two is 450.the first number is half the second number
Answer:
I don't know you're options but the answer could be 30.
If you solve this you are a genius.... Mr.SMITH had 4 daughters , each daughter had a brother... How many children does Mr. Smith have?
Answer:
He has 5 children
Step-by-step explanation:One boy is a brother to all of them
A number is an arithmetic value that is used to represent a quantity and calculate it. The number of children that Mr. Smith will have is 5.
What are Numbers?A number is an arithmetic value that is used to represent a quantity and calculate it. Numericals are written symbols that represent numbers, such as "3."
As it is mentioned that Mr. Smith had 4 daughters and each of the daughters had a brother. Now, if Mr. Smith had just a single son then he will be a brother to every daughter. Therefore, the number of children that Mr. Smith will have is 5.
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Triangle PQR is reflected over the line y = x.
What is the coordinate of the image point R'?
R (2,4)
R (2, -4)
R(-2,4)
Answer:
-2,4
Step-by-step explanation:
i just did it
Answer:
-2,4
Step-by-step explanation:
i just did it on ed 2020
Hotel Rooms and Beyond: Error-and-erasure correction leveraging the CRT (58 pts) Note: for this entire problem, you can use properties of the Chinese Remainder Theorem that we discussed in lecture, notes, homework, and discussion without proof and without having to specify all the details. Alice is staying at a hotel and she wants to share her room number with Bob by leaving a sequence of notes in a list of pre-arranged locations.
i) There are only 100 possible hotel rooms, labeled 0 to 99.
ii) Alice takes her room number p and computes the remainders y; = p mod pi. The specific p; that she uses are pı = 3, p2 = 5,23 = 7,24 = 11, and ps = 13.
iii) She writes y, on the i-th note and places the note in location i. (Both Alice and Bob know which location corresponds to which numbers i and p.) This sequence of notes can be viewed as a codeword y(p). For example, if Alice is in room 51, she sends the codeword y(51)=(0,1,2,7,12) since 51 mod 3=0, 51 mod 5= 1, 51 mod 7=2, 51 mod 11 = 7, 51 mod 13= 12.
(a) Unfortunately, there is a chance that some of Alice's notes get blown away by the wind. Those missing notes will be treated as erasures and denoted by X It turns out that Alice is in room 51 as above, so she sends the codeword (0,1,2,7,12). Bob gets the received symbols (0,1,2,X,X). Explain how Bob can leverage the CRT to identify Alice's room number. (Alt + A)
(b) Generalizing the previous part, prove that in the case of up to two erasures anywhere, your scheme will always correctly identify the room number. For this part, feel free to just use the properties of the CRT along with the following facts. • Pi
Answer:
Step-by-step explanation:
From the information given we know that
[tex]p \equiv 0 \,\,\,\, \text{mod(3)}\\p \equiv 1 \,\,\,\, \text{mod(5)}\\p \equiv 2 \,\,\,\, \text{mod(7)}\\[/tex]
And we know as well that
[tex]p \equiv x \,\,\,\, \text{mod(11)}\\p \equiv x \,\,\,\, \text{mod(13)}[/tex]
Remember what that the Chinese reminder theorem states.
Theorem:
Let p,q be coprimes, then the system of equations
[tex]x \equiv a \,\,\,\, mod(p)\\x \equiv b \,\,\,\, mod(q)[/tex]
has a unique solution [tex]mod(pq)[/tex].
Now, if you read the proof of the theorem you will notice that if
[tex]q_1 = q^{-1} \,\, mod(p) , p_1 = p^{-1} \,\,mod(q)[/tex]
the the solution looks like this.
[tex]x = aqq_1 + bpp_1[/tex]
Now. you can easily generalize what I just stated for multiple equations and you will see that if you apply the theorem for this case it is straightforward that
[tex]p \equiv 0*35*[35^{-1}]_3+1*21*[21^{-1}]_5+2*15[15^{-1}]_7 \,\,\,\,\,\,\,\, mod(3*5*7)\\p \equiv 1*21*1+2*15*1 \,\,\,\,\,\,\,\,mod(105) \\p \equiv 1*21*1+2*15*1 \,\,\,\,\,\,\,\, \\p \equiv 51[/tex]
Therefore, Alice is in room 51.
(b)
Using the Chinese reminder theorem you need less than 2 erasures. The process is very similar.
Answer:
Step-by-step explanation:
From the information given we know that
And we know as well that
Remember what that the Chinese reminder theorem states.
Theorem:
Let p,q be coprimes, then the system of equations
has a unique solution .
Now, if you read the proof of the theorem you will notice that if
the the solution looks like this.
Now. you can easily generalize what I just stated for multiple equations and you will see that if you apply the theorem for this case it is straightforward that
Therefore, Alice is in room 51.
(b)
Using the Chinese reminder theorem you need less than 2 erasures. The process is very similar.
Step-by-step explanation:
16 < b - 8, if b = 22
Answer:
False
Step-by-step explanation:
Substitute 22 in for b
16 < b - 8
16 < 22 - 8
16 < 14
False, 14 is not greater than 16
The solution:
16 < b - 8
Add 8 to both sides
24 < b
3=9-y solve for variable
Answer:
y=6
Step-by-step explanation:
3=9-y
y+3=9
y=6
Answer:
y=6
Step-by-step explanation:
3=9-y
3-9 = -y
-6 = -y
y=6
g Which of the following is NOT true about the distribution function for a normal distribution? It ranges from 0 to 1. It increases as the quantity increases. It generally has a bell shape when graphed. It returns the probability that the outcome from the normal distribution is a certain quantity or lower.
Final answer:
The incorrect statement about the normal distribution is that its distribution function generally has a bell shape when graphed. This shape relates to the probability density function, not the cumulative distribution function, which actually has an S-shaped curve.
Explanation:
When we discuss the properties of a normal distribution, we are dealing with a continuous probability distribution that is widely used across many fields. The statement in question is identifying characteristics of the cumulative distribution function (CDF) of a normal distribution. The correct attributes of this function are that it ranges from 0 to 1, it increases as the quantity increases, and it returns the probability that the outcome from the normal distribution is a certain quantity or lower. However, the statement that the distribution function generally has a bell shape when graphed is incorrect regarding the CDF. The bell shape is a characteristic of the probability density function (PDF) and not the CDF, which increases from 0 to 1 in an S-shaped curve.
The standard normal distribution, which is a special case of the normal distribution with a mean (μ) of zero and a standard deviation (σ) of one, is used for a variety of applications in psychology, business, engineering, and other fields. Understanding the properties of the normal distribution — particularly the standard normal distribution — is crucial for interpreting data and using statistical methods.
It is also important to note that the integral of the normal distribution across its entirety (from ∞ to ∞) is equal to one. This area under the curve represents the total probability of all outcomes and justifies why the CDF ranges from 0 at the minimum to 1 at the maximum end of the distribution.
The number of bald eagles in a state during the winters from 1996 to 2002 can be modeled by the quartic function
f(x)= -3.357x4 + 35.171x3 - 99.868x2 + 41.098x + 177.252
where x is the number of years since 1996. Find the number of bald eagles in the state in the winter of 2001,
Answer:
f(2x+1)sin(9[tex]F(2x+1)sin(90.9292)\pi[/tex]- 4.3784)
Step-by-step explanation:
The number of bald eagles is 26.
what is the area of the shaded region
Given:
Given that the radius of the circle is 12 cm.
The length of the rectangle is 11 cm.
The width of the rectangle is 5 cm.
We need to determine the area of the shaded region.
Area of the rectangle:
The area of the rectangle can be determined using the formula,
[tex]A_1=length \times width[/tex]
Substituting the values, we have;
[tex]A_1=11\times 5[/tex]
[tex]A_1=55 \ cm^2[/tex]
Thus, the area of the rectangle is 55 square cm.
Area of the circle:
The area of the circle can be determined using the formula,
[tex]A_2=\pi r^2[/tex]
Substituting r = 12, we have;
[tex]A_2= (3.14)(12)^2[/tex]
[tex]A_2=452.16 \ cm^2[/tex]
Thus, the area of the circle is 452.16 square cm.
Area of the shaded region:
The area of the shaded region can be determined by subtracting the area of the rectangle from the area of the circle.
Thus, we have;
Area = Area of the circle - Area of the rectangle.
Substituting the values, we have;
[tex]Area=452.16-55[/tex]
[tex]Area=397.16 \ cm^2[/tex]
Thus, the area of the shaded region is 397.16 square cm.
HELPPP ASAPP PLZZ:) Question and answers are in the image.
Answer:
C. all real numbers except 3/4
Step-by-step explanation:
f(x) and g(x) are both defined for all real numbers. However, the ratio f/g will be undefined where g(x) = 0. That occurs when ...
4x -3 = 0
4x = 3 . . . . . add 3
x = 3/4 . . . . . divide by 4
The value of x = 3/4 makes f/g undefined, so must be excluded from the domain.
Find product (7m+8)(4m +1)
Step-by-step explanation:
[tex](7m + 8)(4m + 1) \\ = 7m(4m + 1) + 8(4m + 1) \\ = 28 {m}^{2} + 7m + 32m + 8 \\ \purple { \bold{= 28 {m}^{2} + 39m + 8}}[/tex]
Answer:
You can use foil or the box method.
Step-by-step explanation:
Sorry it looks so bad :P
The maintenance department at the main campus of a large state university receives daily requests to replace fluorecent lightbulbs. The distribution of the number of daily requests is bell-shaped and has a mean of 63 and a standard deviation of 10. Using the 68-95-99.7 rule, what is the approximate percentage of lightbulb replacement requests numbering between 63 and 83
Answer:
47.5% of lightbulb replacement requests numbering between 63 and 83
Step-by-step explanation:
The Empirical Rule(68-95-99.7 rule) states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 63
Standard deviation = 10
What is the approximate percentage of lightbulb replacement requests numbering between 63 and 83
63 is the mean
83 = 63 + 2*20
So 83 is two standard deviations above the mean.
The normal distribution is symmetric, so 50% of the measures are above the mean and 50% below the mean.
Of those above the mean, 95% are within 2 standard deviations of the mean.
So
0.5*95% = 47.5%
47.5% of lightbulb replacement requests numbering between 63 and 83
The area of a rectangle is 100cm2. After a dilation, the area is 25cm2. What is the scale factor?
Answer:
The scale factor is 1/2
Step-by-step explanation:
we know that
A dilation is a non rigid transformation that produces similar figures
If two figures are similar, then the ratio of its areas is equal to the scale factor squared
Let
z ---> the scale factor
x ---> the area after dilation
y ---> the original area
[tex]z^2=\frac{x}{y}[/tex]
we have
[tex]x=25\ cm^2\\y=100\ cm^2[/tex]
substitute
[tex]z^2=\frac{25}{100}[/tex]
[tex]z^2=\frac{1}{4}[/tex]
[tex]z=\frac{1}{2}[/tex]
answer like gauss 1+3+5+7+...=999
1+3+5+7+...+999 =
= 1+2+3+4+...+500
+1+2+3+...+499
= 2·(1+2+3+...+499) + 500
= 2·(499·500)/2 + 500
= 499·500 + 500
= 500·(499 + 1)
= 500·500
= 250.000
To find the sum of the series 1 + 3 + 5 + 7 + ... + 999, we recognize this is the sum of the first 500 odd numbers. Using the property that the sum of the first n odd numbers is n², we find the sum to be 500², which is 250000.
Summation of Odd Numbers:
To find the sum of the series 1 + 3 + 5 + 7 + ... + 999 using Gauss's method, we first recognize that this sequence is a series of the first 500 odd numbers.
An interesting property to note is that the sum of the first n odd numbers is n².
For example:
→ 1 = 1 (which is 1¹)
→ 1 + 3 = 4 (which is 2²)
→ 1 + 3 + 5 = 9 (which is 3³)
And so on...
In general, the [tex]n_{th[/tex] odd number can be expressed as 2n - 1.
For 999, to find its position in the series, note that 999 is the 500th odd number (since 2*500 - 1 = 999).
Thus, the sum of the first 500 odd numbers is:
→ Sum = 500²
= 250000
→ So, the sum of the series 1 + 3 + 5 + 7 + ... + 999 is 250000.
In the United States, 36 percent of the people have a blood type that is A positive. From a random sample of 150 people from Norway, 66 had a blood type that was A positive. Consider a hypothesis test to investigate whether the proportion of people in Norway with a blood type of A positive is different from that in the United States.
Determine the standard deviation used to calculate the test statistic for the one-sample z-test.
Final answer:
The standard deviation used to calculate the test statistic for the one-sample z-test, when investigating the proportion of people in Norway with A positive blood type against the U.S. proportion, is 0.0392.
Explanation:
To calculate the standard deviation used to calculate the test statistic for a one-sample z-test in this scenario, where we are testing whether the proportion of people in Norway with a blood type of A positive is different from that in the United States, we use the formula for the standard deviation of a proportion, which is [tex]\(\sqrt{\frac{p(1-p)}{n}}\)[/tex], where p is the proportion in the population (0.36 in this case, representing 36%), and n is the sample size (150 in this case).
Plugging in the values: [tex]\(\sqrt{\frac{0.36(1-0.36)}{150}}\) = \(\sqrt{\frac{0.36(0.64)}{150}}\) = \(\sqrt{\frac{0.2304}{150}}\) = \(\sqrt{0.001536}\) = 0.0392.[/tex]
So, the standard deviation used to calculate the test statistic for this hypothesis test is 0.0392.
The standard deviation used to calculate the test statistic for the one-sample z-test is approximately 0.0379.
To determine the standard deviation for the one-sample z-test, we use the formula for the standard deviation of a sample proportion, which is given by:
[tex]\[ \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} \][/tex]
Given that the population proportion p of people with A positive blood type in the United States is 0.36, and the sample size n from Norway is 150, we can plug these values into the formula:
[tex]\[ \sigma_{\hat{p}} = \sqrt{\frac{0.36(1-0.36)}{150}} \][/tex]
[tex]\[ \sigma_{\hat{p}} = \sqrt{\frac{0.36 \times 0.64}{150}} \][/tex]
[tex]\[ \sigma_{\hat{p}} \approx 0.0379 \][/tex]
Therefore, the standard deviation used in the calculation of the test statistic for the one-sample z-test is approximately 0.0379.
Traders often buy foreign currency in hope of making money when the currency's value changes. For example, on a particular day, one U.S. dollar could purchase 0.8167 Euros, and one Euro could purchase 145.8038 yen. Let f (x )represent the number of Euros you can buy with x dollars, and let g (x )represent the number of yen you can buy with x Euros.
If you have $100, you can calculate the number of Euros by plugging x = 100 into the f(x) function: f(100) = 0.8167 * 100 = 81.67 Euros. Therefore, with $100, you can buy 81.67 Euros.
If you possess 50 Euros, you can determine the number of yen by substituting x = 50 into the g(x) function: g(50) = 145.8038 * 50 = 7290.19 yen. Hence, with 50 Euros, you can buy 7290.19 yen.
Traders engage in foreign currency trading to potentially profit from fluctuations in currency values. In this example, the exchange rates are as follows: 1 U.S. dollar buys 0.8167 Euros, and 1 Euro buys 145.8038 yen. We are asked to define two functions: f(x) represents the number of Euros you can buy with x dollars, and g(x) represents the number of yen you can buy with x Euros.
1. To determine the number of Euros you can buy with x dollars, multiply the exchange rate between dollars and Euros by x. In this case, the exchange rate is 0.8167 Euros per dollar. Therefore, the function f(x) is f(x) = 0.8167x.
2. To find the number of yen you can buy with x Euros, multiply the exchange rate between Euros and yen by x. In this case, the exchange rate is 145.8038 yen per Euro. The function g(x) is g(x) = 145.8038x.
For example:
- If you have $100, you can calculate the number of Euros by plugging x = 100 into the f(x) function: f(100) = 0.8167 * 100 = 81.67 Euros. Therefore, with $100, you can buy 81.67 Euros.
- If you possess 50 Euros, you can determine the number of yen by substituting x = 50 into the g(x) function: g(50) = 145.8038 * 50 = 7290.19 yen. Hence, with 50 Euros, you can buy 7290.19 yen.
These functions enable traders to evaluate the quantity of foreign currency they can acquire or exchange based on the prevailing exchange rates.
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Write down all of the prime numbers between 15 and 25.
Answer:
17,19,23
that the prime number
Answer:
A prime number is a number whose only factors are one and itself
Prime numbers between 15 and 25:
17, 19, 23
A certain college classifies students according to credits earned. A studentwith less than 7 credits is a Freshman. At least 7 credits are required to bea Sophomore, 16 to be a Junior and 26 to be classified as a Senior. Write aprogram that calculates class standing from the number of credits earned.
Answer and explanation:
Here is code:
credit = int(input("Enter credits : "))
if credit < 7:
print("You are a Freshman")
elif credit >= 7 and credit < 16:
print("You are a Sophomore")
elif credit >= 16 and credit < 26:
print("You are a Junior")
elif credit >= 26:
print("You are a Senior")
Output: check image
To calculate a student's class standing, use conditional statements to determine the appropriate class based on the number of credits earned.
To write a program that calculates a student's class standing based on the number of credits earned, you can use conditional statements. Here's a step-by-step explanation:
Take the input of the number of credits earned from the user.
Use conditional statements (if, else if) to check the number of credits and assign the appropriate class standing.
If the number of credits is less than 7, then the student is a Freshman. If it is at least 7 and less than 16, then the student is a Sophomore.
If it is at least 16 and less than 26, then the student is a Junior. Otherwise, the student is a Senior.
Display the class standing to the user.
Here's an example pseudocode:
credits = input('Enter the number of credits earned: ')
if credits < 7:
print('Class Standing: Freshman')
elif credits < 16:
print('Class Standing: Sophomore')
elif credits < 26:
print('Class Standing: Junior')
else:
print('Class Standing: Senior')
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50 points! ASAP!
Tyler build a dollhouse for his sister shown in the diagram below. Find the volume of the dollhouse. Explain your method for finding the volume of the dollhouse.
(See attached image)
Answer:
V = 18.75
Method: First find the area of the pyramid on top, then find the area of the rectangular prism below.
V of the pyramid = 3.75 ft³
V of the rectangular prism = 15 ft³
3.75 + 15 = 18.75
Hope this helped!! :)
First, find the volume of the pyramid on top:
(2.5 * 3 * 1.5) / 3 = 3.25
Then, find the volume of the rectangular prism:
2.5 * 3 * 2 = 15
Add the two volumes together:
3.25 + 15 = 18.25
To explain: you found the volume of each separate part and added it together.
A health inspector must visit 3 of 13 restaurants on Monday. In how many way can she pick 3 restaurants?
Step-by-step explanation:
A health inspector must visit 3 of 13 restaurants on Monday.
In how many way can she pick 3 restaurants = 13 C 4 !
nCr ! = [tex]\frac{n!}{r!(n-r)!}[/tex]
13 C 4! = [tex]\frac{13!}{3!(10)!}[/tex]
= [tex]\frac{11.12.13}{2.3}[/tex]
= 286
Health inspector can pick 3 restaurants of 13 restaurants on Monday in 286 ways
Final answer:
There are 286 different ways for a health inspector to pick 3 out of 13 restaurants to visit, calculated using the combination formula C(n, k) = n! / (k!(n - k)!).
Explanation:
The student asked how many different ways a health inspector could pick 3 out of 13 restaurants to visit on Monday. This is a problem of combinatorics, specifically concerning combinations without repetition since the order of choosing restaurants does not matter.
To calculate this, we use the combination formula, which is defined as:
C(n, k) = n! / (k!(n - k)!)
Where:
n is the total number of items to choose from (in this case, 13 restaurants)k is the number of items to choose (in this case, 3 restaurants)Applying the formula:
C(13, 3) = 13! / (3!(13 - 3)!) = 13! / (3! * 10!) = (13*12*11) / (3*2*1) = 286
So, the health inspector has 286 different ways to pick 3 restaurants from a list of 13.
What is the slope that passes through points -4 and 1?
On the planet of Mercury, 4-year-olds average 3.2 hours a day unsupervised. Most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.7 hours and the amount of time spent alone is normally distributed. We randomly survey one Mercurian 4-year-old living in a rural area. We are interested in the amount of time X the child spends alone per day.
The question is incomplete! Complete question along with answer and step by step explanation is provided below.
Question:
On the planet of Mercury, 4-year-olds average 3.2 hours a day unsupervised. Most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.7 hours and the amount of time spent alone is normally distributed. We randomly survey one Mercurian 4-year-old living in a rural area. We are interested in the amount of time X the child spends alone per day.
a. In words, define the random variable X
b. What is X ~N(,)
c. Find the probability that the child spends less than 2 hours per day unsupervised.
d. What percent of the children spend over 12 hours per day unsupervised?
Given Information:
Mean = μ = 3.2 hours
Standard deviation = σ = 1.7 hours
Required Information:
a. In words, define the random variable X
b. X ~N(,) = ?
c. P(X < 2) = ?
d. P(X > 12) = ?
Answer:
a) X is the number of hours in a day that a 4-year-old child spends being unsupervised.
b) X ~N(μ,σ) = X ~N(3.2, 1.7)
c) P(X < 2) = 23.88%
d) P(X > 12) = 0%
Explanation:
a)
Let X is the number of hours in a day that a 4-year-old child spends being unsupervised.
b)
X ~N(μ,σ) = X ~N(3.2, 1.7)
Where 3.2 is the average number of hours that 4-year-old child spends being unsupervised and 1.7 is the standard deviation.
c)
We want to find out the probability that a child spends less than 2 hours per day unsupervised.
P(X < 2) = P(Z < (x - μ)/σ)
P(X < 2) = P(Z < (2 - 3.2)/1.7)
P(X < 2) = P(Z < (- 1.2)/1.7)
P(X < 2) = P(Z < -0.71)
The z-score corresponding to -0.71 is 0.2388
P(X < 2) = 0.2388
P(X < 2) = 23.88%
Therefore, the probability that a child spends less than 2 hours per day unsupervised is 23.88%
d)
We want to find out the probability that a child spends over 12 hours per day unsupervised.
P(X > 12) = 1 - P(X < 12 )
P(X > 12) = 1 - P(X < (x - μ)/σ)
P(X > 12) = 1 - P(X < (12 - 3.2)/1.7)
P(X > 12) = 1 - P(X < 8.8/1.7)
P(X > 12) = 1 - P(X < 5.18)
The z-score corresponding to 5.18 is 1
P(X > 12) = 1 - 1
P(X > 12) = 0
Therefore, the probability that a child spends over 12 hours per day unsupervised is 0%
The question discusses unsupervised hours of Mercurian children and it's a statistics problem involving normal distribution where the average unsupervised time is 3.2 hours with a standard deviation of 1.7 hours.
Explanation:This question is related to statistics, specifically about normal distribution. The mean unsupervised time for 4-year-olds on Mercury is 3.2 hours, with a standard deviation of 1.7 hours. If we randomly survey a 4-year-old Mercurian child living in a rural area, the time they likely spend alone, denoted as X, will range according to this distribution.
Normal distributions can be defined by two parameters: the mean (μ) and the standard deviation (σ). In this case, μ = 3.2 hours and σ = 1.7 hours. Therefore, we can say that the time X spent by this sampled child alone is normally distributed with these parameters.
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