Answer:
8x-32
Step-by-step explanation:
Because 8 multiples X and gives 8x and also multiples-4 and gives you -32
:.8x-32
What is the answer to -t+9-4t=59?
Answer:
-10
Step-by-step explanation:
combine like terms -5x+9=59 subtract 9 from both side and get -5x=50 so x=-10
Answer:
-10
Step-by-step explanation:
Step 1: Simplify both sides of the equation.
−t+9−4t=59
−t+9+−4t=59
(−t+−4t)+(9)=59(Combine Like Terms)
−5t+9=59
−5t+9=59
Step 2: Subtract 9 from both sides.
−5t+9−9=59−9
−5t=50
Step 3: Divide both sides by -5.
−5t −5 = 50−5
t=−10
Angle C is inscribed in circle O.
AB is a diameter of circle O.
What is the measure of angle B?
Answer:
The measure of m∠B = 19°.
Step-by-step explanation:
Given
Angle C is inscribed in circle O.m∠A = 71°As we know that an angle inscribed in a semicircle is always a right angle.
So from the given diagram,
m∠C = 90°As we know that the sum of the three interior angles is equal to 180.
so
m∠A + m∠B + m∠C = 180°
71° + m∠B + 90° = 180°
m∠B = 180° - 71° - 90°
m∠B = 19°
Therefore, the measure of m∠B = 19°.
Answer:
Step-by-step explanation: 19
One day, Donnie observes that the wind is blowing at 6 miles per hour. A unladen swallow nesting near Donnie’s house flies three quarters of a mile down the road (in the direction of the wind), turns around, and returns exactly 4 minutes later. What is the airspeed of the unladen swallow? (Here, ‘airspeed’ is the speed that the swallow can fly in still air.) socratic.org
Answer: 11.24 miler per hour
Step-by-step explanation:
Rhada has a 6-pound bag of day . Her craft project requires 6 ounces of clay for each batch of 3 ornaments,
Answer:
Rhada can make 48 ornaments if she uses all of the clay.
Step-by-step explanation:
The complete question is:
Rhada has a 6-pound bag of day . Her craft project requires 6 ounces of clay for each batch of 3 ornaments. If she uses all of the clay, how many ornaments can Rhada make?
Solution:
The amount of clay Rhada has, X = 6-pound.
Convert the weight of clay bag into ounces as follows:
1 pound = 16 ounces
6 pound = 16 × 6
= 96 ounces.
So, Rhada has 96 ounces of clay.
It is provided that her craft project requires 6 ounces of clay for each batch of ornaments.
Compute the number of batches that can be made by 96 ounces of clay as follows:
Number of batches = Total weight of clay ÷ Amount required for each batch
[tex]=\frac{96}{6}\\=16[/tex]
Thus, Rhada can make 16 batches of ornaments.
Now, it is also provided that each batch has 3 ornaments.
Compute the number of ornaments in 16 batches as follows:
Number of ornaments = Number of batches × No. of ornament in 1 batch
[tex]=16\times 3\\=48[/tex]
Thus, Rhada can make 48 ornaments if she uses all of the clay.
The number P, in hundreds of bacteria in a sample, can be modeled by the Equation P= T^4+ 5T^3 + 5T^2 + 6t where t is measured in weeks. explain how to determine the number of weeks which the population would be greater than 10,000
Answer:
The values of T which satisfies the inequality are: [tex](8.8, \infty)[/tex]
Step-by-step explanation:
In the equation [tex]P= T^4+ 5T^3 + 5T^2 + 6T[/tex]
Represent P by 10,000Write an inequality where the expression is greater than 10000: [tex]T^4+ 5T^3 + 5T^2 + 6T>10000[/tex]Get 0 on one side of the inequality.[tex]T^4+ 5T^3 + 5T^2 + 6T-10000>0[/tex]
Graph the polynomial function. We have real x-intercepts of 8.84 and -11.38.Determine intervals where the graph is above the x-axis. Since negative values of x in this situation are irrelevant, the values of T which satisfies the inequality are: [tex](8.8, \infty)[/tex]We now test a value from the set of solution to see if it is valid.Let T=9[tex]T^4+ 5T^3 + 5T^2 + 6T>10000\\9^4+ 5(9)^3 + 5(9)^2 + 6(9)>10000\\10665>10000[/tex]
Since 10665 is grater than 10000, the result is reasonable.
The sodium content of a popular sports drink is listed as 205 mg in a 32-oz bottle. Analysis of 20 bottles indicates a sample mean of 219.2 mg with a sample standard deviation of 18.0 mg. (a) State the hypotheses for a two-tailed test of the claimed sodium content. H0: μ ≥ 205 vs. H1: μ < 205 H0: μ ≤ 205 vs. H1: μ > 205 H0: μ = 205 vs. H1: μ ≠ 205
Answer:
[tex]t=\frac{219.2-205}{\frac{18}{\sqrt{20}}}=3.528[/tex]
[tex]p_v =2*P(t_{19}>3.528)=0.0022[/tex]
If we compare the p value and a significance level for example [tex]\alpha=0.05[/tex] we see that [tex]p_v <\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the lifetime is signficantly different from 205 hours.
Step-by-step explanation:
Data given and notation
[tex]\bar X=219.2[/tex] represent the sample mean
[tex]s=18[/tex] represent the sample standard deviation
[tex]n=20[/tex] sample size
[tex]\mu_o =205[/tex] represent the value that we want to test
[tex]\alpha[/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 apply a two tailed test.
What are H0 and Ha for this study?
Null hypothesis: [tex]\mu = 205[/tex]
Alternative hypothesis :[tex]\mu \neq 205[/tex]
Compute the test statistic
The statistic for this case 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{219.2-205}{\frac{18}{\sqrt{20}}}=3.528[/tex]
Give the appropriate conclusion for the test
The degreed of freedom are:
[tex] df = n-1= 19[/tex]
Since is a two sided test the p value would be:
[tex]p_v =2*P(t_{19}>3.528)=0.0022[/tex]
Conclusion
If we compare the p value and a significance level for example [tex]\alpha=0.05[/tex] we see that [tex]p_v <\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the lifetime is signficantly different from 205 hours.
Answer:
H0: mu equals 205 vs. H1: mu not equals 205.
Step-by-step explanation:
A null hypothesis (H0) is a statement from a population parameter which is either rejected or accepted (fail to reject) upon testing. It expresses equality.
An alternate hypothesis (H1) is also a statement from the population parameter which negates the null hypothesis and is accepted if the null hypothesis is true. It expresses inequality.
A two-tailed test is one in which the alternate hypothesis is expressed using any of the inequality signs below:
not equal to, less than or equal to, greater than or equal to.
The capacity of n elevator is 12 people or 1968 pounds. The capacity will be exceeded if 12 people have weights with a mean greater than 1968/12=164 pounds. Suppose the people have weights that are normally distributed with a mean of 171 lb and a standard deviation of 34 lb.
a. find the probability that if a person is randomly selected, his weight will be greater than 164 pounds.
The probability is approximately ___
b. Find the probability that 12 randomly selected people will have a neam that is greater than 164 pounds.
Answer:
a) 58.32% probability that his weight will be greater than 164 pounds.
b) 76.11% probability that 12 randomly selected people will have a neam that is greater than 164 pounds.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
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.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this problem, we have that:
[tex]\mu = 171, \sigma = 34[/tex]
a. find the probability that if a person is randomly selected, his weight will be greater than 164 pounds.
This is 1 subtracted by the pvalue of Z when X = 164. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{164 - 171}{34}[/tex]
[tex]Z = -0.21[/tex]
[tex]Z = -0.21[/tex] has a pvalue of 0.4168
1 - 0.4168 = 0.5832
58.32% probability that his weight will be greater than 164 pounds.
b. Find the probability that 12 randomly selected people will have a neam that is greater than 164 pounds.
Now [tex]n = 12, s = \frac{34}{\sqrt{12}} = 9.81[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{164 - 171}{9.81}[/tex]
[tex]Z = -0.71[/tex]
[tex]Z = -0.71[/tex] has a pvalue of 0.2389
1 - 0.2389 = 0.7611
76.11% probability that 12 randomly selected people will have a neam that is greater than 164 pounds.
(1 point) Each of the following statements is an attempt to show that a given series is convergent or divergent not using the Comparison Test (NOT the Limit Comparison Test.) For each statement, enter C (for "correct") if the argument is valid, or enter I (for "incorrect") if any part of the argument is flawed. (Note: if the conclusion is true but the argument that led to it was wrong, you must enter I.) I equation editorEquation Editor 1. For all n>2, nn3−2<2n2, and the series 2∑1n2 converges, so by the Comparison Test, the series ∑nn3−2 converges. equation editorEquation Editor 2. For all n>2, 1n2−1<1n2, and the series ∑1n2 converges, so by the Comparison Test, the series ∑1n2−1 converges. I equation editorEquation Editor 3. For all n>2, n+1−−−−−√n>1n, and the series ∑1n diverges, so by the Comparison Test, the series ∑n+1−−−−−√n diverges. I equation editorEquation Editor 4. For all n>2, ln(n)n2>1n2, and the series ∑1n2 converges, so by the Comparison Test, the series ∑ln(n)n2 converges. I equation editorEquation Editor 5. For all n>1, 1nln(n)<2n, and the series 2∑1n diverges, so by the Comparison Test, the series ∑1nln(n) diverges. I equation editorEquation Editor 6. For all n>1, n7−n3<1n2, and the series ∑1n2 converges, so by the Comparison Test, the series ∑n7−n3 converges.
Answer:
I.
CORRECT.
II.
CORRECT.
III.
CORRECT.
IV.
CORRECT.
V.
INCORRECT.
VI.
CORRECT
Step-by-step explanation:
To understand let us restate the comparison test in simple terms.
Comparison test :
Given [tex]\text{Series}_A[/tex] and [tex]\text{Series}_B[/tex] such that [tex]\text{Series}_A < \text{Series}_B[/tex] , then
1. If [tex]\text{Series}_B[/tex] converges then [tex]\text{Series}_A[/tex] converges as well.
2. If [tex]\text{Series}_A[/tex] diverges then [tex]\text{Series}_B[/tex] diverges as well.
Now to give you a more intuitive idea of what is going on, think about it like this. When the series on top converges it is like an "upper bound" for what you have on the bottom, therefore what you have on the bottom has to converge as well.
Similarly if what you have on the bottom explotes, then what you have on top will explote as well.
That's how I like to think about that intuitively.
Now, using those results let us examine the statements.
I.
[tex]\frac{1}{n} < \frac{ln(n)}{n}[/tex]
Since the infinite sum of 1/n diverges in fact the infinite sum of ln(n)/n does not converge.
Therefore, CORRECT.
II.
[tex]\frac{\arctan(n)}{n^3} < \frac{\pi}{2}\frac{1}{n^3}[/tex]
Since the infinite sum of [tex]\frac{\pi}{2}\frac{1}{n^3}[/tex] is in fact convergent then [tex]\frac{\arctan(n)}{n^3}[/tex] converges as well using the comparison theorem. Therefore
CORRECT.
III.
[tex]\frac{n}{2-n^3} < \frac{1}{n^2}[/tex]
Once again [tex]1/n^2[/tex] does converge so what you have on the bottom converges as well. Therefore
CORRECT.
IV.
[tex]\frac{\ln(n)}{n^2} < \frac{1}{n^{1.5}}[/tex]
Once again [tex]\frac{1}{n^{1.5}}[/tex] converges therefore since it is on top what is on the bottom converges as well. Therefore.
CORRECT.
V.
[tex]\frac{\ln(n)}{n} < \frac{2}{n}[/tex]
Now the fact that [tex]\frac{2}{n}[/tex] diverges does not necessarily imply that what you have on the bottom diverges. Therefore
INCORRECT.
VI.
That is correct as well since what you have on top converges therefore what you have on the bottom converges as well.
Suppose a batch of steel rods produced at a steel plant have a mean length of 170170 millimeters, and a standard deviation of 1010 millimeters. If 299299 rods are sampled at random from the batch, what is the probability that the mean length of the sample rods would differ from the population mean by greater than 0.70.7 millimeters? Round your answer to four decimal places.
Answer:
Required probability is 0.2262
Step-by-step explanation:
given data
mean length = 170 millimeters
standard deviation = 10 millimeters
sample n = 299
population mean by greater than = 0.7 millimeters
solution
we consider here batch of steel rod produced = x
probability that mean length sample that differ than the population mean greater than 0.7 required probability is
P ( [tex]\bar {x} -\mu[/tex] > 0.7 ) = 1 - P ( [tex]\bar {x} -\mu[/tex] ≤ 0.7 ) .................1
so we take here value
P ( [tex]\bar {x} -\mu[/tex] > 0.7 ) = 1 - [ P ( -0.7 ≤ [tex]\bar {x} -\mu[/tex] ≤ 0.7 ) ]
P ( [tex]\bar {x} -\mu[/tex] > 0.7 ) = 1 - [ P ( [tex]\frac{-0.7}{\frac{10}{\sqrt{299}}}[/tex] ≤ [tex]\frac{\bar {x} -\mu}{\frac{\sigma }{\sqrt{n}}}[/tex] ≤ [tex]\frac{0.7}{\frac{10}{\sqrt{299}}}[/tex] )
P ( [tex]\bar {x} -\mu[/tex] > 0.7 ) = 1 [ P ( - 1.21 ≤ Z ≤ 1.21 ) ]
P ( [tex]\bar {x} -\mu[/tex] > 0.7 ) = 1 ( P ( Z ≤ 1.21 ) - P ( Z ≤ - 1.21 ) )
we use here excel function that
P ( [tex]\bar {x} -\mu[/tex] > 0.7 ) = 1 - { (=NORMSDiST (1.21) - (=NORMSDiST (-1.21) }
P ( [tex]\bar {x} -\mu[/tex] > 0.7 ) = 1 - ( 0.8869 - 0.1131 )
P ( [tex]\bar {x} -\mu[/tex] > 0.7 ) = 0.2262
so required probability is 0.2262
To find the probability that the sample mean length of steel rods differs from the population mean by more than 0.7 millimeters, we calculate the standard error, convert 0.7 mm to a z-score, and then find the two-tailed probability using the standard normal distribution.
Explanation:To calculate the probability that the mean length of a sample of steel rods will differ from the population mean by more than 0.7 millimeters, we must use the Central Limit Theorem, which states that the sampling distribution of the sample mean will be normally distributed as the sample size gets larger (which holds true in this case since we have 299 rods).
We can apply this theorem to find the standard error of the mean (SEM), which is the standard deviation of the sample mean distribution, calculated as σ/√n (population standard deviation/square root of sample size).
With σ = 10 millimeters and n = 299, the SEM is 10/√299 = 0.5787 millimeters. To find the probability that the sample mean differs from the population mean by more than 0.7 millimeters, we convert 0.7 millimeters to a z-score by dividing by the SEM: z = 0.7/0.5787 = 1.2096.
Using the standard normal distribution, we find the probability for the corresponding z-score and then calculate the two-tailed probability, as we are looking for the probability that the sample mean is either above or below the population mean by 0.7 millimeters. However, to provide the exact figure, we would need a z-table or software to get the precise probability.
Finally, we find that the distribution of the sample mean will have higher probabilities closer to the population mean, as described in the provided information.
A teacher is experimenting with computer-based instruction. In which situation could the teacher use a hypothesis test for a population mean?
A) She gives each student a pretest. Then she teaches a lesson using a computer program. Afterwards, she gives each student a posttest. The teacher wants to see if the difference in scores will show an improvement.
B) She randomly divides the class into two groups. One groups receives computer-based instruction. The other group receives traditional instruction without computers. After instruction, each student has to solve a single problem. The teachers wants to compare the proportion of each group who can solve the problem.
C) The teacher uses a combination of traditional methods and computer-based instruction. She asks students which they liked better. She wants to determine if the majority prefer the computer-based instruction.
A teacher can use a hypothesis test for a population mean in situation A, where pretest scores are compared to posttest scores to determine if there is a significant improvement.
Explanation:A hypothesis test for a population mean can be used in situation A. The teacher can use the pretest scores as the population mean and then compare it to the posttest scores to see if there is a statistically significant improvement. The hypothesis test will determine if the difference in scores is due to chance or if it is a result of the computer-based instruction.
Learn more about Hypothesis test for a population mean here:https://brainly.com/question/31962294
#SPJ12
Suppose a recent plant journal indicated that the mean height of mature plants of a certain species of sunflower is 10.4 ft. A
biologist observing a huge field of mature plants of this particular species of sunflower thinks that the mean height is shorter
than reported. He measures the heights of 45 randomly selected sunflowers and finds the mean height to be 10.1 ft.
The hypotheses of interest are :
(A) H0 : µ = 10.1 vs Ha : µ < 10.1
(B) H0 : µ > 10.4 vs Ha : µ = 10.4
(C) H0 : ¯x = 10.1 vs Ha : ¯x > 10.1
(D) H0 : ¯x = 10.1 vs Ha : ¯x < 10.1
(E) H0 : µ = 10.4 vs Ha : µ < 10.4
Answer: (E) H0 : µ = 10.4 vs Ha : µ < 10.4
Step-by-step explanation:
The null hypothesis is the hypothesis that is assumed to be true. It is an expression that is the opposite of what the researcher predicts.
The alternative hypothesis is what the researcher expects or predicts. It is the statement that is believed to be true if the null hypothesis is rejected.
From the given situation,
The recent plant journal indicated that the mean height of mature plants of a certain species of sunflower is 10.4 ft. This is the null hypothesis.
The biologist sunflower thinks that the mean height is shorter than reported. This is the alternative hypothesis.
Therefore, the correct null and alternative hypotheses are
(E) H0 : µ = 10.4 vs Ha : µ < 10.4
The graphs below have the same shape. Complete the equation of the red graph. Enter exponents using the caret (^); for example, enter x2 as x^2. Do not include "F(x) =" in your answer.
Answer:
f(x) = 1-x^2
Step-by-step explanation:
The graph of f(x) is just shifted down 3 units
A shift down is a subtraction
f(x) = g(x) -3
= 4- x^2 -3
=1-x^2
Answer:
1 - x^2
Step-by-step explanation:
F(x) is 3 units down. So,
G(x) - 2 = 4 - x² - 3
F(x) = 1 - x²
An 8-foot by 5-foot section of wall is to be covered by square tiles that measure 4 inches on each side. If the tiles are not cut, how many of them will needed to cover the wall ?
Answer:
30 tiles
Step-by-step explanation:
SO we need to find the area of the wall so we do
8 feet * 5 feet = 40 feet squared
then we find that area in inches
40*12=480
Now we find the area of the tiles
4*4=16
So now we divide to find how many tiles are needed
480/16=30
Our answer is 30
In 1992, the U.S. Senate was composed of 57 Democrats and 43 Republicans. Of the Democrats, 38 had served in the military, whereas 28 of the Republicans had served. If a senator selected at random is found to have served in the military, what is the probability that the senator is Republican
Answer: 0.66
Step-by-step explanation:
The United States senate had 57 Democrats and 43 Republicans which makes the total numerous of senators 100.
38 Democrats had served in the military while 28 Republicans has served in the military which makes the total number of senators that had served in the military 66.
Probability of a Republican senator that had served in the military will be:
What is the area of the prism below pls help ASAP
Answer:
volume is 3 1/2 units³
Step-by-step explanation:
The question asks for area, but the answers have dimensions of volume.
__
The volume is the product of the length, height, and width dimensions:
V = LWH = (2)(1)(1 3/4) = 2 6/4 = 3 1/2 . . . . units³
The volume is 3 1/2 cubic units.
A carpenter is constructing a straircase in a house. The distance from the first floor to the basement is 10.3 feet. The staircase will be 17.9 feet long. What angle do the stair make with the basement floor?
Answer:
35.1°
Step-by-step explanation:
sin θ = 10.3 / 17.9
θ = sin⁻¹(10.3 / 17.9)
θ = 35.1°
Answer:
157.075 and yes
Step-by-step explanation: turst me
When drawn in standard position, an angle B has a terminal ray that lies in the second quadrant and sin(B)=5/13. Find the value of cos(B), (show your answer as a fraction). Please show all your work
Please a help here.
Thank you
Answer:
[tex]\cos{B} = -\frac{12}{13}[/tex]
Step-by-step explanation:
For any angle B, we have the following trigonometric identity:
[tex]\sin^{2}{B} + \cos^{2}{B} = 1[/tex]
In this problem, we have that:
[tex]\sin{B} = \frac{5}{13}[/tex]
So, applying the trigonometric identity:
[tex]\sin^{2}{B} + \cos^{2}{B} = 1[/tex]
[tex](\frac{5}{13})^{2}) + \cos^{2}{B} = 1[/tex]
[tex]\cos^{2}{B} = 1 - (\frac{5}{13})^{2})[/tex]
[tex]\cos^{2}{B} = 1 - \frac{25}{169}[/tex]
[tex]\cos^{2}{B} = \frac{169}{169} - \frac{25}{169}[/tex]
[tex]\cos^{2}{B} = \frac{144}{169}[/tex]
[tex]\cos{B} = \pm \sqrt{\frac{144}{169}}[/tex]
[tex]\cos{B} = \pm \frac{12}{13}[/tex]
In the second quadrant, the cosine is negative. So
[tex]\cos{B} = -\frac{12}{13}[/tex]
Final answer:
The cosine in the second quadrant is negative, giving cos(B) = -12/13.
Explanation:
To find the value of cos(B) when an angle B is in standard position with its terminal ray in the second quadrant and given that sin(B) = 5/13, we can use the Pythagorean identity:
sin²(B) + cos²(B) = 1.
First, we square the sine value to get sin²(B):
sin²(B) = (5/13)² = 25/169.
Now we use the Pythagorean identity to find cos²(B):
cos²(B) = 1 - sin²(B) = 1 - 25/169 = 144/169.
Since angle B is in the second quadrant, the cosine value must be negative.
Therefore, cos(B) = -√(cos²(B)) = -√(144/169).
The square root of 144/169 is 12/13, and thus:
cos(B) = -12/13.
the sum 27 and a number g
Answer:
27 + g is how you write if u want a solution tell me what G is = to
Step-by-step explanation:
What is the whole number for 6/1?
Answer:
6
Step-by-step explanation:
Answer:
6
Step-by-step explanation:
Because 1/1 = 1 whole and you add 5 more 1/1 and you get 6/1 so their is 6 wholes
Prior Knowledge Questions (Do these BEFORE using the Gizmo.)
Recall that a percent is a ratio of a number to 100. For example 39% means 39 out of 100.
1.Last year, there were 20 students on the track team. Eight of them competed in the long
jump. What percent of the team members competed in the long jump?
Answer:
1.6% should be right.
Step-by-step explanation:
A plane can fly 1200 miles in the same time as it takes a car to go 320 miles. If the car travels 110 mph slower than the
plane, find the speed of the plane.
____mph
Given:
A plane can fly 1200 miles in the same time as it takes a car to go 320 miles. The car travels 110 mph slower than the plane.
We need to determine the speed of the plane.
Speed of the plane:
Let the speed of the plane be x.
Equating the time taken by the plane and the time taken by the car using the formula,
[tex]Time=\frac{Distance}{Speed}[/tex]
Thus, we have;
[tex]\frac{1200}{x}=\frac{320}{x-110}[/tex]
Cross multiplying, we get;
[tex]1200(x-110)=320x[/tex]
Simplifying, we get;
[tex]1200x-132000=320x[/tex]
[tex]880x=132000[/tex]
[tex]x=150[/tex]
Thus, the value of the speed of the plane is 150 mph
Final answer:
The speed of the plane is found by setting up an equation based on the relationship between distance, speed, and time for both the plane and the car. After setting up the equations and equating the times, we find that the plane's speed is 150 mph.
Explanation:
To find the speed of the plane, we can set up an equation based on the relationship between distance, speed, and time. Let's denote the plane's speed as x mph. Then the car's speed would be x - 110 mph. Since they travel for the same amount of time, we can write two equations:
Time for plane = 1200 miles / x mph
Time for car = 320 miles / (x - 110) mph
These two times are equal, so we set the equations equal to each other:
1200 / x = 320 / (x - 110)
To solve for x, cross-multiply and solve the resulting quadratic equation:
1200(x - 110) = 320x
1200x - 132,000 = 320x
1200x - 320x = 132,000
880x = 132,000
x = 132,000 / 880
x = 150 mph
Therefore, the speed of the plane is 150 mph.
A courier service company wishes to estimate the proportion of people in various states that will use its services. Suppose the true proportion is 0.04. If 469 are sampled, what is the probability that the sample proportion will be less than 0.03
Using the normal distribution and the central limit theorem, it is found that there is a 0.1335 = 13.35% probability that the sample proportion will be less than 0.03.
Normal Probability DistributionIn a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
It 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, which is the percentile of X.By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1 - p)}{n}}[/tex], as long as [tex]np \geq 10[/tex] and [tex]n(1 - p) \geq 10[/tex].In this problem:
The true proportion is of p = 0.04.469 people are sampled, hence n = 469.The mean and the standard error are given by:
[tex]\mu = p = 0.04[/tex]
[tex]s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.04(0.96)}{469}} = 0.009[/tex]
The probability that the sample proportion will be less than 0.03 is the p-value of Z when X = 0.03, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.03 - 0.04}{0.009}[/tex]
[tex]Z = -1.11[/tex]
[tex]Z = -1.11[/tex] has a p-value of 0.1335.
0.1335 = 13.35% probability that the sample proportion will be less than 0.03.
To learn more about the normal distribution and the central limit theorem, you can check https://brainly.com/question/24663213
Which two events are independent? A.) go sledding, snow B.) sleep,dream C.) count to 100, fix a bike D.) do chores, earn allowance
Answer:
A and C
Step-by-step explanation:
The meaning of an event being independent is that one event doesn't effect the probability of they other event happening. So B is wrong because whether yo sleep or not can affect whether you dream or not and D is wrong because you earning and allowance is based off whether you do your chores or not.
5) The records of 100 SNHU students show the following courses taken: 53 students took History 41 students took Marketing 48 students took Writing 18 students took History and Marketing 21 students took Marketing and Writing 7 students took all 3 courses 9 students took none of these courses Answer the following questions. Show how you obtained your solution. a) How many students took Marketing and Writing, but not History? b) How many students took only History?.
Answer:
a) How many students took Marketing and Writing, but not History?
14 Students.
b) How many students took only History?
23 Students.
Step-by-step explanation:
We were given the following values in the question
Total number of students = 100
53 students took History = n(H)
41 students took Marketing = n(M)
48 students took Writing = n(W)
18 students took History and Marketing = n(H n M)
21 students took Marketing and Writing = n ( M n W)
7 students took all 3 courses = n ( H n M n W)
9 students took none of these courses
a) How many students took Marketing and Writing, but not History?
This is calculated as
= n(M n W) - n ( H n M n W)
= 21 - 7
= 14
Therefore, only 14 Students took Marketing and Writing but not History.
b) How many students took only History?
Step 1
Subtract the Total number of students from the number of students who did not take any course
100 - 9 = 91 students
Step 2
The number of students who took History and Writing (H n W) was not given, so we find it.
Therefore,
n(H) + n(M) + n(W) -n( H n M) - n( M n W) - n( H n W) + n( H n M n W) = 91
53 + 41 + 48 - 18 - 21 - n( H n W) + 7 = 91
n( H n W) = 149 - 39 -91
n( H n W) = 149 - 130
n( H n W) = 19
Step 3
To find the number of students taking History only, we calculate it as
Number of Students that took History only = n( H) - n(H n W) - n( H n M) + n( H n M n W)
= 53 - 19 - 18 + 7
= 53 + 7 -19 - 18
= 60 - 37
= 23
Therefore, the number of students that took only History is 23 Students.
Which two events are most likely to be independent? A) Being a senior; going to homeroom. B) Registering to vote; being left-handed. C) Having a car accident; having a junior license. D) Doing the Statistics homework; getting an A on the test.
Answer:
It is B) Registering to vote; being left-handed.
Step-by-step explanation:
These two events would be the only situation where the occurrence of one has no effect on the other, and each one of the other examples contain an event that affects the other probability of the other event.
Registering to vote; being left-handed are independent events so option (B) will be correct.
What are dependent and independent events?If the probability of an event affects the other then that will call a dependent event.
For example, getting two red balls from a bag without replacement.
While independent events are events, where the probability of the first event doesn't affect another.
For example tossing a coin.
Given the two events in all options.
In option (A) if you are a senior then it is possible that you can go to homeroom.
In option (B) if you are left-handed then also you can get a mark on the right finger it doesn't matter so that will be an independent event.
Hence "Registering to vote; being left-handed are independent events".
To learn more about dependent and independent events.
https://brainly.com/question/12138721
#SPJ2
A process that fills packages is stopped whenever a package is detected whose weight falls outside the specification. Assume that each package has probability 0.01 of falling outside the specification and that the weights of the packages are independent. Find the mean number of packages that will be filled before the process is stopped.
The mean number of packages that will be filled before the process is stopped is 100
Step-by-step explanation:
Step 1
Given that each package has probability 0.01 of falling outside the specification (Probability of Failure)
The probability of Success is (100-probability of failure)=(100-0.01)=0.99
It is important to note that the weight of the packages are independent
Using the data given in the question we get the following:
P(fail) =Probability of Failure
P(success)=Probability of Success
P(fail) = 0.01 and P(success) = 0.99
Step 2
The mean of the Geometric distribution is:
P = 0.01;
μx = 1/p = 1/0.01 = 100
Step 3
Thus, we can say that the mean number of packages that will be filled before the process is stopped is 100
Solve the equation, 4 - (x + 1) = 6.
Answer:
X=-3
Step-by-step explanation
First of all, remove the parenthesis by distributing the value of the negative sign so it would be 4-x-1=6 , next find the like terms which is 4-1 is 3. so 3-x=6.
transpose the 3 to the other side so it would be -x=6-3
-x=3 divide both sides by -1. X=-3
Find the greatest common factor of 4c and 10c".
Answer:
Step-by-step explanation:
A manufacturer claims that the thickness of the spearmint gum it produces is 7.5 one-hundredths of an inch. A quality control specialist regularly checks this claim. From his experience, the specialist knows the distribution is right skewed and that the standard deviation is 0.4 one-hundredths of an inch. On one production run, he took a random sample of n = 100 pieces of gum and measured their thickness. Which hypothesis test would be most appropriate for this task?
The most appropriate hypothesis test for this task would be a one-sample t-test.
Explanation:For this task, the most appropriate hypothesis test to use would be a one-sample t-test. A one-sample t-test is used to determine whether the mean of a sample is significantly different from a population mean.
In this case, the quality control specialist wants to test whether the mean thickness of the gum produced by the manufacturer is significantly different from the claimed value of 7.5 one-hundredths of an inch. The specialist takes a random sample of n = 100 pieces of gum and measures their thickness.
The one-sample t-test is appropriate because the population distribution is assumed to be right skewed and the standard deviation is known. The t-test allows for the use of a smaller sample size and accommodates the assumption of a non-normal distribution.
Learn more about Hypothesis testing here:
https://brainly.com/question/34171008
#SPJ12
The length Of a rectangular air filter is 1 inches less than twice the width. Find the length and width of the filter if the area is 378 square inches
Answer:52
Step-by-step explanation: