Answer:
[tex]P(\bar X <38)[/tex]
And we can use the z score formula given by:
[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And if we find the z score for 38 we got:
[tex] z = \frac{38-42}{\frac{30}{\sqrt{90}}}= -1.265[/tex]
So then we want to find this probability:
[tex] P(z<-1.265)[/tex]
And we can use the normal standard distribution or excel and we got:
[tex] P(z<-1.265)=0.103[/tex]
Step-by-step explanation:
For this case we define the random variable X as the annual income for people at certain city. And we know the following properties:
[tex] E(X) = 42, Sd(X) =42[/tex]
They select a sample size of n = 90>30. So then we can assume that the central limit can be applied.
From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
We want to calculate this probability:
[tex]P(\bar X <38)[/tex]
And we can use the z score formula given by:
[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And if we find the z score for 38 we got:
[tex] z = \frac{38-42}{\frac{30}{\sqrt{90}}}= -1.265[/tex]
So then we want to find this probability:
[tex] P(z<-1.265)[/tex]
And we can use the normal standard distribution or excel and we got:
[tex] P(z<-1.265)=0.103[/tex]
Check all that apply?
Answer:
–11 and 2
Step-by-step explanation:
observe
x² + 9x – 22 = 0
(x + 11)(x – 2) = 0
x = –11 or x = 2
The divisor of 0.004 is almost zero. What does this tell you about the quotient?
Answer:
Its magnitude will be larger than 0.004.
Step-by-step explanation:
When a divisor is less than 1, the quotient will be greater than the dividend.
When the divisor is "almost zero", the quotient will be much greater than the dividend. Here, the dividend may be considered to be "almost zero", so we cannot say anything about the actual quotient except to say its magnitude will be greater than the dividend.
_____
The dividend is positive, so the quotient will have the same sign as the divisor. (Negative divisors can be "almost zero," too.)
Now focus on the boundary of D, and solve for y2. Restricting f(x,y) to this boundary, we can express f(x,y) as a function of a single variable x. What is this function and its closed interval domain?
Answer:
At critical point in D
a
[tex](x,y) = (0,0)[/tex]
b
[tex]f(x,y) = f(x) =11 -x^2[/tex]
where [tex]-1 \le x \le 1[/tex]
c
maximum value 11
minimum value 10
Step-by-step explanation:
Given [tex]f(x,y) =10x^2 + 11x^2[/tex]
At critical point
[tex]f'(x,y) = 0[/tex]
=> [tex][f'(x,y)]_x = 20x =0[/tex]
=> [tex]x =0[/tex]
Also
[tex][f'(x,y)]_y = 22y =0[/tex]
=> [tex]y =0[/tex]
Now considering along the boundary
[tex]D = 1[/tex]
=> [tex]x^2 +y^2 = 1[/tex]
=> [tex]y =\pm \sqrt{1- x^2}[/tex]
Restricting [tex]f(x,y)[/tex] to this boundary
[tex]f(x,y) = f(x) = 10x^2 +11(1-x^2)^{\frac{2}{1} *\frac{1}{2} }[/tex]
[tex]= 11-x^2[/tex]
At boundary point D = 1
Which implies that [tex]x \le 1[/tex] or [tex]x \ge -1[/tex]
So the range of x is
[tex]-1 \le x \le 1[/tex]
Now along this this boundary the critical point is at
[tex]f'(x) = 0[/tex]
=> [tex]f'(x) = -2x =0[/tex]
=> [tex]x=0[/tex]
Now at maximum point [tex](i.e \ x =0)[/tex]
[tex]f(0) =11 -(0)[/tex]
[tex]= 11[/tex]
For the minimum point x = -1 or x =1
[tex]f(1) = 11 - 1^2[/tex]
[tex]=10[/tex]
[tex]f(-1) = 11 -(-1)^2[/tex]
[tex]=10[/tex]
A teacher used the change of base formula to determine whether the equation below is correct.
(log Subscript 2 Baseline 10) (log Subscript 4 Baseline 8) (log Subscript 10 Baseline 4) = 3
Which statement explains whether the equation is correct?
Answer:
The equation is correct
Step-by-step explanation:
The equation, written as:
[log_2 (10)][log_4 (8)][log_10 (4)] = 3
Consider the change of base formula:
log_a (x) = [log_10 (x)]/ [log_10 (a)]
Applying the change of base formula to change the expressions in base 2 and base 4 to base 10.
(1)
log_2 (10) = [log_10 (10)]/[log_10 (2)]
= 1/[log_10 (2)]
(Because log_10 (10) = 1)
(2)
log_4 (8) = [log_10 (8)]/[log_10 (4)]
Now putting the values of these new logs in base 10 into the left-hand side of original equation to verify if we have 3, we have:
[log_10 (2)][log_8 (4)][log_10 (4)]
= [1/ log_10 (2)][log_10 (8) / log_10 (4)][log_10 (4)]
= [1/log_10 (2)] [log_10 (8)]
= [log_10 (8)]/[log_10 (2)]
= [log_10 (2³)]/[log_10 (2)]
Since log_b (a^x) = xlog_b (a)
= 3[log_10 (2)]/[log_10 (2)]
= 3 as required
Therefore, the left hand side of the equation is equal to the right hand side of the equation.
Answer:
B on E2020.
Step-by-step explanation:
In simplest radical form, what are the solutions to the quadratic equation 6 = x2 – 10x?
Quadratic formula: x =
x = 5
x = 5
x = 5
x = 5
Answer:
Step-by-step explanation:
quadratic equation: ax² + bx + c =0
x' = [-b+√(b²-4ac)]/2a and x" = [-b-√(b²-4ac)]/2a
6 = x² – 10x ; x² - 10x -6 =0
(a=1, b= - 10 and c = - 6
x' = [10+√(10²+4(1)(-6)]/2(1) and x" = [10-√(10²+4(1)(-6)]/2(1)
x' =5+√31 and x' = 5-√31
$1500 is invested at a rate of 3% compounded monthly. Write a compound interest function to model this situation. Then find the
balance after 5 years.
Answer:
Equation: [tex]F=1500(1.0025)^{12t}[/tex]
The balance after 5 years is: $1742.43
Step-by-step explanation:
This is a compound growth problem . THe formula is:
[tex]F=P(1+\frac{r}{n})^{nt}[/tex]
Where
F is future amount
P is present amount
r is rate of interest, annually
n is the number of compounding per year
t is the time in years
Given:
P = 1500
r = 0.03
n = 12 (compounded monthly means 12 times a year)
The compound interest formula modelled by the variables is:
[tex]F=1500(1+\frac{0.03}{12})^{12t}\\F=1500(1.0025)^{12t}[/tex]
Now, we want balance after 5 years, so t = 5, substituting, we get:
[tex]F=1500(1.0025)^{12t}\\F=1500(1.0025)^{12*5}\\F=1500(1.0025)^{60}\\F=1742.43[/tex]
The balance after 5 years is: $1742.43
Kelsey’s bank changed her $17.50
The mean crying time of infants during naptime at a local preschool is 12 mins. The school implements a new naptime routine in a sample of 25 infants and records an average crying time of 8 ± 4.6 (M ± SD) minutes. Test whether this new naptime routine reduced crying time at a 0.05 level of significance.A) The new naptime routine significantly reduced crying time, t(24) = ?4.35, p <0.05.B) The new naptime routine did not reduce crying time, t(24) = ?4.35, p < 0.05.C) The new naptime routine did not reduce crying time, t(24) = 0.92, p > 0.05.D) The new naptime routine significantly reduce crying time, t(24) = 0.92, p < 0.05.
Answer:
Step-by-step explanatio n: ummmoirnd iehcn
A circle with radius four has a sector with a central angle of 8/5 pi radians. what is area of the sector
Answer:
area of sector = 40.192 unit²
Step-by-step explanation:
Area of a sector = ∅/360 × πr²
where
∅ = angle in degree
r = radius
Area of sector when ∅ = radian
area of a sector = 1/2r²∅
where
∅ = radian
r = radius
area of sector = 1/2 × 4² × 8/5 × π
area of sector = 1/2 × 16 × 8/5 × π
area of sector = 128/10 × π
area of sector = 12.8 × π
area of sector = 12.8 × 3.14
area of sector = 40.192 unit²
In how many ways can Susan arrange 7 books into 5 slots on her bookshelf?
Answer:
2520Step-by-step explanation:
This is permutation question
The formula for it:
N = b!/(b-s)!, where N- number of ways, b- number of books, s- number of slotsFinding the answer:
N = 7!/(7-5)! = 7!/2! = 7*6*5*4*3 = 2520The fictional rocket ship Adventure is measured to be 65 m long by the ship's captain inside the rocket.When the rocket moves past a space dock at 0.5c. As rocket ship Adventure passes by the space dock, the ship's captain flashes a flashlight at 2.00-s intervals as measured by space-dock personnel. Part A How often does the flashlight flash relative to the captain
Answer:
a) t₀ = 1.73205 s
b) 1.0 C
Step-by-step explanation:
(A)
The time dilation (t) observed by an observer at rest relative to the time (t₀) measured by observer in motion is;
[tex]t = \frac{t_0}{\sqrt{1 - \frac{V^2}{C^2}}}[/tex]
[tex]t_0 = t \sqrt{1 - \frac{V^2}{C^2}}[/tex] time measured by captain
⇒ [tex]t_0 = 2.0 \sqrt{1 - \frac{0.5^2C^2}{C^2}}[/tex] V = 0.5 c
⇒ t₀ = 1.73205 s
(B)
Speed of the light never exceeds by its real value. The speed of the light in any frame of reference is constant.
∵ It will be "1.0C" or just "C"
The time needed to complete a final examination in a particular college course is normally distributed with a mean of 80 minutes and a standard deviation of 10 minutes. If students have only 90 minutes to complete the exam, what percentage of the class will not finish the exam in time?
Answer:
Hence total of 10 students are not able to complete the exam.
Step-by-step explanation:
Given:
Mean for completing exam =80 min
standard deviation =10 min.
To find:
how much student will not complete the exam?
Solution:
using the Z-table score we can calculate the required probability.
Z=(Required time -mean)/standard deviation.
A standard on an avg class contains:
60 students.
consider for 70 mins and then 90 mins (generally calculate ± standard deviation of mean)(80-10 and 80+10).
1)70 min
Z=(70-80)/10
Z=-1
Now corresponding p will be
P(z=-1)
=0.1587
therefore
Now for required 90 min will be
Z=(90-80)/10
=10/10
z=1
So corresponding value of p is
P(z<1)=0.8413
this means 0.8413 of 60 students are able to complete the exam.
0.8413*60
=50.47
which approximate 50 students,
total number =60
and total number student will able to complete =50
Total number of student will not complete =60-50
=10.
About 15.87% of college students are expected not to finish the final examination within the 90-minute time limit, based on the properties of the normal distribution with a mean of 80 minutes and a standard deviation of 10 minutes.
Explanation:The student's question involves using the properties of the normal distribution to determine the percentage of students who will not finish a final examination in the given time frame.
To compute this, we need to calculate the z-score that corresponds to the 90-minute time limit. The z-score formula is:
Z = (X - μ) / σ
where X is the value of interest, μ (mu) is the mean, and σ (sigma) is the standard deviation. Plugging in the numbers:
Z = (90 - 80) / 10 = 1
A z-score of 1 corresponds to a percentile of approximately 84.13%, meaning about 84.13% of students will finish within 90 minutes. To find the percentage that will not finish in time, we subtract this from 100%:
100% - 84.13% = 15.87%
Therefore, approximately 15.87% of the class will not finish the exam in time.
What is the area of the following circle
Answer:
16*pi=50.24
Step-by-step explanation:
Answer:
16* pi = 50.24
Step-by-step explanation:
Does anyone know this?
Answer:
A. 5 * 1/7
Step-by-step explanation:
When solving division, you can also multiply by the reciprocal of the second number to get the same answer.
100 POINTS HELP ME PLEASE!!!!!! DONT HAVE ALOT OF TIME HURRY PLEASE!!!!!!!!!!!!!
100 POINTS!!!!!
Luke is designing a scale model of a clock tower. The design of the front of the tower is shown below.
A figure can be broken into a triangle and rectangle. The rectangle has a base of 200 millimeters and height of 50 millimeters. The triangle has a base of 50 millimeters and height of 100 millimeters.
What will be the area of the front face of his model?
2,500 square millimeters
10,000 square millimeters
12,500 square millimeters
15,000 square millimeters
Answer:
12,500 square millimeters
Step-by-step explanation:
Answer:
c(12,500)
Step-by-step explanation:
100 POINTS!!!!! HELP ME PLEASE DONT HAVE A LOT OF TIME!!!!! HELP!!!!!
The school wants to order a new counter top for the teacher’s lounge. The shape of the counter top that they are replacing is shown below.
A countertop can be broken into 2 rectangles. 1 rectangle has a base of 70 inches and height of 20 inches. The other rectangle has a base of 20 inches and height of 30 inches.
If the new countertop costs $0.75 per square inch, what is the price of the replacement countertop?
$1,500
$1,800
$2,000
$2,400
Answer:
A. $1500
Step-by-step explanation:
We need to find the countertop area, so let's calculate the areas of the rectangles that the problem broke the countertop into:
1. "1 rectangle has a base of 70 inches and height of 20 inches"
The area of a rectangle is denoted by: A = bh, where b is the base and h is the height. Here, b = 70 and h = 20, so the area is: A = 70 * 20 = 1400 inches squared
2. "The other rectangle has a base of 20 inches and height of 30 inches"
Again, use A = bh: b = 20 and h = 30, so A = 20 * 30 = 600 inches squared
Add up these two areas: 1400 + 600 = 2000 inches squared.
The problem says that the cost is $0.75 per square inch, so multiply this by 2000 to get the total cost of 2000 square inches:
2000 * 0.75 = $1500
Thus, the answer is A.
Hope this helps!
Answer:
$1500
Step-by-step explanation:
You can divide this figure into 2 rectangle with dimensions:
1) 70 × 20
2) 20 × (50-20: 20 × 30
Area:
(70×20) + (20×30)
1400 + 600
2000 in²
Cost per in²: 0.75
2000in² cost:
2000 × 0.75
$1500
-1.9+4+(-1.6) simplify the expression
Answer:
.5
Step-by-step explanation:
-1.9+4=2.1
2.1+(-1.6)=.5
What is the selling price if the original cost is $145 and the markup is 150%? PLEASE HELP!! :(
Answer:
$362.50
Profit: $217.50
Step-by-step explanation:
The locations, given in polar coordinates, for two ships are (8 mi, 639) and (8 mi, 1239). Find the distance between the two
ships,
a. 64 8 mi
C. 11.31 mi
b. 3600.00 mi
d. 4.14 mi
Please select the best answer
Answer:
A. [tex]\sqrt{64}=8[/tex] miles
Step-by-step explanation:
Given two Cartesian coordinates [tex](x_1,y_1)\&(x_2,y_2)[/tex], the distance between the points is given as:
[tex]d = \sqrt{((x_1-x_2)^2+(y_1-y_2)^2)}[/tex]
Converting to polar coordinates
[tex](x_1,y_1) = (r_1 cos \theta_1, r_1 sin \theta_1)\\(x_2,y_2) = (r_2 cos \theta_2, r_2 sin \theta_2)[/tex]
Substitution into the distance formula gives:
[tex]\sqrt{((r_1 cos\theta_1-r_2 cos \theta_2)^2+(r_1 sin \theta_1-r_2 sin \theta_2)^2}\\=\sqrt{(r_1^2+r_2^2-2r_1r_2(cos \theta_1 cos \theta_2+sin\theta_1 sin \theta_2) }\\= \sqrt{r_1^2+r_2^2-2r_1r_2cos (\theta_1 -\theta_2)}[/tex]
In the given problem,
[tex](r_1,\theta_1)=(8 mi, 63^0) \:and\: (r_2,\theta_2)=(8 mi, 123^0 ).[/tex]
[tex]Distance=\sqrt{8^2+8^2-2(8)(8)cos (63 -123)}\\=\sqrt{128-128cos (-60)}\\=\sqrt{64}=8 mile[/tex]
The closest option is A. [tex]\sqrt{64}=8[/tex] miles
Please help :(
There are 10^9 bytes in a gigabyte. There are 10^6 bytes in a megabyte. How manny times greater is the storage capacity of a 1-gigabyte flash drive than a 1-megabyte flash drive?
answer choices above^^^
Answer:
In the screenshot you have the right answer, it is indeed 1000 times greater
Step-by-step explanation:
WILL GIVE BRAINLIEST
Megan finds a bag of 24 craft bows at the store. The bag indicates that 23 of the bows are striped. Megan wants to know the number of bows in the package that are striped. Select ALL the statements that are true. A Megan can divide the number 3 by 24 and then multiply the result by 2 to find the number of striped bows. B Megan can divide the number 24 by 3 and then multiply the result by 2 to find the number of striped bows. C Megan can multiply the number 24 by 2 and then divide the result by 3 to find the number of striped bows. D Megan can multiply the number 24 by 3 and then divide the result by 2 to find the number of striped bows. E The number of striped bows in the package is 36. F The number of striped bows in the package is 16.
Answer:
B
Step-by-step explanation:
Which is a correct first step for solving this equation?
2 + 7 = 2x + 5 – 4x
Step-by-step explanation:
Bringing like terms on one side
2 + 7 - 5 = 2x - 4x
9 - 5 = - 2x
4 = - 2x
4/ - 2 = x
- 2 = x
Many urban zoos are looking at ways to effectively handle animal waste. One zoo has installed a facility that will transform animal waste into electricity. To estimate how many pounds of waste they may have to fuel the new facility they began keeping meticulous records. They discovered that the amount of animal waste they were disposing of daily is approximately Normal with a mean of 348.5 pounds and a standard deviation of 38.2 pounds. Amounts over 350 pounds would generate enough electricity to cover what is needed to for the entire aquarium that day. Approximately what proportion of the days can the zoo expect to obtain enough waste to cover what is needed to run the entire aquarium for the day (A) 0.484 (B) 0.499 (C) 0.516 (D) 0.680 (E) 0.950
Answer:
Option A: 0.484Explanation:
The amount of animal waste one zoo is diposing daily is approximately normal with:
mean, μ = 348.5 lbsstandard deivation, σ = 38.2 lbsThe proportion of waste over 350 lbs may be found using the table for the area under the curve for the cumulative normal standard probability.
First, find the z-score for 350 lbs:
[tex]z-score=\dfrac{X-\mu}{\sigma}[/tex]
[tex]z-score=\dfrac{350lbs-348.5lbs}{38.2lbs}\approx0.04[/tex]
There are tables for the cumulative areas (probabilities) to the left and for the cumulative areas to the right of the z-score.
You want the proportion of the days when the z-score is more than 0.04; then, you can use the table for the values to the rigth of z = 0.04.
From such table, the area or probability is 0.4840.
The attached image shows a portion of the table with that value: it is the cell highlighted in yellow.
Hence, the answer is the option (A) 0.484.
By calculating the z-score for 350 pounds of waste and consulting the standard normal distribution, the proportion of days the zoo can expect to have enough animal waste to power the entire aquarium is approximately 0.484.
To determine the proportion of days the zoo can expect to generate enough animal waste to run the entire aquarium, we can use z-scores in a normal distribution. Given the mean (μ = 348.5) and standard deviation (σ = 38.2), we want to find the proportion of the data that is above 350 pounds.
First, we calculate the z-score for 350 pounds:
z = (X - μ) / σ = (350 - 348.5) / 38.2 ≈ 0.04
Now we need to find the probability that the z-score is greater than 0.04. Consulting a standard normal distribution table or using a calculator, this gives us a probability of approximately 0.484.
Therefore, the proportion of the days the zoo can expect to obtain enough waste to cover the energy demands for the entire aquarium is 0.484.
The measurement of a side of a square is found to be 10 centimeters, with a possible error of 0.07 centimeter. (a) Approximate the percent error in computing the area of the square. % (b) Estimate the maximum allowable percent error in measuring the side if the error in computing the area cannot exceed 2.7%. %
Answer:
a) [tex]\delta = 1.4\,\%[/tex], b) [tex]\delta_{max} = 1.35\,\%[/tex]
Step-by-step explanation:
a) The area formula for a square is:
[tex]A =l^{2}[/tex]
The total differential for the area is:
[tex]\Delta A = \frac{\partial A}{\partial l}\cdot \Delta l[/tex]
[tex]\Delta A = 2\cdot l \cdot \Delta l[/tex]
The absolute error for the area of the square is:
[tex]\Delta A = 2\cdot (10\,cm)\cdot (0.07\,cm)[/tex]
[tex]\Delta A = 1.4\,cm^{2}[/tex]
Thus, the relative error is:
[tex]\delta = \frac{\Delta A}{A}\times 100\,\%[/tex]
[tex]\delta = \frac{1.4\,cm^{2}}{100\,cm^{2}} \times 100\,\%[/tex]
[tex]\delta = 1.4\,\%[/tex]
b) The maximum allowable absolute error for the area of the square is:
[tex]\Delta A_{max} = \left(\frac{\delta}{100} \right)\cdot A[/tex]
[tex]\Delta A_{max} = \left(\frac{2.7}{100} \right)\cdot (100\,cm^{2})[/tex]
[tex]\Delta A_{max} = 2.7\,cm^{2}[/tex]
The maximum allowable absolute error for the length of a side of the square is:
[tex]\Delta l_{max}= \frac{\Delta A_{max}}{2\cdot l}[/tex]
[tex]\Delta l_{max} = \frac{2.7\,cm^{2}}{2\cdot (10\,cm)}[/tex]
[tex]\Delta l_{max} = 0.135\,cm[/tex]
Lastly, the maximum allowable relative error is:
[tex]\delta_{max} = \frac{\Delta l_{max}}{l}\times 100\,\%[/tex]
[tex]\delta_{max} = \frac{0.135\,cm}{10\,cm} \times 100\,\%[/tex]
[tex]\delta_{max} = 1.35\,\%[/tex]
Analyze the diagram below and complete the instructions that follow.
and are similar. Find the value of x.
A.
5
B.
15
C.
60
D.
240
Please select the best answer from the choices provided
A
B
C
D
is 119x10-3 to the power a scientific notation? yes or no? explain
Answer:
Yes, if I'm assuming that it was written as [tex]119*10^-3[/tex], then this is a basis for a scientific notation.
In each part, find the area under the standard normal curve that lies between the specified z-score, sketch a standard normal curve, and shade the area of interest.
a. minus1 and 1
b. minus2 and 2
c. minus3 and 3
Answer:
a) [tex] P(-1<Z<1)= P(Z<1) -P(Z<-1)= 0.841-0.159= 0.682[/tex]
b) [tex] P(-2<Z<2)= P(Z<2) -P(Z<-2)= 0.977-0.0228= 0.954[/tex]
c) [tex] P(-3<Z<3)= P(Z<3) -P(Z<-3)= 0.999-0.0013= 0.998[/tex]
The results are on the fogure attached.
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
Part a
For this case we want to find this probability:
[tex] P(-1<Z<1)[/tex]
And we can find this probability with this difference:
[tex] P(-1<Z<1)= P(Z<1) -P(Z<-1)[/tex]
And if we find the probability using the normal standard distribution or excel we got:
[tex] P(-1<Z<1)= P(Z<1) -P(Z<-1)= 0.841-0.159= 0.682[/tex]
Part b
For this case we want to find this probability:
[tex] P(-2<Z<2)[/tex]
And we can find this probability with this difference:
[tex] P(-2<Z<2)= P(Z<2) -P(Z<-2)[/tex]
And if we find the probability using the normal standard distribution or excel we got:
[tex] P(-2<Z<2)= P(Z<2) -P(Z<-2)= 0.977-0.0228= 0.954[/tex]
Part c
For this case we want to find this probability:
[tex] P(-3<Z<3)[/tex]
And we can find this probability with this difference:
[tex] P(-3<Z<3)= P(Z<3) -P(Z<-3)[/tex]
And if we find the probability using the normal standard distribution or excel we got:
[tex] P(-3<Z<3)= P(Z<3) -P(Z<-3)= 0.999-0.0013= 0.998[/tex]
Final answer:
The question asks to find the area under the standard normal curve for specific z-score ranges. Using the empirical rule, we conclude that respective areas for those ranges are approximately 68%, 95%, and 99.7%. The exact areas can be found using a Z-table.
Explanation:
The question involves finding the area under the standard normal curve between specified z-scores. This is a fundamental concept in statistics, often used to find probabilities related to normally distributed data.
For a z-score between -1 and 1, approximately 68% of the area under the standard normal curve is contained since the empirical rule states that about 68 percent of values lie within one standard deviation of the mean in a normal distribution.For a z-score between -2 and 2, approximately 95% of the area under the curve is contained, as about 95% of the values lie within two standard deviations of the mean.For a z-score between -3 and 3, about 99.7% of the area is contained, reflecting the fact that about 99.7% of values in a normal distribution lie within three standard deviations of the mean.To find the exact areas based on the z-scores, we can refer to the Z-table of Standard Normal Distribution. This table lists the cumulative probabilities from the mean up to a certain z-score. By looking up the area to the left of each positive z-score and doubling it, we can get the approximate area between the negative and positive z-scores mentioned above.
A new car can go 490 miles on 10 gallons of gas. How many miles can it go on 55 gallons of gas?
Answer:
2695 miles
Step-by-step explanation:
The car can travel 2,695 miles on 55 gallons of gas.
To determine how many miles a car can go on 55 gallons of gas if it can go 490 miles on 10 gallons, we need to find the car's miles per gallon (mpg) and then use that to calculate the distance for 55 gallons.
First, calculate the miles per gallon (mpg):
mpg = 490 miles / 10 gallons = 49 miles per gallon
Now, use the mpg to find the distance the car can travel on 55 gallons:
Distance = 49 miles per gallon * 55 gallons = 2,695 miles
Therefore, the car can go 2,695 miles on 55 gallons of gas.
a survey amony freshman at a certain university revealed that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 15. a sample of 36 students was selected. what is the probability that the average time spent studying for the sample was between 29.0 and 30 hours studying?
Answer:
Probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.
Step-by-step explanation:
We are given that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 15.
A sample of 36 students was selected.
Let [tex]\bar X[/tex] = sample average time spent studying
The z-score probability distribution for sample mean is given by;
Z = [tex]\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean hours spent studying = 25 hours
[tex]\sigma[/tex] = standard deviation = 15 hours
n = sample of students = 36
The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.
Now, Probability that the average time spent studying for the sample was between 29 and 30 hours studying is given by = P(29 hours < [tex]\bar X[/tex] < 30 hours)
P(29 hours < [tex]\bar X[/tex] < 30 hours) = P([tex]\bar X[/tex] < 30 hours) - P([tex]\bar X[/tex] [tex]\leq[/tex] 29 hours)
P([tex]\bar X[/tex] < 30 hours) = P( [tex]\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }[/tex] < [tex]\frac{ 30-25}{\frac{15}{\sqrt{36} } }} }[/tex] ) = P(Z < 2) = 0.97725
P([tex]\bar X[/tex] [tex]\leq[/tex] 29 hours) = P( [tex]\frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }[/tex] [tex]\leq[/tex] [tex]\frac{ 29-25}{\frac{15}{\sqrt{36} } }} }[/tex] ) = P(Z [tex]\leq[/tex] 1.60) = 0.94520
So, in the z table the P(Z [tex]\leq[/tex] x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2 and x = 1.60 in the z table which has an area of 0.97725 and 0.94520 respectively.
Therefore, P(29 hours < [tex]\bar X[/tex] < 30 hours) = 0.97725 - 0.94520 = 0.0321
Hence, the probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.