Step-by-step explanation:
Let the two-digit number be 10t+u where t is the tens digit and u the units digit.
-------------------
EQUATION:
t+u = 11
10u+t = 10t+u + 45
-------------
Rearrange the equations:
t+u = 11
9t-9u = -45
-------------
Simplify:
t+u = 11
t-u = -5
---------
Add the equations to solve for "t":
2t = 6
t = 3
--------
Substitute to solve for "u":
3+u = 11
u = 8
A car manufacturer sent out survey cards to owners who had purchased new cars. The survey card only had boxes to check for replies, and no places to write additional comments. What is a possible result of the design of the comment cards?
A.
The results cannot be examined because comment cards were used.
B.
The owners who filled out the cards may not express their true opinion.
C.
People would refuse to fill out the cards.
D.
Not enough cards would be returned.
Answer: I think the answer is B. I don't fully understand the question but it seems like that would be the answer. You might double check though.
Step-by-step explanation:
What is x in this geometry equation?
Answer:
x = 7
Step-by-step explanation:
[tex]m\overset{\frown} {PQ} = m\overset{\frown} {MN}... (given) \\
\therefore Chord \:PQ \cong Chord\: MN\\
\therefore \: 4x + 10 = 38 \\ \therefore \: 4x= 38 - 10\\ \therefore \: 4x= 28\\ \therefore \: x= \frac{28}{4} \\ \huge \red{ \boxed{\therefore \: x= 7}}[/tex]
The net of a triangular pyramid. The base is D, and the sides are A, B, and C.
The net of the triangular pyramid shows the four faces of the pyramid labeled A, B, C, and D. Use the drop-down menus to complete the statements.
The base area is the total area of .
The lateral area is the total area of .
The surface area is the total area of .
Answer:
The base area is the total area of D
The lateral area is the total area of A, B, and C
The surface area is the total area of A, B, C, and D.
Answer:
(d) the base area is the total area
Explain: because i got it right on the test
The circular base of the cone has a radius of 5 cm. The height of the cone is 12 cm and the slant height is 13 cm. What is approximate surface area of the cone? Use 3.14 as pi and round to the nearest whole number!
Answer: 283cm
Step-by-step explanation:
Jordan spends $9.90 for 5 pounds of apples find the unit cost 7 dollars
Answer:
$1.98/ lbs
Step-by-step explanation:
To find the unit cost, take the total cost and divide the the number of units purchased
$9.90/5 lbs
$1.98/ lbs
A researcher is interested in the lengths of brook trout, which are known to be approximately Normally distributed with mean 80 centimeters and standard deviation 5 centimeters. To help preserve brook trout populations, some regulatory standards need to be set for limiting the size of fish that can be caught. What is the probability of catching a brook trout less than 72 centimeters in length
Answer:
[tex]P(X<72)=P(\frac{X-\mu}{\sigma}<\frac{72-\mu}{\sigma})=P(Z<\frac{72-80}{5})=P(z<-1.6)[/tex]
And we can find this probability with the normal standard distirbution or excel and we got:
[tex]P(z<-1.6)=0.055[/tex]
Step-by-step explanation:
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".
Solution to the problem
Let X the random variable that represent the lenghts of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(80,5)[/tex]
Where [tex]\mu=80[/tex] and [tex]\sigma=5[/tex]
We are interested on this probability
[tex]P(X<72)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X<72)=P(\frac{X-\mu}{\sigma}<\frac{72-\mu}{\sigma})=P(Z<\frac{72-80}{5})=P(z<-1.6)[/tex]
And we can find this probability with the normal standard distirbution or excel and we got:
[tex]P(z<-1.6)=0.055[/tex]
The function f(x) = -3x + 200 shows the height in feet of Complete the statements to show the difference in the
a feather dropped from an initial height of 200 feet above mathematical and reasonable domain and range for the
the moon's surface after x seconds.
function. Round any decimals to the nearest integer.
Mathematical range:ys
Reasonable range:
Answer:
Mathematical 200
reasonable 0,200
Step-by-step explanation:
i got it right on the test
You select a marble without looking and then put it back. If you do this 60 times, what is the best prediction possible for the number of times you will pick a blue or a purple marble?
The best prediction for selecting a blue marble 60 times, with replacement, from a bag of 4 blue marbles and 3 white marbles is approximately 34.29 times.
To predict the number of times you will select a blue or purple marble, we need to know the total number of marbles, the number of blue marbles, the number of purple marbles, and the number of trials (which is 60 in this case). Since no information about purple marbles is provided, let's assume you meant blue marbles only for this explanation. In the scenario presented in the question, the probability of drawing a blue marble is 4/7 each time since there are 4 blue marbles and 3 white marbles, totaling 7 marbles.
The probability of drawing a blue marble is 4/7.Each draw is independent because the marble is replaced each time.The predicted number of times a blue marble is drawn in 60 trials is calculated by multiplying the probability by the number of trials which is (4/7) * 60.Therefore, the best prediction for the number of times a blue marble will be selected is approximately 34.29 times (which you might round to 34 times).
write 7.630 in word form.
Answer:
seven and six hundred thirty thousandths
Step-by-step explanation:
the decimal point is when you say and when reading it
The representation of 7.630 in word form is; Seven and six hundred thirty thousandths.
How to identify the place value of a digit in a number?The place values on left of decimal point start as ones, tens, hundreds, and so on.
The place value on right of decimal point starts from point and goes to right as tenths, hundreths and so on
The tens means multiply by 10
The tenth means tenth part of that digit which is that digit divided by 10
Place value of decimal numbers;
The given number is; 7.630
The given number can be written as;
7 and 0.630
Hence, the number can be pronounced as Seven and 630 thousandths.
However, we have,
Seven and six hundred thirty thousandths.
Hence, The representation of 7.630 in word form is; Seven and six hundred thirty thousandths.
Read more on place value;
brainly.com/question/627658
#SPJ5
What is the area of the sector that is not shaded?
Answer:
24
Step-by-step explanation:
A survey of an urban university (population of 25,450) showed that 883 of 1,112 students sampled supported a fee increase to fund improvements to the student recreation center. Using the 95% level of confidence, what is the confidence interval for the proportion of students supporting the fee increase
Answer:
The confidence interval for the proportion of students supporting the fee increase
( 0.77024, 0.81776)
Step-by-step explanation:
Explanation:
Given data a survey of an urban university (population of 25,450) showed that 883 of 1,112 students sampled supported a fee increase to fund improvements to the student recreation center.
Given sample size 'n' = 1112
Sample proportion 'p' = [tex]\frac{883}{1112} = 0.7940[/tex]
q = 1 - p = 1- 0.7940 = 0.206
The 95% level of confidence intervals
The confidence interval for the proportion of students supporting the fee increase
[tex](p-z_{\alpha } \sqrt{\frac{pq}{n} } ,p + z_{\alpha } \sqrt{\frac{pq}{n} } )[/tex]
The Z-score at 95% level of significance =1.96
[tex](0.7940-1.96\sqrt{\frac{0.7940 X 0.206}{1112} } ,0.7940 + 1.96 \sqrt{\frac{0.7940 X 0.206}{1112} } )[/tex]
(0.7940-0.02376 , 0.7940+0.02376)
( 0.77024, 0.81776)
Conclusion:-
The confidence interval for the proportion of students supporting the fee increase
( 0.77024, 0.81776)
A shipping company handles containers in three different sizes: (1) 27 ft3 (3 × 3 × 3), (2) 125 ft3, and (3) 512 ft3. Let Xi (i = 1, 2, 3) denote the number of type i containers shipped during a given week. With μi = E(Xi) and σi2 = V(Xi), suppose that the mean values and standard deviations are as follows:
μ1 = 230 μ2 = 240 μ3 = 120
σ1 = 11 σ2 = 12 σ3 = 7
Assuming that X1, X2, X3 are independent, calculate the expected value and variance of the total volume shipped.
Answer:
Expected value of the total volume shipped = 97,650 ft³
Variance of the total volume shipped = 15,183,265
Standard deviation = 3896.6 ft³
Step-by-step explanation:
The mean of number of type 1, 2 and 3 containers in a week
μ₁ = 230, μ₂ = 240, μ₃ = 120
The standard deviations for the number of type 1, 2 and 3 containers in a week
σ₁ = 11, σ₂ = 12, σ₃ = 7
When independent distributions are combined, the combined mean and combined variance are given through the relation
Combined mean = Σ λᵢμᵢ
(summing all of the distributions in the manner that they are combined)
Combined variance = Σ λᵢ²σᵢ²
(summing all of the distributions in the manner that they are combined)
Volume of each container type
λ₁ = 27 ft³
λ₂ = 125 ft³
λ₃ = 512 ft³
Distribution of total volume shipped
= 27X₁ + 125X₂ + 512X₃
Expected value = Combined Mean = 27μ₁ + 125μ₂ + 512μ₃
= (27×230) + (125×240) + (512×120) = 590
Combined Variance = 27²σ₁² + 125²σ₂² + 512²σ₃²
= (27² × 11²) + (125² × 12²) + (512² × 7²)
= 88,209 + 2,250,000 + 12,845,056
= 15,183,265
Standard deviation = √(15,183,265) = 3896.6 ft³
Hope this Helps!!!
In 2017, the entire fleet of light‑duty vehicles sold in the United States by each manufacturer must emit an average of no more than 92 milligrams per mile (mg/mi) of nitrogen oxides (NOX) and non methane organic gas (NMOG) over the useful life ( 150,000 miles of driving) of the vehicle. NOX + NMOG emissions over the useful life for one car model vary Normally with mean 88 mg/mi and standard deviation 4 mg/mi. (a) What is the probability that a single car of this model emits more than 86 mg/mi of NOX + NMOG? (Enter your answer rounded to four decimal places.)
Answer:
0.6915
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 88, \sigma = 4[/ex]
What is the probability that a single car of this model emits more than 86 mg/mi of NOX + NMOG?
This is 1 subtracted by the pvalue of Z when X = 86. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{86 - 88}{4}[/tex]
[tex]Z = -0.5[/tex]
[tex]Z = -0.5[/tex] has a pvalue of 0.3085
1 - 0.3085 = 0.6915
The answer is 0.6915
mike is flying a kite the horizontal distance to the kite a is about 19 feet the kite is 31 feet from the ground how long is the string
Answer:36.36feet (that rounded but if you want you can round it to 36.4
Step-by-step explanation:
Final answer:
The problem involves using the Pythagorean theorem to find the length of a kite string given the horizontal and vertical distance, which would be the square root of the sum of the squares of 19 feet and 31 feet.
Explanation:
The student's question concerns the length of the string used to fly a kite when given the horizontal distance and the vertical distance from the ground to the kite. To solve this, we apply the Pythagorean theorem, which states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In this problem, the kite string functions as the hypotenuse, the horizontal distance to the kite is one leg, and the vertical distance from the ground to the kite is the other leg.
For Mike's kite:
The horizontal distance (leg a) to the kite = 19 feetThe vertical distance (leg b) from the ground to the kite = 31 feetAccording to the Pythagorean theorem:
[tex]\( c^2 = a^2 + b^2 \\( c^2 = 19^2 + 31^2 \\( c^2 = 361 + 961 \\( c^2 = 1322 \\( c = \\( c \\) = \\( c \\) = \\) is the length of the kite string[/tex].
Therefore, Mike's kite string is approximately \\( c \\) long.
A cheeseburger franchise uses 11 different buns, 4 different cheeses, 3 types of lettuce, and 4 types of tomatoes. Takes 10 minutes to eat each cheeseburger. How long will it take to eat all of the possible cheeseburgers
Answer:
2.2
Step-by-step explanation:
It would take 15840 min to eat all of them that means 264 hours or 11 days
lol why you need this info?
A test of H0: p = 0.5 versus Ha: p > 0.5 has the test statistic z = 1.15. Part A: What conclusion can you draw at the 5% significance level? At the 1% significance level? (6 points) Part B: If the alternative hypothesis is Ha: p ≠ 0.5, what conclusion can you draw at the 5% significance level? At the 1% significance level? (4 points)
Answer:
Part A: The null hypothesis failed to be rejected.
Part B: The null hypothesis failed to be rejected.
Step-by-step explanation:
We have an hypothesis test with null and alternative hypothesis H0: p = 0.5 versus Ha: p > 0.5, which has the test statistic z=1.15.
Part A: If the significance level is 0.05, the conclusion depends on the P-value.
If the P-value is below 5%, the null hypothesis is rejected.
The P-value for this right-tailed tes and z=1.15 is:
[tex]P-value=P(z>1.15)=0.125[/tex]
The P-value is bigger than the significance level, so the effect is not significant and the null hypothesis is failed to be rejected.
Part B: In this case, the significance level is 0.01 and, as the alternative hypothesis is defined with an unequal sign, the test is two-tailed.
This changes the way we calculate the P-value, as we need to compute the two tails.
The P-value is:
[tex]P-value=2P(z>1.15)=0.25[/tex]
The P-value is bigger than the significance level, so the effect is not significant and the null hypothesis is failed to be rejected.
Null hypothesis of a sample suggest that there is no statistical relationship exists in a set of provided single observed variable. The null hypothesis is failed to be rejected for both part A and part B of the question.
Given-Using the hypothesis test we have,
[tex]H_o; p=0.5[/tex]
[tex]H_ap>0.5[/tex]
[tex]z=1.15[/tex]
Null hypothesisNull hypothesis of a sample suggest that there is no statistical relationship exists in a set of provided single observed variable.
Part A: The conclusion to draw at the 5% significance level-
In the right tailed test or upper test the p value for the z score can be given form z table which is,
p-value[tex]=0.125[/tex]
Here the p value 0.125 is bigger than the significance level 0.05. Thus the effect is not significant and the null hypothesis is failed to be rejected.
Part B: If the alternative hypothesis is Ha: p ≠ 0.5, the conclusion to draw at the 5% significance level-
Here for the unequal sign 2-test is performed. The p value for the 2-test is,
p- value[tex]=0.25[/tex]
Here the p value 0.25 is bigger than the significance level 0.05. Thus the effect is not significant and the null hypothesis is failed to be rejected.
Hence, the null hypothesis is failed to be rejected for both part A and part B of the question.
For more about the null hypothesis, follow the link below-
https://brainly.com/question/4454077
Which of the following statements are correct? A. The statement that the random variable X is normally distributed with parameters is often abbreviated . B. If the random variable X is normally distributed with parameters , then . C. The graph of any normal probability density function is symmetric about the mean and bell-shaped, so the center of the bell (point of symmetry) is both the mean of the distribution and the median. D. If the random variable X is normally distributed with parameters , then a large implies that a value of X far from may well be observed, whereas such a value is quite unlikely when is small. E. All of the above statements are correct
Answer:
A. The statement that the random variable X is normally distributed with parameters is often abbreviated.
Step-by-step explanation:
The X is normally distributed with parameters u and is often distributed as X ≈N (u 2 ). The correct answer is a. The statement that the random variable is normally distributed.
The radius of a sphere X is 3 inches, and radius of sphere Y is 9 inches. How many
times larger is the volume of sphere Y compared to sphere X?
The volume of the sphere Y is 27 times the volume of the sphere X
Step-by-step explanation:
Sphere X:
Radius of sphere X = 3 inches
Volume = (4/3) π(27)
Sphere Y:
Radius of sphere Y = 9 inches
Volume = (4/3) π(729)
Volume of sphere Y /volume of sphere X = (4/3) π(729) /(4/3) π(27)
= (729) /(27)
= 27
The volume of the sphere Y is 27 times the volume of the sphere X
3.- In a certain desert region the average number of persons who become seriously ill each year from eating a certain poisonous plant each year is 3.2, determine what is the probability that at least 5 people will become seriously ill in two years (hint: this is a rare event and can be modeled using a Poisson distribution)
Answer:
[tex]P(X\geq 5)=1-P(X<5)=1-P(X\leq 4)=1-[P(X=0)+P(X=1)+P(X=2) +P(X=3) +P(X=4)][/tex]
Using the pmf we can find the individual probabilities like this:
[tex]P(X=0)=\frac{e^{-6.4} 6.4^0}{0!}=0.001662[/tex]
[tex]P(X=1)=\frac{e^{-6.4} 6.4^1}{1!}=0.010634 [/tex]
[tex]P(X=2)=\frac{e^{-6.4} 6.4^2}{2!}=0.034029 [/tex]
[tex]P(X=3)=\frac{e^{-6.4} 6.4^3}{3!}=0.072595 [/tex]
[tex]P(X=4)=\frac{e^{-6.4} 6.4^4}{4!}=0.116151 [/tex]
And replacing we got:
[tex] P(X \geq 5) =0.76493[/tex]
Step-by-step explanation:
Previous concepts
Let X the random variable that represent the number of people that will become sereiosly ill in two years. We know that [tex]X \sim Poisson(\lambda)[/tex]
The probability mass function for the random variable is given by:
[tex]f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...[/tex]
And f(x)=0 for other case.
For this case the value for [tex]\lambda[/tex] would be:
[tex]\lambda = 3.2 \frac{ills}{year} *2 years = 6.4[/tex]
For this distribution the expected value is the same parameter [tex]\lambda[/tex]
[tex]E(X)=\mu =\lambda[/tex]
On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:
[tex]P(X\geq 5)=1-P(X<5)=1-P(X\leq 4)=1-[P(X=0)+P(X=1)+P(X=2) +P(X=3) +P(X=4)][/tex]
Using the pmf we can find the individual probabilities like this:
[tex]P(X=0)=\frac{e^{-6.4} 6.4^0}{0!}=0.001662[/tex]
[tex]P(X=1)=\frac{e^{-6.4} 6.4^1}{1!}=0.010634 [/tex]
[tex]P(X=2)=\frac{e^{-6.4} 6.4^2}{2!}=0.034029 [/tex]
[tex]P(X=3)=\frac{e^{-6.4} 6.4^3}{3!}=0.072595 [/tex]
[tex]P(X=4)=\frac{e^{-6.4} 6.4^4}{4!}=0.116151 [/tex]
And replacing we got:
[tex] P(X \geq 5) =0.76493[/tex]
Answer:
The probability that at least 5 people will become seriously ill in two years is 0.7649.
Step-by-step explanation:
The question mentions that this problem can be modeled using the Poisson distribution so, we will use the formula:
P(X=x) = [(e^-λt)*(λt^x)]/x!
where λ = average number of occurrences per year
t = no. of years
x = number of people
We need to determine P(X≥5) so first we will calculate the probabilities at X=0,1,2,3,4 and subtract them from the total probability i.e. 1 to find P(X≥5).
We have λ = 3.2, t=2 years hence λt = (3.2)(2) = 6.4. So,
P(X=0) = [(e^(-6.4)*(6.4^0)]/0! = 0.00166
P(X=1) = [(e^(-6.4)*(6.4^1)]/1! = 0.01063
P(X=2) = [(e^(-6.4)*(6.4^2)]/2! = 0.03403
P(X=3) = [(e^(-6.4)*(6.4^3)]/3! = 0.07259
P(X=4) = [(e^(-6.4)*(6.4^4)]/4! = 0.011615
P(X≥5) = 1 - P(X<5)
= 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)]
= 1 - (0.00166 + 0.01063 + 0.03403 + 0.07259 + 0.011615)
= 1 - 0.23506
P(X≥5) = 0.7649
The probability that at least 5 people will become seriously ill in two years is 0.7649.
Solve for c
7c + 3 - 7c = 3c - 9
Answer:
4=c
Step-by-step explanation:
7c + 3 - 7c = 3c - 9
Combine like terms
7c-7c+3 = 3c-9
3 = 3c-9
Add 9 to each side
3+9 = 3c-9+9
12 = 3c
Divide each side by 3
12/3 = 3c/3
4 =c
Answe
c=−4/3
Step-by-step explanation:
The mean time required to repair breakdowns of a certain copying machine is 93 minutes. The company which manufactures the machines claims that breakdowns of its new, improved model are easier to fix. To test this claim, sample of 73 breakdowns of the new model were observed, resulting in a mean repair time of 88.8 minutes and a standard deviation of 26.6 minutes. Use significance level α = 0.05. State clearly what are your null and alternative hypothesis, show your work, and state your conclusion.
Answer:
[tex]t=\frac{88.8-93}{\frac{26.6}{\sqrt{73}}}=-1.349[/tex]
[tex]p_v =P(t_{(72)}<-1.349)=0.0908[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, and we can't concluce that the true mean is less than 93 min at 5% of signficance.
Step-by-step explanation:
Data given and notation
[tex]\bar X=88.8[/tex] represent the sample mean
[tex]s=26.6[/tex] represent the sample standard deviation for the sample
[tex]n=73[/tex] sample size
[tex]\mu_o =93[/tex] represent the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the true mean i lower than 93 min, the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 93[/tex]
Alternative hypothesis:[tex]\mu < 93[/tex]
If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{88.8-93}{\frac{26.6}{\sqrt{73}}}=-1.349[/tex]
P-value
The first step is calculate the degrees of freedom, on this case:
[tex]df=n-1=73-1=72[/tex]
Since is a one side test the p value would be:
[tex]p_v =P(t_{(72)}<-1.349)=0.0908[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, and we can't concluce that the true mean is less than 93 min at 5% of signficance.
The null hypothesis is that the mean repair time for the new model is equal to the mean repair time for the previous model. The alternative hypothesis is that the mean repair time for the new model is less than the mean repair time for the previous model. By performing a one-sample t-test, we compare the sample mean repair time to the population mean repair time. Using the calculated t-statistic and the critical t-value, we determine whether to reject or fail to reject the null hypothesis.
Explanation:The null hypothesis, denoted as H0, states that the mean repair time for the new model of copying machine is equal to the mean repair time for the previous model (93 minutes). The alternative hypothesis, denoted as H1, states that the mean repair time for the new model is less than 93 minutes.
To test these hypotheses, we can perform a one-sample t-test. Using the given sample data, we calculate the t-statistic as:
t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))
Using the t-distribution table or a calculator, we find the critical t-value at a significance level of 0.05 and degrees of freedom (sample size - 1). If the calculated t-statistic is less than the critical t-value, we reject the null hypothesis and conclude that the new model has a lower mean repair time. Otherwise, we fail to reject the null hypothesis.
In this case, the calculated t-statistic is:
t = (88.8 - 93) / (26.6 / sqrt(73)) ≈ -1.34
With 72 degrees of freedom, the critical t-value at α = 0.05 is -1.666. Since the calculated t-statistic (-1.34) is greater than the critical t-value (-1.666), we fail to reject the null hypothesis. Therefore, we do not have sufficient evidence to conclude that the new model of copying machine has a significantly lower mean repair time than the previous model.
Learn more about Hypothesis testing here:https://brainly.com/question/34171008
#SPJ11
Which statement is true
A. (X,y) (x-7,y+2) represents a translation 7 units down and 2 units to the right
B. (X,y) (-x,-y) represents a rotation 180° clockwise
C. (X,y) (x+3.5,y+3.5) represents a dilation wuth a scale factor of 3.5
D. (X,y) (-x, y) represents a reflection over the x axis
Answer:
B
Step-by-step explanation:
this is true because if you rotate 180 degrees on a graph both the x and y will always become their opposite.
One side of a right triangle is 10 centimeters. The longest side of the triangle is 26 centimeters. What is the length, in centimeters, of the other side of the triangle?
Solve 2 + 3ex + 2 = 7 for x.
Answer:
The answer is C. x=In(5/3)-2
Step-by-step explanation:
Answer:
C
Step-by-step explanation:
Identify the correlation in the scatterplot.
Positive correlation
Negative correlation
No correlation
Answer:
B
Step-by-step explanation:
Shaun needs to make mortar for brick-laying. It is made in the ratio 7:3 of sand to cement.
If 210 kg of sand is used, how much cement is needed?
{83,73,83,59,24,12,43, 83, 59}
What is the mean, median, mode, and range for the set
Look at the attached picture⤴
Hope it will help uh...:)
Answer:
Mean: 57.667
Median: 59
Mode: 83
Range: 71
Explanation
Mean = sum of values / number of values
[tex]\frac{x1+x2...xn}{n}[/tex] = [tex]\frac{83+73+83+59+24+12+43+83+59}{9}[/tex] = 57.667Median = middle number
values sorted smallest to largest: 12, 24, 43, 59, 59, 73, 83, 83, 83 = 59Mode = number that occurs most often
There are 3 83s, so 83 is the mode = 83Range = largest number minus smallest number
83 - 12 = 71Helena is in charge of selling tickets for the upcoming
play at Willowbrook School. Each drama club member
has been asked to sell tickets to friends and family
members, and Helena wants to know approximately how
many tickets have been sold so far.
Homeroom Data of Tickets Sold:
20, 15, 19, 16, 10
Calculate the mean
Answer:
16
Step-by-step explanation:
You want the mean of {20, 15, 19, 16, 10}.
MeanThe mean is the sum of the numbers, divided by the number of numbers.
(20 +15 +19 +16 +10)/5 = 80/5 = 16
The mean is 16.
Final answer:
The mean of tickets sold by Helena for the Willowbrook School play is calculated by adding the total tickets sold (80) and dividing by the number of homerooms (5), resulting in a mean of 16 tickets.
Explanation:
To calculate the mean of tickets sold for the Willowbrook School play, we need to add up the number of tickets sold in each homeroom and then divide by the number of homerooms. The data given is:
20 tickets15 tickets19 tickets16 tickets10 ticketsAdding these together, we get 20 + 15 + 19 + 16 + 10 = 80 tickets sold in total. As there are 5 sets of data, we divide the total by 5 to find the mean.
Therefore, the mean number of tickets sold is 16 tickets.
A videotape store has an average weekly gross of $1,158 with a standard deviation of $120. Let x be the store's gross during a randomly selected week. If this is a normally distributed random variable, then the number of standard deviations from $1,158 to $1,360 is:
Answer:
The number of standard deviations from $1,158 to $1,360 is 1.68.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 1158, \sigma = 120[/tex]
The number of standard deviations from $1,158 to $1,360 is:
This is Z when X = 1360. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1360 - 1158}{120}[/tex]
[tex]Z = 1.68[/tex]
The number of standard deviations from $1,158 to $1,360 is 1.68.
A certain organization recommends the use of passwords with the following format: consonant commaconsonant, consonant comma vowel comma consonant comma vowel comma consonant comma numberconsonant, vowel, consonant, vowel, consonant, number (for example, prinay 6prinay6). Assume that repeats are allowed. Complete parts (a) and (b). (a) Assuming passwords are not case sensitive, how many such passwords are possible (assume that there are 5 vowels and 21 consonants)?
Answer: (a). Passwords = 48620250
(b). Passwords = 3111696000
Step-by-step explanation:
The question reads that 'A certain organization recommends the use of passwords with the following format':
Consonant, Vowel, Consonant, Consonant, Vowel, Consonant, Number, Number (for example, pinray45).
The number of consonants is 21, the number of vowels is 5 and the numbers are 10 (0 to 9).
Since, there are 21 consonants and 5 vowels and repetition is allowed.
Use the general multiplication rule for each characters in the sequence.
Multiply by 21 consonants, 5 for vowels and 10 for numbers.
Assuming that the passwords are not case sensitive.
i.e. upper case & lower case letters are same.
Thus, the number of possible passwords is:
Passwords = Consonant * Vowel * Consonant * Consonant * Vowel * Consonant * Number * Number
Passwords = 5 x 21 x 21 x 21 x 5 x21 x 10 = 48620250
(b). Assuming that the passwords are case sensitive
Passwords = 10 x 42 x 42 x 42 x 10 x 42 x 10 = 42 x 420³ = 3111696000
cheers i hope this helps!!!!
The total number of possible passwords is 95,887,125,000.
Explanation:To calculate the number of possible passwords, we need to consider the number of choices for each position in the password. In the given format, we have 5 positions for consonants, 2 positions for vowels, and 1 position for a number. Let's break it down:
For the 1st consonant position, we have 21 choices (since there are 21 consonants).
For the 1st vowel position, we have 5 choices (since there are 5 vowels).
For the 2nd consonant position, we have 21 choices.
For the 2nd vowel position, we have 5 choices.
For the 3rd consonant position, we have 21 choices.
For the number position, we have 10 choices (since there are 10 numbers).
For the 4th consonant position, we have 21 choices.
For the 3rd vowel position, we have 5 choices.
For the 5th consonant position, we have 21 choices.
Finally, for the last consonant and the last vowel positions, we have 21 and 5 choices respectively.
To find the total number of possible passwords, we multiply the number of choices for each position:
Total number of passwords = 21 * 5 * 21 * 5 * 21 * 10 * 21 * 5 * 21 * 5 = 95,887,125,000.