Answer:
[tex] P(X>14)= 1-P(X<14) =1- F(14)[/tex]
And replacing we got:
[tex] P(X>14)= 1- \frac{14-8}{22-8}= 0.5714[/tex]
The probability that a randomly selected barber needs at least 14 minutes to complete the haircut is 0.5714
Step-by-step explanation:
We define the random variable of interest as x " time it takes a barber to complete a haircuts" and we know that the distribution for X is given by:
[tex] X \sim Unif (a= 8, b=22)[/tex]
And for this case we want to find the following probability:
[tex] P(X>14)[/tex]
We can find this probability using the complement rule and the cumulative distribution function given by:
[tex] P(X<x) = \frac{x-a}{b-a} ,a \leq x \leq b[/tex]
Using this formula we got:
[tex] P(X>14)= 1-P(X<14) =1- F(14)[/tex]
And replacing we got:
[tex] P(X>14)= 1- \frac{14-8}{22-8}= 0.5714[/tex]
The probability that a randomly selected barber needs at least 14 minutes to complete the haircut is 0.5714
Need help with this . The picture is enclosed
Answer: (fоg)(24)=5
Step-by-step explanation:
(fоg)(24) is f of g of 24. This means you plug in g(24) into f(x).
[tex]g(24)=\sqrt{24-8}[/tex]
[tex]g(24)=\sqrt{16}[/tex]
[tex]g(24)=4[/tex]
Now that we know g(24), we can plug it into f(x).
f(4)=2(4)-3
f(4)=8-3
f(4)=5
HELP ASAP!!!The first picture is what each variables equal too
Answer:
Just replace the variables with the number
d5
c4 (uh oh)
a2
b-3
f-7
d-c = 5 - 4 = 1
1/3 - 4(ab+f)
2 x -3 = -6
-6 + -7 = -13
-13 x 4 = -52
1/3 - -52 = 1/3 + 52 =
52 1/3
Hope this helps
Step-by-step explanation:
Which of the following is the solution to 9|x-1|=-45
Answer:
No solutions.
Step-by-step explanation:
9|x-1|=-45
Divide 9 into both sides.
|x-1| = -45/9
|x-1| = -5
Absolute value cannot be less than 0.
Answer:
No solution
Step-by-step explanation:
=> 9|x-1| = -45
Dividing both sides by 9
=> |x-1| = -5
Since, this is less than zero, hence the equation has no solutions
Number of multiples of 7 between 200 and 1000
Answer:
114
Step-by-step explanation:
Answer:
144Step-by-step explanation:
The figure shows a square floor plan with a smaller square area that will accommodate a combination fountain and pool.The floor with the fountain pool area removed has an area of 33 Square meters and a perimeter of 36 meters. Find the dimensions of the floor and the dimensions of the square that will accommodate the fountain and pool.
Answer:
(x, y) = (7, 4) meters
Step-by-step explanation:
The area of the floor without the removal is x^2, so with the smaller square removed, it is x^2 -y^2.
The perimeter of the floor is the sum of all side lengths, so is 4x +2y.
The given dimensions tell us ...
x^2 -y^2 = 33
4x +2y = 36
From the latter equation, we can write an expression for y:
y = 18 -2x
Substituting this into the first equation gives ...
x^2 -(18 -2x)^2 = 33
x^2 -(324 -72x +4x^2) = 33
3x^2 -72x + 357 = 0 . . . . write in standard form
3(x -7)(x -17) = 0 . . . . . factor
Solutions to this equation are x=7 and x=17. However, for y > 0, we must have x < 9.
y = 18 -2(7) = 4
The floor dimension x is 7 meters; the inset dimension y is 4 meters.
Consider the set of sequences of seven letters chosen from W and L. We may think of these sequences as representing the outcomes of a match of seven games, where W means the first team wins the game and L means the second team wins the game. The match is won by the first team to win four games (thus, some games may never get played, but we need to include their hypothetical outcomes in the points in order that we have a probability space of equally likely points).A. What is the probability that a team will win the match, given that it has won the first game?B. What is the probability that a team will win the match, given that it has won the first two games? C. What is the probability that a team will win the match, given that it has won two out of the first three games?
Answer:
a) Probability that a team will win the match given that it has won the first game = 0.66
b) Probability that a team will win the match given that it has won the first two games= 0.81
c) Probability that a team will win the match, given that it has won two out of the first three games = 0.69
Step-by-step explanation:
There are a total of seven games to be played. Therefore, W and L consists of 2⁷ equi-probable sample points
a) Since one game has already been won by the team, there are 2⁶ = 64 sample points left. If the team wins three or more matches, it has won.
Number of ways of winning the three or more matches left = [tex]6C3 + 6C4 + 6C5 + 6C6[/tex]
= 20 + 15 + 6 + 1 = 42
P( a team will win the match given that it has won the first game) = 42/64 = 0.66
b) Since two games have already been won by the team, there are 2⁵ = 32 sample points left. If the team wins two or more matches, it has won.
Number of ways of winning the three or more matches left = [tex]5C2 + 5C3 + 5C4 + 5C5[/tex] = 10 + 10 + 5 +1 = 26
P( a team will win the match given that it has won the first two games) = 26/32 = 0.81
c) Probability that a team will win the match, given that it has won two out of the first three games
They have played 3 games out of 7, this means that there are 4 more games to play. The sample points remain 2⁴ = 16
They have won 2 games already, it means they have two or more games to win.
Number of ways of winning the three or more matches left = [tex]4C2 + 4C3 + 4C4[/tex] = 6 + 4 + 1 = 11
Probability that a team will win the match, given that it has won two out of the first three games = 11/16
Probability that a team will win the match, given that it has won two out of the first three games = 0.69
Consider random samples of size 900 from a population with proportion 0.75 . Find the standard error of the distribution of sample proportions. Round your answer for the standard error to three decimal places. standard error
Answer:
[tex] SE =\sqrt{\frac{p(1-p)}{n}}[/tex]
And replacing we got:
[tex] SE=\sqrt{\frac{0.75*(1-0.75)}{900}}= 0.014[/tex]
Step-by-step explanation:
For this case we have the following info given:
[tex] n=900[/tex] represent the sample size selected
[tex]p = 0.75[/tex] represent the population proportion
We want to find the standard error and we can use the distribution for the sample proportion and for this case since the sample size is large enough and we satisfy np>10 and n(1-p) >10 we have:
[tex] \hat p \sim N (p,\sqrt{\frac{p(1-p)}{n}})[/tex]
And the standard error is given;
[tex] SE =\sqrt{\frac{p(1-p)}{n}}[/tex]
And replacing we got:
[tex] SE= \sqrt{\frac{0.75* (1-0.75)}{900}}= 0.014[/tex]
Black walnut trees contain chemicals that inhibit the growth of other plants. In a simple experiment to test whether this is true, you grow several tomato plants in soil with and without decomposing leaves from a black walnut tree. You collect data on plant height as a measure of growth. In this experiment, __________ is the independent variable, __________ is the dependent variable, and __________ is the control.
Answer:
Height of tomato plant is the dependent variable
Presence of walnut leaves in the soil is the independent variable
Tomato plants grown without walnut leaves is the control
Step-by-step explanation:
An independent variable is the variable in an experiment that can be altered to test for a certain result. It is independent, or does not change with change in other factors in the experiment. In this case, the presence or absence, or quantity of walnut available in the soil is the independent variable in the experiment.
A dependent variable varies, and depends on the independent variable. It is what is measured in the experiment. In this case, the height of the tomato plants is the dependent variable that depends on the presence, absence or quantity of walnut in the soil.
A control in an experiment, is a replicate experiment, that is manipulated in order to be able to test a single variable at a time. Controls are variables are held constant so as to minimize their effect on the system under study. In this case, some of the tomato plants are planted without walnut in the soil, to test the effect of the absence of the walnut in the soil.
Insurance companies track life expectancy information to assist in determining the cost of life insurance policies. AIB Insurance randomly sampled 100 recently paid policies and determined the average age of clients in this sample to be 77.7 years with a standard deviation of 3.6. The 90% confidence interval for the true mean age of its life insurance policy holders is
A. (76.87, 80.33)
B. (72.5, 82.9)
C. (77.1, 78.3)
D. (74.1, 81.3)
E. (74.5, 80)
Answer:
[tex]77.7-1.66\frac{3.6}{\sqrt{100}} =77.102[/tex]
[tex]77.7+1.66\frac{3.6}{\sqrt{100}} =78.30[/tex]
And the best option would be:
C. (77.1, 78.3)
Step-by-step explanation:
Information given
[tex]\bar X=77.7[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
s=3.6 represent the sample standard deviation
n=100 represent the sample size
Confidence interval
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
The degrees of freedom are given by:
[tex]df=n-1=100-1=99[/tex]
Since the Confidence is 0.90 or 90%, the significance would be [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and the critical value for this case would be [tex]t_{\alpha/2}=1.66[/tex]
And replacing we got:
[tex]77.7-1.66\frac{3.6}{\sqrt{100}} =77.10[/tex]
[tex]77.7+1.66\frac{3.6}{\sqrt{100}} =78.30[/tex]
And the best option would be:
C. (77.1, 78.3)
An Undergraduate Study Committee of 6 members at a major university is to be formed from a pool of faculty of 18 men and 6 women. If the committee members are chosen randomly, what is the probability that precisely half of the members will be women?
Answer:
5/33649= approx 0.00015
Step-by-step explanation:
Total number of outcomes are C24 6= 24!/(24-6)!/6!=19*20*21*22*23*24/(2*3*4*5*6)= 19*14*22*23
Half of the Committee =3 persons. That mens that number of the women in Commettee=3. 3 women from 6 can be elected C6 3 ways ( outputs)=
6!/3!/3!=4*5*6*/2/3=20
So the probability that 3 members of the commettee are women is
P(women=3)= 20/(19*14*22*23)=5/(77*19*23)=5/33649=approx 0.00015
The probability that precisely half of the members will be women is;
P(3 women) = 0.1213
This question will be solved by hypergeometric distribution which has the formula;
P(x) = [S_C_s × (N - S)_C_(n - s)]/(NC_n)
where;
S is success from population
s is success from sample
N is population size
n is sample size
We are give;
s = 3 women (which is precisely half of the members selected)
S = 6 women
N = 24 men and women
n = 6 people selected
Thus;
P(3 women) = (⁶C₃ * ⁽¹⁸⁾C₍₃₎)/(²⁴C₆)
P(3 women) = (20 * 816)/134596
P(3 women) = 0.1213
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What is the result of −18⋅16 2/3? Enter the result as an improper fraction and as a mixed number.
Answer:
-30000/100
300 0/1
Step-by-step explanation:
We have the following numbers -18 and 16 2/3, the first is an integer and the second is a mixed number, the first thing is to pass the mixed number to a decimal number.
16 2/3 = 16.67
We do the multiplication:
−18⋅16 2/3 = -300
We have an improper fraction is a fraction in which the numerator (top number) is greater than or equal to the denominator (bottom number), therefore it would be:
-30000/100
How mixed number would it be:
300 0/1
segment AB is dilated from the origin to create segment A prime B prime at A' (0, 6) and B' (6, 9). What scale factor was segment AB dilated by?
1/2
2
3
4
Answer:
the answer is 3
Step-by-step explanation:
i took the test
The amount of pollutants that are found in waterways near large cities is normally distributed with mean 8.5 ppm and standard deviation 1.4 ppm. 18 randomly selected large cities are studied. Round all answers to two decimal places.
A. xBar~ N( ____) (____)
B. For the 18 cities, find the probability that the average amount of pollutants is more than 9 ppm.
C. What is the probability that one randomly selected city's waterway will have more than 9 ppm pollutants?
D. Find the IQR for the average of 18 cities.Q1 =
Q3 =
IQR:
2. X ~ N(30,10). Suppose that you form random samples with sample size 4 from this distribution. Let xBar be the random variable of averages. Let ΣX be the random variable of sums. Round all answers to two decimal places.
A. xBar~ N(___) (____)
B. P(xBar<30) =
C. Find the 95th percentile for the xBar distribution.
D. P(xBar > 36)=
E. Q3 for the xBar distribution =
Answer:
1)
A) [tex]\frac{}{X}[/tex] ~ N(8.5;0.108)
B) P([tex]\frac{}{X}[/tex] > 9)= 0.0552
C) P(X> 9)= 0.36317
D) IQR= 0.4422
2)
A) [tex]\frac{}{X}[/tex] ~ N(30;2.5)
B) P( [tex]\frac{}{X}[/tex]<30)= 0.50
C) P₉₅= 32.60
D) P( [tex]\frac{}{X}[/tex]>36)= 0
E) Q₃: 31.0586
Step-by-step explanation:
Hello!
1)
The variable of interest is
X: pollutants found in waterways near a large city. (ppm)
This variable has a normal distribution:
X~N(μ;σ²)
μ= 8.5 ppm
σ= 1.4 ppm
A sample of 18 large cities were studied.
A) The sample mean is also a random variable and it has the same distribution as the population of origin with exception that it's variance is affected by the sample size:
[tex]\frac{}{X}[/tex] ~ N(μ;σ²/n)
The population mean is the same as the mean of the variable
μ= 8.5 ppm
The standard deviation is
σ/√n= 1.4/√18= 0.329= 0.33 ⇒σ²/n= 0.33²= 0.108
So: [tex]\frac{}{X}[/tex] ~ N(8.5;0.108)
B)
P([tex]\frac{}{X}[/tex] > 9)= 1 - P([tex]\frac{}{X}[/tex] ≤ 9)
To calculate this probability you have to standardize the value of the sample mean and then use the Z-tables to reach the corresponding value of probability.
Z= [tex]\frac{\frac{}{X} - Mu}{\frac{Sigma}{\sqrt{n} } } = \frac{9-8.5}{0.33}= 1.51[/tex]
Then using the Z table you'll find the probability of
P(Z≤1.51)= 0.93448
Then
1 - P([tex]\frac{}{X}[/tex] ≤ 9)= 1 - P(Z≤1.51)= 1 - 0.93448= 0.0552
C)
In this item, since only one city is chosen at random, instead of working with the distribution of the sample mean, you have to work with the distribution of the variable X:
P(X> 9)= 1 - P(X ≤ 9)
Z= (X-μ)/δ= (9-8.5)/1.44
Z= 0.347= 0.35
P(Z≤0.35)= 0.63683
Then
P(X> 9)= 1 - P(X ≤ 9)= 1 - P(Z≤0.35)= 1 - 0.63683= 0.36317
D)
The first quartile is the value of the distribution that separates the bottom 2% of the distribution from the top 75%, in this case it will be the value of the sample average that marks the bottom 25% symbolically:
Q₁: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₁)= 0.25
Which is equivalent to the first quartile of the standard normal distribution. So first you have to identify the first quartile for the Z dist:
P(Z≤z₁)= 0.25
Using the table you have to identify the value of Z that accumulates 0.25 of probability:
z₁= -0.67
Now you have to translate the value of Z to a value of [tex]\frac{}{X}[/tex]:
z₁= ([tex]\frac{}{X}[/tex]₁-μ)/(σ/√n)
z₁*(σ/√n)= ([tex]\frac{}{X}[/tex]₁-μ)
[tex]\frac{}{X}[/tex]₁= z₁*(σ/√n)+μ
[tex]\frac{}{X}[/tex]₁= (-0.67*0.33)+8.5= 8.2789 ppm
The third quartile is the value that separates the bottom 75% of the distribution from the top 25%. For this distribution, it will be that value of the sample mean that accumulates 75%:
Q₃: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₃)= 0.75
⇒ P(Z≤z₃)= 0.75
Using the table you have to identify the value of Z that accumulates 0.75 of probability:
z₃= 0.67
Now you have to translate the value of Z to a value of [tex]\frac{}{X}[/tex]:
z₃= ([tex]\frac{}{X}[/tex]₃-μ)/(σ/√n)
z₃*(σ/√n)= ([tex]\frac{}{X}[/tex]₃-μ)
[tex]\frac{}{X}[/tex]₃= z₃*(σ/√n)+μ
[tex]\frac{}{X}[/tex]₃= (0.67*0.33)+8.5= 8.7211 ppm
IQR= Q₃-Q₁= 8.7211-8.2789= 0.4422
2)
A)
X ~ N(30,10)
For n=4
[tex]\frac{}{X}[/tex] ~ N(μ;σ²/n)
Population mean μ= 30
Population variance σ²/n= 10/4= 2.5
Population standard deviation σ/√n= √2.5= 1.58
[tex]\frac{}{X}[/tex] ~ N(30;2.5)
B)
P( [tex]\frac{}{X}[/tex]<30)
First you have to standardize the value and then look for the probability:
Z= ([tex]\frac{}{X}[/tex]-μ)/(σ/√n)= (30-30)/1.58= 0
P(Z<0)= 0.50
Then
P( [tex]\frac{}{X}[/tex]<30)= 0.50
Which is no surprise since 30 y the value of the mean of the distribution.
C)
P( [tex]\frac{}{X}[/tex]≤ [tex]\frac{}{X}[/tex]₀)= 0.95
P( Z≤ z₀)= 0.95
z₀= 1.645
Now you have to reverse the standardization:
z₀= ([tex]\frac{}{X}[/tex]₀-μ)/(σ/√n)
z₀*(σ/√n)= ([tex]\frac{}{X}[/tex]₀-μ)
[tex]\frac{}{X}[/tex]₀= z₀*(σ/√n)+μ
[tex]\frac{}{X}[/tex]₀= (1.645*1.58)+30= 32.60
P₉₅= 32.60
D)
P( [tex]\frac{}{X}[/tex]>36)= 1 - P( [tex]\frac{}{X}[/tex]≤36)= 1 - P(Z≤(36-30)/1.58)= 1 - P(Z≤3.79)= 1 - 1 = 0
E)
Q₃: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₃)= 0.75
⇒ P(Z≤z₃)= 0.75
z₃= 0.67
z₃= ([tex]\frac{}{X}[/tex]₃-μ)/(σ/√n)
z₃*(σ/√n)= ([tex]\frac{}{X}[/tex]₃-μ)
[tex]\frac{}{X}[/tex]₃= z₃*(σ/√n)+μ
[tex]\frac{}{X}[/tex]₃= (0.67*1.58)+30= 31.0586
Q₃: 31.0586
Please answer this question I give brainliest thank you! Number 8
Answer:
The third options
Step-by-step explanation:
Counting we can see that 10 students went to two or less states, and 10 went to three or more
Find the Laplace transform F(s)=L{f(t)} of the function f(t)=sin2(wt), defined on the interval t≥0. F(s)=L{sin2(wt)}= help (formulas) Hint: Use a double-angle trigonometric identity. For what values of s does the Laplace transform exist? help (inequalities)
The Laplace transform of the function [tex]\frac{1}{2} (\frac{1}{s} - \frac{s}{s^2 + 4w^2} )[/tex] .
The Laplace transform exist when s > 0 .
Here, the given function is f(t) = sin²(wt) .
The Laplace transform of the the function f(t),
F(s) = f(t) = { [tex]{\frac{1}{2} \times 2sin^2(wt) }[/tex] }
F(s) = { [tex]\frac{1}{2} \times (1- cos2wt)[/tex] }
F(s) = { [tex]\frac{1}{2} - \frac{1}{2} \times cos(2wt)\\[/tex] }
F(s) = [tex]\frac{1}{2} (\frac{1}{s} - \frac{s}{s^2 + 4w^2} )[/tex]
Next,
The above Laplace transform exist if s > 0 .
Know more about Laplace transform,
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Please answer this correctly I want genius expert or ace people to answer this correctly as soon as possible as my work is due today
Answer:
25%
Step-by-step explanation:
The last percentile always contains 25% of the observations.
Find the area of the smaller sector.
A
6 in
030°
Area = [? ]in?
B
Round your answer to the nearest hundredth.
Answer:
9.42 in²
Step-by-step explanation:
The area of whole circle S=pi*R² , where pi is appr. 3.14, R= 6 in
S= 3.14*6² =113.04 in²
The area of smaller sector is Ssec=S/360*30=113,04/12=9.42 in²
The area of the smaller sector with a central angle of 30 degrees and a radius of 6 inches is 9.42478 square inches.
To find the area of a sector, you can use the formula:
Area of sector = (θ/360) × π × r²
where θ is the central angle in degrees, r is the radius of the sector.
The central angle is 30 degrees and the radius is 6 inches.
Plugging these values into the formula:
Area of sector = (30/360) × π × 6²
= (1/12) × π × 36
= (1/12) × 3.14159 × 36
= 9.42478 square inches
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Identify the Type II error if the null hypothesis, H0, is: The capacity of Anna's car gas tank is 10 gallons. And, the alternative hypothesis, Ha, is: Anna believes the capacity of her car's gas tank is not 10 gallons.
Answer:
20gallons
Step-by-step explanation:
Pleassseee hhheeelllppp
Answer/Step-by-step explanation:
When solving problems like this, remember the following:
1. + × + = +
2. + × - = -
3. - × + = -
4. - × - = +
Let's solve:
a. (-4) + (+10) + (+4) + (-2)
Open the bracket
- 4 + 10 + 4 - 2
= - 4 - 2 + 10 + 4
= - 6 + 14 = 8
b. (+5) + (-8) + (+3) + (-7)
= + 5 - 8 + 3 - 7
= 5 + 3 - 8 - 7
= 8 - 15
= - 7
c. (-19) + (+14) + (+21) + (-23)
= - 19 + 14 + 21 - 23
= - 19 - 23 + 14 + 21
= - 42 + 35
= - 7
d. (+5) - (-10) - (+4)
= + 5 + 10 - 4
= 15 - 4 = 11
e. (-3) - (-3) - (-3)
= - 3 + 3 + 3
= - 3 + 9
= 6
f. (+26) - (-32) - (+15) - (-8)
= 26 + 32 - 15 + 8
= 26 + 32 + 8 - 15
= 66 - 15
= 51
Terry has a number cube that is numbered from 1 to 6. She rolls the cube 50 times. Which equation can be used to predict the number of times that she will roll a number that is greater than 4? P (number greater than 4) = StartFraction 1 over 6 EndFraction (50) P (number greater than 4) = StartFraction 2 over 6 EndFraction (50) P (number greater than 4) = StartFraction 3 over 6 EndFraction (50) P (number greater than 4) = StartFraction 4 over 6 EndFraction (50)
Answer:
Step-by-step explanation:
Answer:
B
Step-by-step explanation:
Pet Place sells pet food and supplies including a popular bailed hay for horses. When the stock of this hay drops to 20 bails, a replenishment order is placed. The store manager is concerned that sales are being lost due to stock outs while waiting for a replenishment order. It has been previously determined that demand during the lead-time is normally distributed with a mean of 15 bails and a standard deviation of 6 bails. The manager would like to know the probability of a stockout during replenishment lead-time. In other words, what is the probability that demand during lead-time will exceed 20 bails
Answer:
The probability that demand during lead-time will exceed 20 bails is 0.2033.
Step-by-step explanation:
We are given that it has been previously determined that demand during the lead-time is normally distributed with a mean of 15 bails and a standard deviation of 6 bails.
Let X = demand during the lead-time
So, X ~ Normal([tex]\mu=15, \sigma^{2} = 6^{2}[/tex])
The z-score probability distribution for the normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu=[/tex] population mean demand = 15 bails
[tex]\sigma[/tex] = standard deviation = 6 bails
Now, the probability that demand during lead-time will exceed 20 bails is given by = P(X > 20 bails)
P(X > 20 bails) = P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{20-15}{6}[/tex] ) = P(Z > 0.83) = 1 - P(Z [tex]\leq[/tex] 0.83)
= 1 - 0.7967 = 0.2033
A 2011 survey, by the Bureau of Labor Statistics, reported that 91% of Americans have paid leave. In January 2012, a random survey of 1000 workers showed that 89% had paid leave. The resulting p-value is .0271; thus, the null hypothesis is rejected. It is concluded that there has been a decrease in the proportion of people, who have paid leave from 2011 to January 2012. What type of error is possible in this situation?
Answer:
Is possible to make a Type I error, where we reject a true null hypothesis.
Step-by-step explanation:
We have a hypothesis test of a proportion. The claim is that the proportion of paid leave has significantly decrease from 2011 to january 2012. The P-value for this test is 0.0271 and the nunll hypothesis is rejected.
As the conclusion is to reject the null hypothesis, the only error that we may have comitted is rejecting a true null hypothesis.
The null hypothesis would have stated that there is no significant decrease in the proportion of paid leave.
This is a Type I error, where we reject a true null hypothesis.
3z/10 - 4 = -6
someone help?
Answer:
[tex]z=-\frac{20}{3}[/tex]
Step-by-step explanation:
[tex]\frac{3z}{10}-4=-6\\\\\frac{3z}{10}-4+4=-6+4\\\\\frac{3z}{10}=-2\\\\\frac{10\cdot \:3z}{10}=10\left(-2\right)\\\\3z=-20\\\\\frac{3z}{3}=\frac{-20}{3}\\\\z=-\frac{20}{3}[/tex]
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ali's typing rate between 8:00 am and noon is 48 words per minute . after lunch a lunch break, Ali's typing rate between 1:00 pm and 4:00 pm is 2,040 words per hour . what is Ali's average typing rate per minute for the whole time she works?
Answer:
41 word/min
Step-by-step explanation:
Before noon Ali works:
4 hours= 4*60 min= 240 minShe types:
240*48= 11520 wordsAfter lunch she works:
4 hoursShe types:
4*2040= 8160 wordsTotal Ali works= 4+4= 8 hours= 480 min
Total Ali types= 11520+8160= 19680 words
Average typing rate= 19680 words/480 min= 41 word/min
angle ∠DAC= angle ∠BAD. What is the length of BD? Round to one decimal place.
Answer: 3.9
Step-by-step explanation: Khan Academy
The length of BD if The angle ∠ DAC is equal to the angle ∠ BAD is 3.92.
What is the triangle?Three straight lines coming together create a triangle. There are three sides and three corners on every triangle (angles). A triangle's vertex is the intersection of two of its sides. Any one of a triangle's three sides can serve as its base, however typically the bottom side is used.
Given:
The angle ∠ DAC = angle ∠ BAD
As we can see that the triangle BAD and triangle DAC are similar By SAS similarity,
AC / AB = CD / BD (By the proportional theorem of similarity)
5.6 / 5.1 = 4.3 / BD
1.09 = 4.3 / BD
BD = 4.3 / 1.09
BD = 3.92
Thus, the length of BD is 3.92.
To know more about Triangles:
https://brainly.com/question/16886469
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Find the area of a triangle that has the base of 5 inches and a height of 3 3/4 inches
Answer:
9.375 in^2
Step-by-step explanation:
5/a - 4/b as a single fraction
Answer:
I'm not completely sure what you mean by a, "single fraction," but I'm pretty sure the answer you are looking for is [tex]\frac{5-4}{a-b}[/tex]
Step-by-step explanation:
HELP ASAP GIVING BRANLIST!!
Answer:
Question 1: 3 - 5 hours.
Question 2: 0 - 1 hour
Step-by-step explanation:
Question 1: As you can see in the diagram, the guy is moving really slowly and is almost stuck, therefore, it is 3 - 5 hours.
Question 2: In hours 0 - 1, you can see that the graph is the closest to vertical as it gets.
What is the slope of the line with the two
points A(-4, 8) and B(-9, 12)?
Answer:
slope = -4/5
Step-by-step explanation:
A line passes two points (x1, y1) and (x2, y2).
The slope of this line can be calculate by the formula:
s = (y2 - y1)/(x2 - x1)
=>The line that passes A(-4, 8) and B(-9, 12) has the slope:
s = (12 - 8)/(-9 - -4) = 4/(-5) = -4/5
Hope this helps!
Need help ASAP!! thank you sorry if u can’t see it good :(
Answer/Step-by-step explanation:
==>Given:
=>Rectangular Pyramid:
L = 5mm
W = 3mm
H = 4mm
=>Rectangular prism:
L = 5mm
W = 3mm
H = 4mm
==>Required:
a. Volume of pyramid:
Formula for calculating volume of a rectangular pyramid us given as L*W*H
V = 5*3*4
V = 60 mm³
b. Volume of prism = ⅓*L*W*H
thus,
Volume of rectangular prism given = ⅓*5*3*4
= ⅓*60
= 20mm³
c. Volume of the prism = ⅓ x volume of the pyramid
Thus, 20 = ⅓ × 60
As we can observe from our calculation of the solid shapes given, the equation written above is true for all rectangular prism and rectangular pyramid of the same length, width and height.