Answer:
The different single-scoop ice-cream cones can you buy from this vendor=12
Step-by-step explanation:
Given that a vendor sells ice cream from a cart on the boardwalk. He offers vanilla, chocolate, strawberry, blueberry, and pistachio ice cream, served on either a waffle, sugar, or plain cone.
Different types of icecreams are 4
Types of cones are 3
For each ice cream say we can have either waffle, sugar or plain cone.
i.e. for each type of ice cream, we have 3 different cones.
Hence for 4 types of ice creams we have 3*4=12 different single scoop ice creamcones.
There are 15 different single-scoop ice-cream cones that can be bought from the vendor.
Explanation:To calculate the number of different single-scoop ice-cream cones that can be bought from the vendor, we need to multiply the number of ice cream flavors with the number of cone types. The vendor offers 5 flavors and 3 cone types, so the total number of different single-scoop ice-cream cones that can be bought is 5 x 3 = 15.
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A Chevrolet was purchased for $23, 750 and a 20% down payment was made. Find the amount financed.
They put 20% down so that means 80% was financed ( 100% - 20% = 80%).
Multiply the purchase price by 80%:
23750 x 0.80 = $19,000 was financed.
the units digit of a two digit number is 10. If the digits are reversed, the new number is 9 less than the original number. find the original number
Answer: the units digit has to be a 1 digit number not 2
Step-by-step explanation:
A company uses a combination of three components- A, B and C to create three different drone designs. The first design Glider uses 3 parts of component A and 2 parts of components B. Design Blimp uses 2 parts of component B and C, and the last design, Pilot uses one part of each component. A sample of 75 components, 25 A, 25 B, 25 C, will be used to make prototypes for the various designs. If 30 components are selected at random, what is the likelihood two prototypes of each design can be made?
Answer:
Consider the following calculations
Step-by-step explanation:
Since 1 Blimp uses 2 components of B and C each
=> choosing 2 components of B(remaining after using in other prototypes) for 1st model= 22C2
choosing 2 components of B(remaining after using in other prototypes) for 2nd model= 21C2
choosing 2 components of B(remaining after using in other prototypes) for 3rd model= 20C2
choosing 2 components of B(remaining after using in other prototypes) for 4th model= 19C2
choosing 2 components of B(remaining after using in other prototypes) for 5th model= 18C2
and choosing 2 components of C(remaining after using in other prototypes) = 24C2
Similarly for C
P(5 prototypes of Blimp created)=[(22C2 / 25C2 )*(24C2 / 25C2 )] + [(21C2 / 25C2 )*(23C2 / 25C2 )]+[(20C2 / 25C2 )*(22C2 / 25C2 )]+[(19C2 / 25C2 )*(21C2 / 25C2 )]+[(18C2 / 25C2 )*(20C2 / 25C2 )]
Eight-year-old Terry's performance on an intelligence test is at a level characteristic of an average 5-year-old. In other words, Terry was average for his age group. Which of the following is most likely Terry’s IQ score?
6
4
5
8
Answer:
5
Step-by-step explanation:
Assuming the intelligence test had 10 questions, since Terry was average for his age group, his most likely IQ score is 5
Northern Light Health Partners offers routine physical examinations as part of a health service program for its employees. The exams showed that 8% of the employees needed corrective shoes, 15% needed major dental work, and 3% needed both corrective shoes and major dental work. Based o n this information what is the probability that an employee selected at random will need either corrective shoes or major dental work?
Answer: Our required probability is 20%.
Step-by-step explanation:
Since we have given that
Probability of the employees needed corrective shoes = 8%
Probability of the employees needed major dental work = 15%
Probability of employees needed both corrective shoes and major dental work = 3%
So, we need to find the probability that an employee will need either corrective shoes or major dental work.
So, it becomes,
[tex]P(S\cup D)=P(S)+P(D)-P(S\cap D)\\\\P(S\cup D)=8+15-3\\\\P(S\cup D)=23-3\\\\P(S\cup D)=20\%[/tex]
Hence, our required probability is 20%.
Helen plays basketball. For free throws, she makes the shot 78% of the time. Helen must now attempt two free throws. C = the event that Helen makes the first shot. P(C) = 0.78. D = the event Helen makes the second shot. P(D) = 0.78. The probability that Helen makes the second free throw given that she made the first is 0.86. What is the probability that Helen makes both free throws?
Answer:
0.6708 or 67.08%
Step-by-step explanation:
Helen can only make both free throws if she makes the first. The probability that she makes the first free throw is P(C) = 0.78, now given that she has already made the first one, the probability that she makes the second is P(D|C) = 0.86. Therefore, the probability of Helen making both free throws is:
[tex]P(C+D) = P(C) *P(D|C) = 0.78*0.86\\P(C+D) = 0.6708[/tex]
There is a 0.6708 probability that Helen makes both free throws.
5) The Queen Mary is a large cruise ship that has 8, 123, 700 cubic feet of cargo space. Express this number in scientific notation.
Explain in words how you converted the number of cubic feet into scientific notation?
Help!!!
Given the function f(x) = x/3 + 7. Find the input for which f(x) = 13
[tex]
\begin{cases}
f(x)=x/3+7\\
f(x)=13
\end{cases}
[/tex]
We find that
[tex]13=x/3+7\Longrightarrow x=\boxed{18}[/tex]
Hope this helps.
a math teacher claims that she has developed a review course that increases the score of students on the math portion of a college entrance exam. based on the data from the administrator of the exam, scores are normally distributed with u=514. the teacher obtains a random sample of 2000 students. puts them through the review class and finds that the mean math score of the 2000 is 520 with a standard deviation of 119. A.state the null and alternative hypotheses. B.test the hypothesis at the a=.10 level of confidence. is a mean math score of 520 significantly higher than 514? find the test statistic, find P-value. is the sample statistic significantly higher? C. do you think that a mean math score of 520 vs 514 will affect the decision of a school admissions adminstrator? in other words does the increase in the score have any practical significance? D. test the hypothesis at the a=.10 level of confidence with n=350 students. assume that the sample mean is still 520 and the sample standard deviation is still 119. is a sample mean of 520 significantly more than 514? find test statistic, find p value, is the sample mean statisically significantly higher? what do you conclude about the impact of large samples on the p-value?
Answer:
A.
[tex]H_0: \mu\leq514\\\\H_1: \mu>514[/tex]
B. Z=2.255. P=0.01207.
C. Although it is not big difference, it is an improvement that has evidence. The scores are expected to be higher in average than without the review course.
D. P(z>0.94)=0.1736
Step-by-step explanation:
A. state the null and alternative hypotheses.
The null hypothesis states that the review course has no effect, so the scores are still the same. The alternative hypothesis states that the review course increase the score.
[tex]H_0: \mu\leq514\\\\H_1: \mu>514[/tex]
B. test the hypothesis at the a=.10 level of confidence. is a mean math score of 520 significantly higher than 514? find the test statistic, find P-value. is the sample statistic significantly higher?
The test statistic Z can be calculated as
[tex]Z=\frac{M-\mu}{s/\sqrt{N}} =\frac{520-514}{119/\sqrt{2000}}=\frac{6}{2.661}=2.255[/tex]
The P-value of z=2.255 is P=0.01207.
The P-value is smaller than the significance level, so the effect is significant. The null hypothesis is rejected.
Then we can conclude that the score of 520 is significantly higher than 514, in this case, specially because the big sample size.
C. do you think that a mean math score of 520 vs 514 will affect the decision of a school admissions adminstrator? in other words does the increase in the score have any practical significance?
Although it is not big difference, it is an improvement that has evidence. The scores are expected to be higher in average than without the review course.
D. test the hypothesis at the a=.10 level of confidence with n=350 students. assume that the sample mean is still 520 and the sample standard deviation is still 119. is a sample mean of 520 significantly more than 514? find test statistic, find p value, is the sample mean statisically significantly higher? what do you conclude about the impact of large samples on the p-value?
In this case, the z-value is
[tex]Z=\frac{520-514}{s/\sqrt{n}} =\frac{6}{119/\sqrt{350}} =\frac{6}{6.36} =0.94\\\\P(z>0.94)=0.1736>\alpha[/tex]
In this case, the P-value is greater than the significance level, so there is no evidence that the review course is increasing the scores.
The sample size gives robustness to the results of the sample. A large sample is more representative of the population than a smaller sample. A little difference, as 520 from 514, but in a big sample leads to a more strong evidence of a change in the mean score.
Large samples give more extreme values of z, so P-values that are smaller and therefore tend to be smaller than the significance level.
A math teacher claims to have developed a review course that increases the score of students on the math portion of a college entrance exam. The null hypothesis is that there is no difference in scores before and after the course, while the alternative hypothesis is that the course increases the scores. The hypothesis is tested at the 0.10 level of confidence by calculating the z-score and finding the p-value. The impact of the score increase on the decision of a school admissions administrator depends on their criteria. The hypothesis is then tested again with a larger sample size, following the same steps as earlier.
Explanation:A. The null hypothesis (H0) is that there is no difference in scores before and after the review course, while the alternative hypothesis (H1) is that the review course increases the scores on the math portion of the college entrance exam.
B. To test the hypothesis at the 0.10 level of confidence, we calculate the z-score and find the p-value. The test statistic is z = (sample mean - population mean) / (population standard deviation / sqrt(sample size)). The p-value is the probability of obtaining a test statistic as extreme as the observed value, assuming the null hypothesis is true. If the p-value is less than the significance level, we reject the null hypothesis and conclude that the mean math score is significantly higher than 514.
C. Whether a mean math score increase from 514 to 520 has any practical significance depends on the requirements set by the school admissions administrator. If the administrator determines that a small increase in score does not significantly impact their decision, then the increase may not have practical significance. However, if the administrator considers even small improvements in score as valuable, then the increase may have practical significance.
D. To test the hypothesis at the 0.10 level of confidence with a sample size of 350 students, we follow the same steps as in part B to calculate the test statistic and p-value. If the p-value is less than the significance level, we reject the null hypothesis and conclude that the sample mean of 520 is significantly higher than 514. Large sample sizes can result in smaller p-values, indicating stronger evidence against the null hypothesis.
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A line has a slope of -5/8. What is the slope of the line parallel to it; and what is the slope of the line perpendicular to it?
Final answer:
A line parallel to one with a slope of -5/8 has the same slope, -5/8, while the slope of a line perpendicular to it is the negative reciprocal, which is 8/5.
Explanation:
The slope of a line that is parallel to a given line is identical to that of the given line. Therefore, a line parallel to one with a slope of -5/8 would also have a slope of -5/8. However, a line that is perpendicular to the given line would have a slope that is the negative reciprocal of the original slope.
To find the negative reciprocal, you flip the fraction and change the sign. So, the slope of a line perpendicular to a line with a slope of -5/8 would be 8/5.
If you roll a pair of fair dice, what is the probability of each of the following? (round all answers to 4 decimal places, .XXXX)
a) getting a sum of 1?
b) getting a sum of 5?
c) getting a sum of 12?
Answer: The required probabilities are
[tex](a)~0\\\\(b)\dfrac{1}{9},\\\\(c)~\dfrac{1}{36}.[/tex]
Step-by-step explanation: Given that a pair of fair dice is rolled.
We are to find the probability of getting
(a) getting a sum of 1.
b) getting a sum of 5.
c) getting a sum of 12.
Let S be the sample space for the experiment of rolling a pair of fair dice.
Then, S = {(1,1), (1,2), (1,3), (1, 4), (1,5), (1,6), . . . , (6,5), (6,6)}.
And, n(S) =36.
(a) Let E denote the event of getting a sum of 1.
Since the sum of the numbers on two dice is minimum 2, so
E = { } ⇒ n(E) = 0.
Therefore, the probability of event E is
[tex]P(E)=\dfrac{n(E)}{n(S)}=\dfrac{0}{36}=0.[/tex]
(b) Let F denote the event of getting a sum of 5.
Then,
F = {(1,4), (2,3), (3,2), (4,1)} ⇒ n(F) = 4.
Therefore, the probability of event F is
[tex]P(F)=\dfrac{n(F)}{n(S)}=\dfrac{4}{36}=\dfrac{1}{9}.[/tex]
(c) Let G denote the event of getting a sum of 12.
Then,
G = {(6,6)} ⇒ n(G) = 1.
Therefore, the probability of event G is
[tex]P(G)=\dfrac{n(G)}{n(S)}=\dfrac{1}{36}.[/tex]
Thus, the required probabilities are
[tex](a)~0\\\\(b)\dfrac{1}{9},\\\\(c)~\dfrac{1}{36}.[/tex]
The probability of getting a sum of 1, the sum of 5, and the sum of 12 are 0, 1/9, and 1/36 respectively
Probability is the likelihood or chance that an event will occur.
For a pair of rolled dice, the total outcome will be 6² = 36
Probability = Expected outcome/Total outcome
a) Pr(getting a sum of 1) = 0/36 = 0
Note that since we have a pair of dice, the least sum we can have is 2
b) The event for getting a sum of 5 are (1, 4), (4, 1), (3, 2), (2, 3)
n(E) = 4
Pr( getting a sum of 5) = 4/36 = 1/9
c) The event for getting a sum of 12 are (6, 6)
n(E) = 1
Pr( getting a sum of 12) = 1/36
Hence the probability of getting a sum of 1, the sum of 5, and the sum of 12 are 0, 1/9, and 1/36 respectively
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Let U represent the set of people in a certain community who were asked if they subscribe to an information source. Let D ={ x ∈ U | x subscribes to The Daily Informer } and N ={ x ∈ U | x subscribes to News Magazine } (Assume these sets are not disjoint.) Write the set that represents the set of people surveyed who subscribe to exactly one of the two news sources given.
a. N ∪ D
b. ( N c ∩ D ) ∪ ( N ∩ D c )
c. N c ∩ D
d. ( N ∩ D )c
e. ( N c ∪ D ) ∩ ( N ∪ D c )
Answer:
b. [tex]( N^c \cap D ) \cup ( N \cap D^c )[/tex]
Step-by-step explanation:
Here,
N = { x ∈ U | x subscribes to News Magazine },
D = { x ∈ U | x subscribes to The Daily Informer },
Such that N∩D ≠ ∅,
Now, the number of the set of people surveyed who subscribe only News Magazine
= Subscribe News Magazine and not Daily Informer
= [tex]N\cap D^c[/tex]
Similarly, the number of the set of people surveyed who subscribe only Daily informer
= Subscribe Daily informer and not subscribe News Magazine
= [tex]D\cap N^c[/tex]
Hence, the set of people surveyed who subscribe to exactly one of the two news sources given
= Subscribe News Magazine and not subscribe Daily Informer or subscribe Daily informer and not subscribe News Magazine
[tex]= (N\cap D^c)\cup (D\cap N^c)[/tex]
[tex]=( N^c \cap D ) \cup ( N \cap D^c )[/tex]
The set of people who subscribe to exactly one of the two news sources is represented by ( Nc ∩ D ) ∪ ( N ∩ Dc ). This includes people who subscribe to The Daily Informer but not News Magazine and those who subscribe to News Magazine but not The Daily Informer.
Explanation:The set that represents the set of people surveyed who subscribe to exactly one of the two news sources, The Daily Informer (D) or News Magazine (N), can be written as ( Nc ∩ D ) ∪ ( N ∩ Dc ). The notation Nc represents the set of all people who do not subscribe to News Magazine and Dc represents the set of all people who do not subscribe to The Daily Informer. Thus, Nc ∩ D represents those who subscribe to The Daily Informer but not News Magazine, while N ∩ Dc represents those who subscribe to News Magazine but not The Daily Informer. The union of these two sets, denoted by ∪, represents the complete set of people who subscribe to exactly one of the two news sources.
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A plane traveled from California and back. It took one hour on the way out than it did on the way back. The plane's average speed out wad 300 mph. The speed on the way back was 350mph. How many hours did the trip out take?
Answer:
6 hours
Step-by-step explanation:
You can set up a chart:
R T D
O 300 1+x ~ O=out B=back R=Rate/Speed T=time D=distance
B 350 x ~
The distance is the same. On the way out the speed was 300 mph, so you can see where that is on the chart. And the speed back was 350 mph. It took an extra hour on the way out than on the way back, so you can call the way back's time x and then say the way out's time was x+1.
It's important to know that R times T = D
Therefore we can say 300(x+1)=D and 350(x)=D
Since the distance is the same, 300(x+1)=350(x)
Solve:
300(x+1)=350x
300x+300=350x
50x=300
x=6
Therefore, the trip out took 6 hours.
I hope that helps! :)
Answer:the trip out took 7 hours
Step-by-step explanation:
Distance travelled = speed × time
Let t represent the time it took the plane to travel from California.
It took one hour on the way out than it did on the way back. This means that the time it took the plane to return would be t - 1
The plane's average speed out ward 300 mph. Distance travelled when going out would be
300 × t = 300 = 300t
The speed on the way back was 350mph. Distance travelled when returning would be
350(t - 1) = 350t - 350
Since the distance is the same, it means that
300t = 350t - 350
350t - 300t = 350
50t = 350
t = 350/50 = 7
Time taken to go out is 7 hours
Search the Internet to find the density of surface seawater in kilograms/cubic meter.
Answer:
1020–1029 kg/m3
Step-by-step explanation:
The density of seawater at the surface of the ocean varies from 1,020 to 1,029 kilograms per cubic meter.
The density of seawater is a function of temperature, salinity, and pressure.
To measure the density of ocean water a sample of sea water is collected and brought into the laboratory to be measured. Salinity, temperature and pressure are measured to find density. The formula for measuring density is p(density) = m(mass) / V(volume).
Answer:
1020 to 1029.
Step-by-step explanation:
I searched it like it says in the unit actinity course Geometry semester B.
Type the correct answer in each box. A circle is centered at the point (5, -4) and passes through the point (-3, 2). The equation of this circle is (x + )2 + (y + )2 = .
Answer:
[tex](x+ \boxed{-5})^2+(y+\boxed4)^2=\boxed{100}[/tex]
Step-by-step explanation:
Given:
Center of circle is at (5, -4).
A point on the circle is [tex](x_1,y_1)=(-3, 2)[/tex]
Equation of a circle with center [tex](h,k)[/tex] and radius 'r' is given as:
[tex](x-h)^2+(y-k)^2=r^2[/tex]
Here, [tex](h,k)=(5,-4)[/tex]
Radius of a circle is equal to the distance of point on the circle from the center of the circle and is given using the distance formula for square of the distance as:
[tex]r^2=(h-x_1)^2+(k-y_1)^2[/tex]
Using distance formula for the points (5, -4) and (-3, 2), we get
[tex]r^2=(5-(-3))^2+(-4-2)^2\\r^2=(5+3)^2+(-6)^2\\r^2=8^2+6^2\\r^2=64+36=100[/tex]
Therefore, the equation of the circle is:
[tex](x-5)^2+(y-(-4))^2=100\\(x-5)^2+(y+4)^2=100[/tex]
Now, rewriting it in the form asked in the question, we get
[tex](x+ \boxed{-5})^2+(y+\boxed4)^2=\boxed{100}[/tex]