Answer:
Part a ) The degrees of freedom for the given two sample non-pooled t test is 24
Part b ) The degrees of freedom for the given two sample non-pooled t test is 30
Part c ) The degrees of freedom for the given two sample non-pooled t test is 30
Part d ) The degrees of freedom for the given two sample non-pooled t test is 25
Step-by-step explanation:
Degrees of freedom for a non-pooled two sample t-test is given by;
Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}
Now given the information;
a) :- m = 12, n = 15, s₁ = 4.0, s₂ = 6.0
we substitute
Δf = {[ 4²/12 + 6²/15 ]²} / {[( 4²/12)²/12-1] + [(6²/15)²/15-1]}
Δf = 30184 / 1241
Δf = 24.3223 ≈ 24 (down to the nearest whole number)
b) :- m = 12, n = 21, s₁ = 4.0, s₂ = 6.0
we substitute using same formula
Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}
Δf = {[ 4²/12 + 6²/21 ]²} / {[( 4²/12)²/12-1] + [(6²/21)²/21-1]}
Δf = 56320 / 1871
Δf = 30.1015 ≈ 30 (down to the nearest whole number)
c) :- m = 12, n = 21, s₁ = 3.0, s₂ = 6.0
we substitute using same formula
Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}
Δf = {[ 3²/12 + 6²/21 ]²} / {[( 3²/12)²/12-1] + [(6²/21)²/21-1]}
Δf = 29095 / 949
Δf = 30.6585 ≈ 30 (down to the nearest whole number)
d) :- m = 10, n = 24, s₁ = 4.0, s₂ = 6.0
we substitute using same formula
Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}
Δf = {[ 4²/10 + 6²/24 ]²} / {[( 4²/10)²/10-1] + [(6²/24)²/24-1]}
Δf = 1044 / 41
Δf = 25.4634 ≈ 25 (down to the nearest whole number).
A box contains 40 identical discs which are either red or white if probably picking a red disc is 1/4. Calculate the number of;
1. White disc.
2. red disc that should be added such that the probability of picking a red disc will be 1/4
Let X denote the day she gets enrolled in her first class and let Y denote the day she gets enrolled in both the classes. What is the distribution of X
Answer:
X is uniformly distributed.
Step-by-step explanation:
Uniform Distribution:
This is the type of distribution where all outcome of a certain event have equal likeliness of occurrence.
Example of Uniform Distribution is - tossing a coin. The probability of getting a head is the same as the probability of getting a tail. The have equal likeliness of occurrence.
602/100 into a decimal describe plz
Answer:
6.02
six point zero two
Step-by-step explanation:
Answer:
602 / 100= 6,02
Step-by-step explanation:
602 to divide 100 = 6,02
The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100, and the standard deviation is 2. You wish to test H0: μ = 100 versus H1: μ ≠ 100 with a sample of n = 9 specimens.
A. If the acceptance region is defined as 98.5 le x- 101.5, find the type I error probability alpha.
B. Find beta for the case where the true mean heat evolved is 103.
C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?
Answer:
A.the type 1 error probability is [tex]\mathbf{\alpha = 0.0244 }[/tex]
B. β = 0.0122
C. β = 0.0000
Step-by-step explanation:
Given that:
Mean = 100
standard deviation = 2
sample size = 9
The null and the alternative hypothesis can be computed as follows:
[tex]\mathtt{H_o: \mu = 100}[/tex]
[tex]\mathtt{H_1: \mu \neq 100}[/tex]
A. If the acceptance region is defined as [tex]98.5 < \overline x > 101.5[/tex] , find the type I error probability [tex]\alpha[/tex] .
Assuming the critical region lies within [tex]\overline x < 98.5[/tex] or [tex]\overline x > 101.5[/tex], for a type 1 error to take place, then the sample average x will be within the critical region when the true mean heat evolved is [tex]\mu = 100[/tex]
∴
[tex]\mathtt{\alpha = P( type \ 1 \ error ) = P( reject \ H_o)}[/tex]
[tex]\mathtt{\alpha = P( \overline x < 98.5 ) + P( \overline x > 101.5 )}[/tex]
when [tex]\mu = 100[/tex]
[tex]\mathtt{\alpha = P \begin {pmatrix} \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{\overline 98.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} + \begin {pmatrix}P(\dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} > \dfrac{101.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} }[/tex]
[tex]\mathtt{\alpha = P ( Z < \dfrac{-1.5}{\dfrac{2}{3}} ) + P(Z > \dfrac{1.5}{\dfrac{2}{3}}) }[/tex]
[tex]\mathtt{\alpha = P ( Z <-2.25 ) + P(Z > 2.25) }[/tex]
[tex]\mathtt{\alpha = P ( Z <-2.25 ) +( 1- P(Z < 2.25) })[/tex]
From the standard normal distribution tables
[tex]\mathtt{\alpha = 0.0122+( 1- 0.9878) })[/tex]
[tex]\mathtt{\alpha = 0.0122+( 0.0122) })[/tex]
[tex]\mathbf{\alpha = 0.0244 }[/tex]
Thus, the type 1 error probability is [tex]\mathbf{\alpha = 0.0244 }[/tex]
B. Find beta for the case where the true mean heat evolved is 103.
The probability of type II error is represented by β. Type II error implies that we fail to reject null hypothesis [tex]\mathtt{H_o}[/tex]
Thus;
β = P( type II error) - P( fail to reject [tex]\mathtt{H_o}[/tex] )
∴
[tex]\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }[/tex]
Given that [tex]\mu = 103[/tex]
[tex]\mathtt{\beta = P( \dfrac{98.5 -103}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-103}{\dfrac{2}{\sqrt{9}}}) }[/tex]
[tex]\mathtt{\beta = P( \dfrac{-4.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-1.5}{\dfrac{2}{3}}) }[/tex]
[tex]\mathtt{\beta = P(-6.75 \leq Z \leq -2.25) }[/tex]
[tex]\mathtt{\beta = P(z< -2.25) - P(z < -6.75 )}[/tex]
From standard normal distribution table
β = 0.0122 - 0.0000
β = 0.0122
C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?
[tex]\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }[/tex]
Given that [tex]\mu = 105[/tex]
[tex]\mathtt{\beta = P( \dfrac{98.5 -105}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-105}{\dfrac{2}{\sqrt{9}}}) }[/tex]
[tex]\mathtt{\beta = P( \dfrac{-6.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-3.5}{\dfrac{2}{3}}) }[/tex]
[tex]\mathtt{\beta = P(-9.75 \leq Z \leq -5.25) }[/tex]
[tex]\mathtt{\beta = P(z< -5.25) - P(z < -9.75 )}[/tex]
From standard normal distribution table
β = 0.0000 - 0.0000
β = 0.0000
The reason why the value of beta is smaller here is that since the difference between the value for the true mean and the hypothesized value increases, the probability of type II error decreases.
Kenji earned the test scores below in English class.
79, 91, 93, 85, 86, and 88
What are the mean and median of his test scores?
Answer:
mean=87
median=87
Step-by-step explanation:
mean=sum of test score/number of subject
mean=79+91+93+85+86+88/6
mean=522/6
mean=87
Literal meaning of median is medium.
To find the number which lies in the medium, we must rearrange the number in ascending.
79, 91, 93, 85, 86, 88
79, 85, 86, 88, 91, 93
86+88/2=87
Hope this helps ;) ❤❤❤
Let me know if there is an error in my answer.
Find usubscript10 in the sequence -23, -18, -13, -8, -3, ...
Step-by-step explanation:
utilise the formula a+(n-1)d
a is the first number while d is common difference
Answer:
22
Step-by-step explanation:
Using the formular, Un = a + (n - 1)d
Where n = 10; a = -23; d = 5
U10 = -23 + (9)* 5
U10 = -23 + 45 = 22
The cost of a daily rental car is as follows: The initial fee is $39.99 for the car, and it costs $0.20 per mile. If Julie's final bill was $100.00 before taxes, how many miles did she drive?
Answer:
300.05 miles
Step-by-step explanation:
initial fee= $39.99
final bill = $ 100
cost =$ 0.20 per mile
remaining amount = $ 60.01
solution,
she drive = remaining amount / cost
=60.01/0.20
=300.05 miles
Answer:
500 miles
Step-by-step explanation:
Let us use cross multiplication to find the unknown amount.
Given:
1) Cost for 1 mile=$0.20
2)Cost for x miles=$100
Solution:
No of miles Cost
1) 1 $0.20
2)x $100
By cross multiplying,
100 x 1= 0.20x
x=100/0.20
x=500 miles
Thank you!
Match the base to the corresponding height.
Base (b)
Height (h)
b
h
h
b
The base 1 is matched with height 2, base 2 is matched with height 3 and base 3 is matched with height 1. The base to the corresponding height is matched in the attached figure.
What is a triangle?Triangle is the closed shaped polygon which has 3 sides and 3 interior angles. The height of the triangle is the dimension of the elevation from the opposite peak to the length of the base.
Thus, the base 1 is matched with height 2, base 2 is matched with height 3 and base 3 is matched with height 1. The base to the corresponding height is matched in the attached figure.
In the given figure, three triangles is shown with base and height. Here,
The base 1 is matched with height 2, as the height shown in figure 2 is the dimension of the elevation from the opposite peak to the length of the base 1.Similarly, base 2 is matched with height 3.Base 3 is matched with height 1.
Thus, the base 1 is matched with height 2, base 2 is matched with height 3 and base 3 is matched with height 1. The base to the corresponding height is matched in the attached figure.
Learn more about the base and height of the triangle here;
https://brainly.com/question/26043588
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Max believes that the sales of coffee at his coffee shop depend upon the weather. He has taken a sample of 5 days. Below you are given the results of the sample.
Cups of Coffee Sold Temperature
350 50
200 60
210 70
100 80
60 90
40 100
A. Which variable is the dependent variable?
B. Compute the least squares estimated line.
C. Compute the correlation coefficient between temperature and the sales of coffee.
D. Predict sales of a 90 degree day.
Answer:
1. cups of coffee sold
2.Y = 605.7 - 5.943x
3. -0.952
4. 70.84
Step-by-step explanation:
1. the dependent variable in this question is the cups of coffee sold
2. least square estimation line
Y = a+bx
we have y as the cups of coffee sold
x as temperature.
first we will have to solve for a and then b
∑X = 450
∑Y = 960
∑XY = 61600
∑X² = 35500
∑Y² = 221800
a = ∑y∑x²-∑x∑xy/n∑x²-(∑x)²
a = 960 * 35500-450*61600/6*35500-450²
a = 6360000/10500
= 605.7
b = n∑xy - ∑x∑y/n∑x²-(∑x)²
= 6*61600 - 450*960/6*35500 - 450²
= -5.943
the regression line
Y = a + bx
Y = 605.7 - 5.943x
3. we are to find correlation coefficient
r = n∑xy - ∑x∑y multiplied by√(n∑x²-(∑x)² * (n∑y² - (∑y)²)
= 6*61600 -960*450/√(6*35500 - 450²)*(6*221800 - 960²)
=-62400/√4296600000
= -62400/65548.5
= -0.952
4. we have to predict sales of a 90 degree day fro the regression line
Y = 605.7 - 5.943x
y = 605.7 - 5.943(90)
y = 605.7 - 534.87
= 70.84
If the normality requirement is not satisfied (that is, np(1p) is not at least 10), then a 95% confidence interval about the population proportion will include the population proportion in ________ 95% of the intervals. (This is a reading assessment question. Be certain of your answer because you only get one attempt on this question.)
Answer:
less than
Step-by-step explanation:
If the normality requirement is not satisfied (that is, np(1 - p) is not at least 10), then a 95% confidence interval about the population proportion will include the population proportion in _less than__ 95% of the intervals.
The confidence interval consist of all reasonable values of a population mean. These are value for which the null hypothesis will not be rejected.
So, let assume that If the 95% confidence interval contains the value for the hypothesized mean, then the sample mean is reasonably close to the hypothesized mean. The effect of this is that the p- value is going to be greater than 0.05, so we fail to reject the null hypothesis.
On the other hand,
If the 95% confidence interval do not contains the value for the hypothesized mean, then the sample mean is far away from the hypothesized mean. The effect of this is that the p- value is going to be lesser than 0.05, so we reject the null hypothesis.
please help me in these question ????
A school bag contains 12 pens of which 5 are red and the other are black. 4 pens are selected from the bag.
(a) How many different samples of size 4 pens are possible?
(b) How many samples have 3 red pens and 1 black pen?
(c) How many samples of size 4 contain at least two red pens?
(d) How many samples of size 4 contain
If the average yield of cucumber acre is 800 kg, with a variance 1600 kg, and that the amount of the cucumber follows the normal distribution.
1- What percentage of a cucumber give the crop amount between and 834 kg?
2- What the probability of cucumber give the crop exceed 900 kg ?
Answer:
Step-by-step explanation:
A school bag contains 12 pens of which 5 are red and the other are black. 4 pens are selected from the bag.
(a) How many different samples of size 4 pens are possible?
12C4=12!/(4!*8!)=495
(b) How many samples have 3 red pens and 1 black pen?
5C3*7C1
5C3=5!/(3!*2!)=10
7C1=7!/(1!*6!)=7
=>5C3*7C1=10*7=70
(c) How many samples of size 4 contain at least two red pens?
(5C2*7C2)+(5C3*7C1)+(5C4*7C0)
5C2=5!/(2!*3!)=10
7C2=7!/(2!*5!)=21
5C3=5!/(3!*2!)=10
7C1=7!/(1!*6!)=7
5C4=5!/(4!*1!)=5
7C0=7!/(0!*7!)=1
=>(5C2*7C2)+(5C3*7C1)+(5C4*7C0)=285
(d) How many samples of size 4 contain at most one black pen?
(7C1*5C3)+(7C0*5C4)
7C1=7!/(1!*6!)=7
7C0=7!/(0!*7!)=1
5C3=5!/(3!*2!)=10
5C4=5!/(4!*1!)=5
=>(7C1*5C3)+(7C0*5C4)=(7*10)+(1*5)=75
Time
(minutes)
Water
(gallons)
1
16.50
1.5
24.75
2
33
find the constant of proportionality for the second and third row
Answer:
16.50
Step-by-step explanation:
Constant of proportionality = no of gallons of water per 1 minute.
In the first row, we have 16.50 gallons of water per 1 minute.
In the 2nd row, we have 24.75 gallons of water in 1.5 minutes. In 1 minute, we will have 24.75 ÷ 1.5 = 16.50 gallons
In the 3rd row, we have 33 gallons in 2 minutes. In 1 minute, we will have 33 ÷ 2 = 16.50 gallons.
We can see that there seems to be the same constant of proportionality for the 2nd and 3rd row, which is 16.50.
Thus, a relationship between gallons of water (w) and time (t), considering the constant, 16.50, can be written as: [tex] w = 16.50t [/tex]
This means the constant of proportionality, 16.50, is same for all rows.
The cost, C, in United States Dollars ($), of cleaning up x percent of an oil spill along the Gulf Coast of the United States increases tremendously as x approaches 100. One equation for determining the cost (in millions $) is:
Complete Question
On the uploaded image is a similar question that will explain the given question
Answer:
The value of k is [tex]k = 214285.7[/tex]
The percentage of the oil that will be cleaned is [tex]x = 80.77\%[/tex]
Step-by-step explanation:
From the question we are told that
The cost of cleaning up the spillage is [tex]C = \frac{ k x }{100 - x }[/tex] [tex]x \le x \le 100[/tex]
The cost of cleaning x = 70% of the oil is [tex]C = \$500,000[/tex]
Now at [tex]C = \$500,000[/tex] we have
[tex]\$ 500000 = \frac{ k * 70 }{100 - 70 }[/tex]
[tex]\$ 500000 = \frac{ k * 70 }{30 }[/tex]
[tex]\$ 500000 = \frac{ k * 70 }{30 }[/tex]
[tex]k = 214285.7[/tex]
Now When [tex]C = \$900,000[/tex]
[tex]x = 80.77\%[/tex]
Salaries of 42 college graduates who took a statistics course in college have a mean, , of . Assuming a standard deviation, , of $, construct a % confidence interval for estimating the population mean .
Answer:
The 99% confidence interval for estimating the population mean μ is ($60,112.60, $68087.40).
Step-by-step explanation:
The complete question is:
Salaries of 42 college graduates who took a statistics course in college have a mean, [tex]\bar x[/tex] of, $64, 100. Assuming a standard deviation, σ of $10,016 construct a 99% confidence interval for estimating the population mean μ.
Solution:
The (1 - α)% confidence interval for estimating the population mean μ is:
[tex]CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]
The critical value of z for 99% confidence interval is:
[tex]z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.57[/tex]
Compute the 99% confidence interval for estimating the population mean μ as follows:
[tex]CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]
[tex]=64100\pm 2.58\times\frac{10016}{\sqrt{42}}\\\\=64100+3987.3961\\\\=(60112.6039, 68087.3961)\\\\\approx (60112.60, 68087.40)[/tex]
Thus, the 99% confidence interval for estimating the population mean μ is ($60,112.60, $68087.40).
HELP ASAP ROCKY!!! will get branliest.
Answer:
work pictured and shown
Answer:
Last one
Step-by-step explanation:
● [ ( 3^2 × 5^0) / 4 ]^2
5^0 is 1 since any number that has a null power is equal to 1.
●[ (3^2 ×1 ) / 4 ]^2
● (9/4)^2
● 81 / 16
5x+4(-x-2)=-5x+2(x-1)+12
Answer:
x=9/2
Step-by-step explanation:
Let's solve your equation step-by-step.
5x+4(−x−2)=−5x+2(x−1)+12
Step 1: Simplify both sides of the equation.
5x+4(−x−2)=−5x+2(x−1)+12
5x+(4)(−x)+(4)(−2)=−5x+(2)(x)+(2)(−1)+12 (Distribute)
5x+−4x+−8=−5x+2x+−2+12
(5x+−4x)+(−8)=(−5x+2x)+(−2+12) (Combine Like Terms)
x+−8=−3x+10
x−8=−3x+10
Step 2: Add 3x to both sides.
x−8+3x=−3x+10+3x
4x−8=10
Step 3: Add 8 to both sides.
4x−8+8=10+8
4x=18
Step 4: Divide both sides by 4.
4x/4=18/4
x=9/2
The quotient of 8 and the difference of three and a number.
Answer: 8÷(3-x)
Answer:
Below
Step-by-step explanation:
● 8 ÷ (3-x)
Dividing by 3-x is like multiplying by 1/(3-x)
● 8 × (1/3-x)
● 8 /(3-x)
The following shape is based only on squares, semicircles, and quarter circles. Find the area of the shaded part.
Answer:
this? hope it helps ........
Answer:
The answer is area=32pi-64 and the perimeter is 8pi
Step-by-step explanation:
one third multiplied by the sum of a and b
Answer:
1/3(a+b)
hope it helps :>
find the area of square whose side is 2.5 cm
Answer:
6.25
Step-by-step explanation:
2.5 *2.5=6.25
Answer:
6.25cm^2.
Step-by-step explanation:
To find the area of a square, you multiply the two sides, 2.5✖️2.5.
This gives the area of 6.25cm^2.
Hope this helped!
Have a nice day:)
Two sides of a triangle are equal length. The length of the third side exceeds the length of one of the other sides by 3 centimeters. The perimeter of the triangle is 93 centimeters. Find the length of each of the shorter sides of the triangle
Answer:
30 cm
Step-by-step explanation:
let x be the lenght of the two sides of equal lenghts, so the other is x+3
and the perimeter is x+x +x +3
P=3x+3
P=3(x+1)
93=3(x+1)
31=x+1
x=30
so the shorter sides are of 30 centimeters and the longest is 33
A mass of 5 kg stretches a spring 10 cm. The mass is acted on by an external force of 10sin( t ) N(newtons) and moves in a medium that imparts a viscous force of 2 N
when the speed of the mass is 4 cm/s. If the mass is set in motion from its equilibrium position with an initial velocity of 3 cm/s, formulate the initial value problem describing the motion of the mass.
A)Find the solution of the initial value problem in the above problem.
B)Plot the graph of the steady state solution
C)If the given external force is replaced by a force of 2 cos(ωt) of frequency ω , find the value of ω for which the amplitude of the forced response is maximum.
Answer:
A) C1 = 0.00187 m = 0.187 cm, C2 = 0.0062 m = 0.62 cm
B) A sample of how the graph looks like is attached below ( periodic sine wave )
C) w = [tex]\sqrt[4]{3}[/tex] is when the amplitude of the forced response is maximum
Step-by-step explanation:
Given data :
mass = 5kg
length of spring = 10 cm = 0.1 m
f(t) = 10sin(t) N
viscous force = 2 N
speed of mass = 4 cm/s = 0.04 m/s
initial velocity = 3 cm/s = 0.03 m/s
Formulating initial value problem
y = viscous force / speed = 2 N / 0.04 m/s = 50 N sec/m
spring constant = mg/ Length of spring = (5 * 9.8) / 0.1 = 490 N/m
f(t) = 10sin(t/2) N
using the initial conditions of u(0) = 0 m and u"(0) = 0.03 m/s to express the equation of motion
the equation of motion = 5u" + 50u' + 490u = 10sin(t/2)
A) finding the solution of the initial value
attached below is the solution and
B) attached is a periodic sine wave replica of how the grapgh of the steady state solution looks like
C attached below
a
A solid metal cone of base radius a cm and height 2a cm is melted and solid
spheres of radius are made without wastage. How many such spheres can be
made?
volume of a cone
.
.
.
volume of sphere
.
.
number of spheres that can be made......
.
.
hence a hemisphere can be formed
Find a cubic polynomial with integer coefficients that has $\sqrt[3]{2} + \sqrt[3]{4}$ as a root.
Find the powers [tex]a=\sqrt{2}+\sqrt{3}[/tex]
$a^{2}=5+2 \sqrt{6}$
$a^{3}=11 \sqrt{2}+9 \sqrt{3}$
The cubic term gives us a clue, we can use a linear combination to eliminate the root 3 term $a^{3}-9 a=2 \sqrt{2}$ Square $\left(a^{3}-9 a\right)^{2}=8$ which gives one solution. Expand we have $a^{6}-18 a^{4}-81 a^{2}=8$ Hence the polynomial $x^{6}-18 x^{4}-81 x^{2}-8$ will have a as a solution.
Note this is not the simplest solution as $x^{6}-18 x^{4}-81 x^{2}-8=\left(x^{2}-8\right)\left(x^{4}-10 x^{2}+1\right)$
so fits with the other answers.
Answer:
[tex]y^3 -6y-6[/tex]
Foram prescritos 500mg de dipirona para uma criança com febre.Na unidade tem disponivel ampola de 1g/2ml.Quantos g vão ser administrados no paciente
De acordo com a disponibilidade da unidade, há apenas a seguinte dosagem: 1g/2mL - ou seja, uma grama de dipirona a cada 2mL
O enunciado está meio mal formulado, pois é dito que foram prescritos 500mg de dipirona e é essa quantidade de farmaco que a criança tem que tomar. Deseja-se saber quantos mL deverao ser administrados.
Fazendo a classica regra de 3, podemos chegar no volume desejado:
(atentar que 500mg = 0,5g)
g mL
1 --------- 2
0,5 --------- X
1 . X = 0,5 . 2
X = 1mLThe quotient of 3 and the cube
of y+2
Answer:
[tex]\dfrac{3}{(y+2)^3}[/tex]
Step-by-step explanation:
Maybe you want this written using math symbols. It will be ...
[tex]\boxed{\dfrac{3}{(y+2)^3}}[/tex]
If f(x)=x/2-3and g(x)=4x^2+x-4, find (f+g)(x)
Step-by-step explanation:
(f+g)(x) = f(x) + g(x)
= x/2-3 + 4x²+x+4
= ..........
Findℒ{f(t)}by first using a trigonometric identity. (Write your answer as a function of s.)f(t) = 12 cost −π6
Answer:
[tex]L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ][/tex]
Step-by-step explanation:
Given that:
[tex]f(t) = 12 cos (t- \dfrac{\pi}{6})[/tex]
recall that:
cos (A-B) = cos AcosB + sin A sin B
∴
[tex]f(t) = 12 [cos\ t \ cos \dfrac{\pi}{6}+ sin \ t \ sin \dfrac{\pi}{6}][/tex]
[tex]f(t) = 12 [cos \ t \ \dfrac{3}{2}+ sin \ t \ sin \dfrac{1}{2}][/tex]
[tex]f(t) = 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t)[/tex]
[tex]L(f(t)) = L ( 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t) ][/tex]
[tex]L(f(t)) = 6 \sqrt{3} \ L [cos \ (t) ] + 6\ L [ sin \ (t) ][/tex]
[tex]L(f(t)) = 6 \sqrt{3} \dfrac{S}{S^2 + 1^2}+ 6 \dfrac{1}{S^2 +1^2}[/tex]
[tex]L(f(t)) = \dfrac{6 \sqrt{3} +6 }{S^2+1}[/tex]
[tex]L(f(t)) = \dfrac{6( \sqrt{3} \ S +1 }{S^2+1}[/tex]
[tex]L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ][/tex]
88 feet/second = 60 miles/hour. How many feet per second is 1 mile/hour? (Hint: divide both sides of the equation
by the same amount.)
Round to the nearest thousandth.
One mile per hour is equivalent to
ao feet/second
A machine used to fill gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of ounces and a standard deviation of ounce. You randomly select cans and carefully measure the contents. The sample mean of the cans is ounces. Does the machine need to be reset? Explain your reasoning. ▼ Yes No , it is ▼ very unlikely likely that you would have randomly sampled cans with a mean equal to ounces, because it ▼ lies does not lie within the range of a usual event, namely within ▼ 1 standard deviation 2 standard deviations 3 standard deviations of the mean of the sample means.
Complete question is;
A machine used to fill gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of 128 ounces and a standard deviation of 0.20 ounce. You randomly select 35 cans and carefully measure the contents. The sample mean of the cans is 127.9 ounces. Does the machine need to be? reset? Explain your reasoning.
(yes/no)?, it is (very unlikely/ likely) that you would have randomly sampled 35 cans with a mean equal to 127.9 ?ounces, because it (lies/ does not lie) within the range of a usual? event, namely within (1 standard deviation, 2 standard deviations 3 standard deviations) of the mean of the sample means.
Answer:
Yes, we should reset the machine because it is unusual to have a mean equal to 127.9 from a random sample of 35 as the mean of 127.9 doesn't fall within range of a usual event with 2 standard deviations of the mean of the sample means.
Step-by-step explanation:
We are given;
Mean: μ = 128
Standard deviation; σ = 0.2
n = 35
Now, formula for standard error of mean is given as;
se = σ/√n
se = 0.2/√35
se = 0.0338
Normally, the range of values should be within 2 standard deviations of mean. In this case, normal range of values will be;
μ ± 2se = 128 ± 0.0338
This gives; 127.9662, 128.0338
So, Yes, we should reset the machine because it is unusual to have a mean equal to 127.9 from a random sample of 35 as the mean of 127.9 doesn't fall within range of a usual event with 2 standard deviations of the mean of the sample means.
