Answer:
[tex]V(m) = (2 + 5m)^3[/tex]
Step-by-step explanation:
Given
Solid Shape = Cube
Edge = 2 feet
Increment = 5 feet per minute
Required
Determine volume as a function of minute
From the question, we have that the edge of the cube increases in a minute by 5 feet
This implies that,the edge will increase by 5m feet in m minutes;
Hence,
[tex]New\ Edge = 2 + 5m[/tex]
Volume of a cube is calculated as thus;
[tex]Volume = Edge^3[/tex]
Substitute 2 + 5m for Edge
[tex]Volume = (2 + 5m)^3[/tex]
Represent Volume as a function of m
[tex]V(m) = (2 + 5m)^3[/tex]
Which of the following is equal to the rational expression below when x=-1
or -8?
11(x+8)
/(x + 1)(x+8)
Answer:
11/(x + 1) thus d: is the answer
Step-by-step explanation:
Simplify the following:
(11 (x + 8))/((x + 1) (x + 8))
(11 (x + 8))/((x + 1) (x + 8)) = (x + 8)/(x + 8)×11/(x + 1) = 11/(x + 1):
Answer: 11/(x + 1)
The diameter of steel rods manufactured on two different extrusion machines is being investigated. Two random samples of sizes n1"=15 and n2"=17 are selected, and the sample means and sample variances are x1 =8.73, s2=0.35, x =8.68, and s2=0.40, respectively. Assume that σ1^2 = σ2^2 that the data are drawn from a normal distribution.
Required:
a. Is there evidence to support the claim that the two machines produce rods with different mean diameters? Use alpha=0.05 in arriving at this conclusion.
b. Find the P-value for thet-statistic you calculated in part (a).
c. Construct a 95% confidence interval for the difference in mean rod diameter. Interpret this interval.
Answer:
a) No sufficient evidence to support the claim that the two machines produce rods with different mean diameters.
b) P-value is 0.80
c) −0.3939 <μ< 0.4939
Step-by-step explanation:
Given Data:
sample sizes
n1 = 15
n2 = 17
sample means:
x1 = 8.73
x2 = 8.68
sample variances:
s1² = 0.35
s2² = 0.40
Hypothesis:
H₀ : μ₁ = μ₂
H₁ : μ₁ ≠ μ₂
Compute the pooled standard deviation:
[tex]s_{p} = \sqrt{\frac{(n_{1} - 1)s_{1}^{2} + (n_{2} - 1)s_{2}^{2}}{n_{1} +n_{2} -2} }[/tex]
[tex]= \sqrt{\frac{(15-1)0.35+(17-1)0.40}{15+7-2}}[/tex]
[tex]= \sqrt{\frac{(14)0.35+(16)0.40}{30}}[/tex]
[tex]= \sqrt{\frac{4.9+6.4}{30}}[/tex]
[tex]= \sqrt{\frac{11.3}{30}}[/tex]
[tex]= \sqrt{0.376667}[/tex]
= 0.613732
= 0.6137
Compute the test statistic:
[tex]t = \frac{x_{1} -x_{2} }{s_{p} \sqrt{\frac{1}{n_{1} }+\frac{1}{n_{2} } } }[/tex]
[tex]= \frac{8.73-8.68}{0.6137\sqrt{\frac{1}{15}+\frac{1}{17} } }[/tex]
[tex]= \frac{0.05}{0.6137\sqrt{0.06667+0.05882} } }[/tex]
[tex]= \frac{0.05}{0.6137\sqrt{0.12549} } }[/tex]
[tex]= \frac{0.05}{0.6137(0.354246)} } }[/tex]
[tex]= \frac{0.05}{0.6137(0.354246)} } }[/tex]
= 0.05 / 0.217401
= 0.22999
t = 0.230
Compute degree of freedom:
df = n1 + n2 -2 = 15 + 17 - 2 = 30
Compute the P-value from table using df = 30
P > 2 * 0.40 = 0.80
P > 0.05 ⇒ Fail to reject H₀
Null hypothesis is rejected when P-value is less than or equals to level of significance. But here the P-value = 0.80 and level of significance = 0.05. So P-value is greater than significance level. Hence there is not sufficient evidence to support the claim that population means are different.
Construct a 95% confidence interval for the difference in mean rod diameter:
confidence = c = 95% = 0.95
α = 1 - c
= 1 - 0.95
α = 0.05
Compute degree of freedom:
df = n1 + n2 -2 = 15 + 17 - 2 = 30
Compute [tex]t_{\alpha /2}[/tex] with df = 30 using table:
t₀.₀₂₅ = 2.042
Compute confidence interval:
= [tex](x_{1}-x_{2})-t_{\alpha/2} ( s_{p} )\sqrt{\frac{1}{n_{1} }+\frac{1}{n_{2} } }[/tex]
= (8.73 - 8.68) - 2.042 ( 0.6137 ) [tex]\sqrt{\frac{1}{15} +\frac{1}{17} }[/tex]
= 0.05 - 2.042 ( 0.6137 ) [tex]\sqrt{\frac{1}{15} +\frac{1}{17} }[/tex]
= 0.05 - 1.253175 [tex]\sqrt{0.06667+0.05882} } }[/tex]
= 0.05 - 1.253175 [tex]\sqrt{0.12549} } }[/tex]
= 0.05 - 1.253175 (0.35424))
= 0.05 - 0.443925
= −0.393925
= −0.3939
[tex](x_{1}-x_{2})+t_{\alpha/2} ( s_{p} )\sqrt{\frac{1}{n_{1} }+\frac{1}{n_{2} } }[/tex]
= (8.73 - 8.68) + 2.042 ( 0.6137 ) [tex]\sqrt{\frac{1}{15} +\frac{1}{17} }[/tex]
= 0.05 + 2.042 ( 0.6137 ) [tex]\sqrt{\frac{1}{15} +\frac{1}{17} }[/tex]
= 0.05 + 1.253175 [tex]\sqrt{0.06667+0.05882} } }[/tex]
= 0.05 + 1.253175 [tex]\sqrt{0.12549} } }[/tex]
= 0.05 + 1.253175 (0.35424))
= 0.05 + 0.443925
= 0.493925
= 0.4939
−0.3939 <μ₁ - μ₂< 0.4939
Julissa gave out an equal number of oranges to each of the 6 apartments on her floor. if she gave each apartment 5 oranges, how many oranges did Julissa give out in all?
julissa gave equal oranges in 6 apartments
she gave each apartment 5 oranges
so total no. of oranges are = 6×5 = 30
Answer:
D. 30
Step-by-step explanation:
Find the value of x. A: 15 B: 12 C: 10 D: 8
Answer:
[tex]\boxed{\sf C. \ 10}[/tex]
Step-by-step explanation:
[tex]\sf The \ intersecting \ chord \ theorem \ states \ that \ the \ products[/tex]
[tex]\sf of \ the \ lengths \ of \ the \ line \ segments \ on \ each \ chord \ are \ equal.[/tex]
[tex]NH \times HT = MH \times HY[/tex]
[tex](x+20) \times 8=12 \times 20[/tex]
[tex]\sf Expand \ brackets \ and \ multiply.[/tex]
[tex]8x+160=240[/tex]
[tex]\sf Subtract \ 160 \ from \ both \ sides.[/tex]
[tex]8x+160-160=240-160[/tex]
[tex]8x=80[/tex]
[tex]\sf Divide \ both \ sides \ by \ 8.[/tex]
[tex]\displaystyle \frac{8x}{8} =\frac{80}{8}[/tex]
[tex]x=10[/tex]
The value of x is 10.
We have a circle and inside it two chords MY and NT intersect at point H.
We have to find the value of x in the figure.
What is intersecting chord theorem?According to the intersecting chord theorem, when two chords say AB and CD intersect at point O, then
AO x OB = CO x OD
Applying the chord intersecting theorem to the figure in the question, we get -
MH x HY = NH x HT
12 x 20 = (x+20) x 8
240 = 8x + 160
8x = 80
x = 10
Hence the value of x is 10.
To solve more questions on Circles and chords, visit the link below -
https://brainly.com/question/15568573
#SPJ5
Factor 4(20) + 84. 4(20 + 21) 4(21 + 20) 20(4 + 84) 20(4 + 4)
Answer:
[tex]\huge\boxed{4 ( 20 + 21)}[/tex]
Step-by-step explanation:
4(20) + 84
Resolve Parenthesis
80 + 84
Taking 4 common as both are the multiples of 4
4 ( 20 + 21)
You look over the songs in a jukebox and determine that you like of the songs. (a) What is the probability that you like the next four songs that are played? (Assume a song cannot be repeated.) (b) What is the probability that you do not like the any of the next four songs that are played? (Assume a song cannot be repeated.) (a) The probability that you like the next four songs that are played is nothing. (Round to three decimal places as needed.) (b) The probability that you do not like any of the next four songs that are played is nothing. (Round to three decimal places as needed.)
Complete Question
You look over the songs in a jukebox and determine that you like 18 of 59 songs.
(a) What is the probability that you like the next four songs that are played? (Assume a song cannot be repeated) Round to three decimal places as needed)
(b) What is the probability that you do not like the next four songs that are played? (Assume a song cannot be repeated.) Round to three decimal places as needed
Answer:
a
[tex]P = 0.0067[/tex]
b
[tex]Q = 0.222[/tex]
Step-by-step explanation:
From the question we are told that
The total number of songs is [tex]n = 59[/tex]
The number of songs you liked is [tex]k = 18[/tex]
The probability that you like the next four songs that are played? (Assume a song cannot be repeated) is mathematically represented as
[tex]P = \frac{ ^{k} C _4 }{ ^{n} C _4}[/tex]
=> [tex]P = \frac{ ^{18} C _4 }{ ^{59} C _4}[/tex]
Now using a combination calculator
[tex]P= \frac{ 3060}{ 455126}[/tex]
[tex]P = 0.0067[/tex]
The probability that you do not like the next four songs that are played? (Assume a song cannot be repeated.) is mathematically evaluated as
[tex]Q = \frac{ ^{n- k} C _4 }{ ^{n} C _4}[/tex]
=> [tex]Q = \frac{ ^{59- 18} C _4 }{ ^{n} C _4}[/tex]
=> [tex]Q = \frac{ ^{41} C _4 }{ ^{59} C _4}[/tex]
Now using a combination calculator
[tex]Q = \frac{ 101270}{ 455126}[/tex]
[tex]Q = 0.222[/tex]
Find the minimum sample size n needed to estimate for the given values of c, , and E. c, , and E Assume that a preliminary sample has at least 30 members.
Answer:
hello your question is incomplete below is the complete question
Find the minimum sample size n needed to estimate μ For the given values of c, σ, and E. c=0.98, σ=6.5, and E=22 Assume that a preliminary sample has at least 30 members.
Answer : 48
Step-by-step explanation:
Given data:
E = 2.2,
std ( σ ) = 6.5
c ( level of confidence ) = 0.98
To find the minimum sample size
we have to first obtain the value of [tex]Z_{a/2}[/tex]
note : a can be found using this relation :
( 1 - a ) = 0.98 ----- equation 1
a = 1 - 0.98 = 0.02
hence: a/2 = 0.01
This means that P( Z ≤ z ) = 0.99 the value of z can be found using the table of standard normal distribution. from the table the value of z = 2.33
P( Z ≤ 2.33 ) = 0.99
To obtain the sample size n
[tex]n = (\frac{std*z}{E} )^{2}[/tex]
n = [tex](\frac{6.5*2.33}{2.2} )^2[/tex] = (6.88409)^2
Therefore n ≈ 48
Records indicate that x years after 2008, the average property tax on a three bedroom home in a certain community was T(x) =20x^2+40x+600 dollars.
Required:
a. At what rate was the property tax increasing with respect to time in 2008?
b. By how much did the tax change between the years 2008 and 2012?
Answer:
a) 40 dollars
b) 480 dollars
Step-by-step explanation:
Given the average property tax on a three bedroom home in a certain community modelled by the equation T(x) =20x²+40x+600, the rate at which the property tax is increasing with respect to time in 2008 can be derived by solving for the function T'(x) at x=0
T'(x) = 2(20)x¹ + 40x° + 0
T'(x) = 40x+40
At x = 0,
T'(0) = 40(0)+40
T'(0) = 40
Hence the property tax was increasing at a rate of 40dollars with respect to the initial year (2008).
b) There are 4 years between 2008 and 2012. To know how much that the tax change between the years 2008 and 2012, we will find T(4) - T(0)
Given T(x) =20x²+40x+600
T(4) =20(4)²+40(4)+600
T(4) = 320+160+600
T(4) = 1080 dollars
Also T(0) =20(0)²+40(0)+600
T(0) = 0+0+600
T(0)= 600 dollars
T(4) - T(0) = 1080 - 600
T(4) - T(0) = 480 dollars
Hence, the tax has changed by $480 between 2008 and 2012
The coffee cups can hold 7/9 of a pint of liquid. If Emily pours 2/3 of a pint of coffee into a cup,how much milk can a customer add? PLZ HELP!
Answer:
1/9
Step-by-step explanation:
easy 2/3 is equivalent to 6/9. So there is 1/9 of a pint left
Pattern A: 0, 5, 10, 15, 20,... Pattern B: 0, 20, 40, 60, 80,... Which statement is true about the relationship between the corresponding terms of Pattern A and Pattern B? A. The terms in Pattern B is 4 times the corresponding terms in Pattern A. B. The terms in Pattern A is 1/2 times the corresponding terms in Pattern B. C. The terms in Pattern B is 20 more than the corresponding terms in Pattern A. D. The terms in Pattern A is 5 more than the corresponding terms in Pattern B.
Answer:
Option 1: The terms in Pattern B is 4 times the corresponding terms of Pattern A
Step-by-step explanation:
Answer:
Pattern B has more then pattern A so option 2
Step-by-step explanation:
Given the function, Calculate the following values:
Answer:
[tex]f(-2)=33\\f(-1)=12\\f(0)=1\\f(1)=0\\f(2)=9[/tex]
Step-by-step explanation:
[tex]f(x)=5x^{2} -6x+1\\f(-2)=5(-2)^{2} -6(-2)+1\\f(-2)=5(4)+12+1\\f(-2)=20+13\\f(-2)=33[/tex]
[tex]f(x)=5x^{2}-6x+1\\f(-1)=5(-1)^{2} -6(-1)+1\\f(-1)=5(1)+6+1\\f(-1)=5+7\\f(-1)=12[/tex]
[tex]f(x)=5x^{2}-6x+1\\f(0)=5(0)^{2}-6(0)+1\\f(0)=5(0)-0+1\\f(0)=0+1\\f(0)=1[/tex]
[tex]f(x)=5x^{2}-6x+1\\f(1)=5(1)^{2}-6(1)+1\\f(1)=5(1)-6+1\\f(1)=5-5\\f(1)=0[/tex]
[tex]f(x)=5x^{2}-6x+1\\f(2)=5(2)^{2}-6(2)+1\\f(2)=5(4)-12+1\\f(2)=20-11\\f(2)=9[/tex]
Hey market sales six cans of food for every seven boxes of food the market sold a total of 26 cans and boxes today how many of each kind did the market sale
Answer:
It sold 14 cans boxes of food and 12 cans of food.
Step-by-step explanation:
The factor for the food cans depend upon every seven food boxes .So, the same no. of sets of food cans will be sold.
Let the no. of sets of food boxes be x.
According to the question,
6x+7x=26
13x=26
x=26/13
x=2
No. of food cans =6x=6×2=12 cans
No. of food boxes=7x=7×2=14 boxes
Please mark brainliest ,if it is truly the best ! Thank you!
Daniel and Jack together sell 96 tickets to a raffle. Daniel sold 12 more tickets than his friend. How many raffle tickets each friend sell?
Answer:
Daniel sold 54 and Jack sold 42
Step-by-step explanation:
D = number of tickets that Daniel sold
J = number of tickets that Jack sold
D + J = 96
D = 12+ J
Substitute the second equation into the first equation
12 + J + J = 96
Combine like terms
12 + 2J = 96
Subtract 12 from each side
2J = 84
Divide by 2
J = 42
D = J+12
D = 54
Daniel sold 54 and Jack sold 42
Answer:
Jack sold 42 & Daniel sold 54.
Step-by-step explanation:
96 - 12 = 84
84 / 2 = 42
Jack sold 42.
42 + 12 = 54
Daniel sold 54.
42 + 54 = 96
which statement correctly describes the relation between the variable in the equation C = nd
Answer:
nd is c
Step-by-step explanation:
It has been reported that 20.4% of incoming freshmen indicate that they will major in business or a related field. A random sample of 400 incoming college freshmen was asked their preference, and 95 replied that they were considering business as a major. Estimate the true proportion of freshman business majors with 98% confidence. Does your interval contain 20.4%?
Answer:
The 98% confidence interval
[tex]0.1884 < p < 0.2876[/tex]
The confidence interval contains 20.4%
Step-by-step explanation:
From the question we are told that
The sample size is n = 400
The number that replied that they were considering business as a major [tex]x = 95[/tex]
The sample proportion is mathematically evaluated as
[tex]\r p = \frac{95}{400}[/tex]
[tex]\r p = 0.238[/tex]
Given that the confidence level 98% then the level of significance is evaluated as
[tex]\alpha = 100 - 98[/tex]
[tex]\alpha = 2 \%[/tex]
[tex]\alpha = 0.02[/tex]
Next we obtain the critical value of [tex]\frac{ \alpha }{2}[/tex] from the normal distribution table is
[tex]Z_{\frac{ \alpha }{2} } = 2.33[/tex]
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{ \alpha }{2} } * \sqrt{ \frac{ p (1 - p )}{n} }[/tex]
[tex]E = 2.33 * \sqrt{ \frac{ 0.238 (1 - 0.238 )}{400} }[/tex]
[tex]E = 0.0496[/tex]
The 98% confidence interval is mathematically represented
[tex]\r p - E < p < \r p + E[/tex]
=> [tex]0.238 - 0.0496 < p <0.238 + 0.0496[/tex]
=> [tex]0.1884 < p < 0.2876[/tex]
PLSSSS!!! (10points)
Answer:
angle B is 62 Degress angle A is 87 degress D is 87 degress C is 28 degress.
Step-by-step explanation:
I am in geometry btw so i know this stuff and 65 plus 28 is 93 and 180 -93 is 87 so a is 87 and d is 87 too becuase of vertical angles and b is 62 becuase 90 -28 is 62 and c is 28 becuase of vertical angles your wellcome kid good luck!!!!
Find the distance between the points. Give an exact answer and an approximation to three decimal places.
(3.1,0.3) and (2.7. - 4.9)
The exact distance is
(Simplify your answer. Type an exact ans
Answer: sqrt(27.2) =approx 5.215
Step-by-step explanation:
The distance between 2 points can be calculated using Phitagor theorem
L= sqrt( (x1-x2)²+(y1-y2)²)
Where x1, y1 are the coordinates of the first point and x2, y2 are the coordinates of the 2-nd point.
L=sqrt((3.1-2.7)²+(0.3-(-4.9))²)= sqrt(0.4²+5.2²)=sqrt(27.2) - this is exact answer.
sqrt(27.2)=5.21536...=approx 5.215
The sum of the product of a number x and 14, and 13
Answer:
ax+182
Step-by-step explanation:
a*x+14*13
ax+182
Answer this will give 10 points
Answer:
maximum --> 62
median --> 46.5
upper quartile --> 60
lower quartile --> 37
minimum --> 32
Step-by-step explanation:
Forgive me on the explanation as I'm a bit rusty on these types of problems.
First, we need to put the set of numbers in order -->
from: 34, 37, 39, 32, 48, 45, 53, 62, 58, 61, 60, 41 -->
to: 32, 34, 37, 39, 41, 45, 48, 53, 58, 60, 61, 62
maximum = biggest number => thus, 62
median = middle number in a sense => (45+48)/2 => thus, 46.5
upper quartile = median over the median => thus, 60
lower quartile = median under the median => thus, 37
minimum = lowest number => thus, 32
And there we have our 5 answers.
Hope this helps!
Power +, Inc. produces AA batteries used in remote-controlled toy cars. The mean life of these batteries follows the normal probability distribution with a mean of 35.0 hours and a standard deviation of 5.5 hours. As a part of its quality assurance program, Power +, Inc. tests samples of 25 batteries.
A) What can you say about the shape of the distribution of the sample mean?
B) What is the standard error of the distribution of the sample mean?
C) What proportion of the samples will have a mean useful life of more than 36 hours?
D) What proportion of the sample will have a mean useful life greater than 34.5 hours?
E) What proportion of the sample will have a mean useful life between 34.5 and 36.0 hours?
Answer:
(A) The shape of the distribution of the sample mean is bell-shaped.
(B) The standard error of the distribution of the sample mean is 1.1.
(C) The proportion of the samples that have a mean useful life of more than 36 hours is 0.1814.
(D) The proportion of the sample that has a mean useful life greater than 34.5 hours is 0.6736.
(E) The proportion of the sample that has a mean useful life between 34.5 and 36.0 hours is 0.4922.
Step-by-step explanation:
We are given that Power +, Inc. produces AA batteries used in remote-controlled toy cars. The mean life of these batteries follows the normal probability distribution with a mean of 35.0 hours and a standard deviation of 5.5 hours.
As a part of its quality assurance program, Power +, Inc. tests samples of 25 batteries.
Let [tex]\bar X[/tex] = sample mean life of these batteries
(A) The shape of the distribution of the sample mean will be bell-shaped because the sample mean also follows the normal distribution as it is taken from the population data only.
(B) The standard error of the distribution of the sample mean is given by;
Standard error = [tex]\frac{\sigma}{\sqrt{n} }[/tex]
Here, [tex]\sigma[/tex] = standard deviation = 5.5 hours
n = sample of batteries = 25
So, the standard error = [tex]\frac{5.5}{\sqrt{25} }[/tex] = 1.1.
(C) The z-score probability distribution for the 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 life of battery = 35.0 hours
[tex]\sigma[/tex] = standard deviation = 5.5 hours
n = sample of batteries = 25
Now, the proportion of the samples that will have a mean useful life of more than 36 hours is given by = P([tex]\bar X[/tex] > 36 hours)
P([tex]\bar X[/tex] > 36 hours) = P( [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{36-35}{\frac{5.5}{\sqrt{25} } }[/tex] ) = P(Z > 0.91) = 1 - P(Z [tex]\leq[/tex] 0.91)
= 1 - 0.8186 = 0.1814
(D) The proportion of the samples that will have a mean useful life of more than 34.5 hours is given by = P([tex]\bar X[/tex] > 34.5 hours)
P([tex]\bar X[/tex] > 34.5 hours) = P( [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{34.5-35}{\frac{5.5}{\sqrt{25} } }[/tex] ) = P(Z > -0.45) = P(Z [tex]\leq[/tex] 0.45)
= 0.6736
(E) The proportion of the samples that will have a mean useful life between 34.5 and 36.0 hours is given by = P(34.5 hrs < [tex]\bar X[/tex] > 36 hrs)
P(34.5 hrs < [tex]\bar X[/tex] < 36 hrs) = P([tex]\bar X[/tex] < 36 hrs) - P([tex]\bar X[/tex] [tex]\leq[/tex] 34.5 hrs)
P([tex]\bar X[/tex] < 36 hours) = P( [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{36-35}{\frac{5.5}{\sqrt{25} } }[/tex] ) = P(Z < 0.91) = 0.8186
P([tex]\bar X[/tex] [tex]\leq[/tex] 34.5 hours) = P( [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] [tex]\leq[/tex] [tex]\frac{34.5-35}{\frac{5.5}{\sqrt{25} } }[/tex] ) = P(Z [tex]\leq[/tex] -0.45) = 1 - P(Z [tex]\leq[/tex] 0.45)
= 1 - 0.6736 = 0.3264
Therefore, P(34.5 hrs < [tex]\bar X[/tex] < 36 hrs) = 0.8186 - 0.3264 = 0.4922.
distance between 2,-5 and 3,-7
Answer:
√5
Step-by-step explanation:
[tex](2 ,-5) = (x_1,y_1)\\(3,-7)=(x_2,y_2)\\\\d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ \\d = \sqrt{(3-2)^2 +(-7-(-5))^2}\\ \\d = \sqrt{(1)^2+(-7+5)^2}\\ \\d = \sqrt{(1)^2 + (-2)^2}\\ \\d = \sqrt{1 +4}\\ \\d = \sqrt{5}[/tex]
The mean weight of newborn infants at a community hospital is 6.6 pounds. A sample of seven infants is randomly selected and their weights at birth are recorded as 9.0, 7.3, 6.0, 8.8, 6.8, 8.4, and 6.6 pounds. Does the sample data show a significant increase in the average birthrate at a 5% level of significance?
A. Fail to reject the null hypothesis and conclude the mean is 6.6 lb.
B. Reject the null hypothesis and conclude the mean is lower than 6.6 lb.
C. Reject the null hypothesis and conclude the mean is greater than 6.6 lb.
D. Cannot calculate because the population standard deviation is unknown
Answer:
The correct option is A
Step-by-step explanation:
From the question we are told that
The population is [tex]\mu = 6.6[/tex]
The level of significance is [tex]\alpha = 5\% = 0.05[/tex]
The sample data is 9.0, 7.3, 6.0, 8.8, 6.8, 8.4, and 6.6 pounds
The Null hypothesis is [tex]H_o : \mu = 6.6[/tex]
The Alternative hypothesis is [tex]H_a : \mu > 6.6[/tex]
The critical value of the level of significance obtained from the normal distribution table is
[tex]Z_{\alpha } = Z_{0.05 } = 1.645[/tex]
Generally the sample mean is mathematically evaluated as
[tex]\=x = \frac{\sum x_i }{n}[/tex]
substituting values
[tex]\=x = \frac{9.0 + 7.3 + 6.0+ 8.8+ 6.8+ 8.4+6.6 }{7}[/tex]
[tex]\=x = 7.5571[/tex]
The standard deviation is mathematically evaluated as
[tex]\sigma = \sqrt{\frac{\sum [ x - \= x ]}{n} }[/tex]
substituting values
[tex]\sigma = \sqrt{\frac{ [ 9.0-7.5571]^2 + [7.3 -7.5571]^2 + [6.0-7.5571]^2 + [8.8- 7.5571]^2 + [6.8- 7.5571]^2 + [8.4 - 7.5571]^2+ [6.6- 7.5571]^2 }{7} }[/tex][tex]\sigma = 1.1774[/tex]
Generally the test statistic is mathematically evaluated as
[tex]t = \frac{\= x - \mu } { \frac{\sigma }{\sqrt{n} } }[/tex]
substituting values
[tex]t = \frac{7.5571 - 6.6 } { \frac{1.1774 }{\sqrt{7} } }[/tex]
[tex]t = 1.4274[/tex]
Looking at the value of t and [tex]Z_{\alpha }[/tex] we see that [tex]t < Z_{\alpha }[/tex] hence we fail to reject the null hypothesis
What this implies is that there is no sufficient evidence to state that the sample data show as significant increase in the average birth rate
The conclusion is that the mean is [tex]\mu = 6.6 \ lb[/tex]
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.
cSuppose you are standing such that a 45-foot tree is directly between you and the sun. If you are standing 200 feet away from the tree and the tree casts a 225-foot shadow, how tall could you be and still be completely in the shadow of the tree? x 225 ft 200 ft 45 ft Your height is ft (If needed, round to 1 decimal place.)
Answer:
you could stand at 5.0 ft and still be completely in the shadow of the tree
Step-by-step explanation:
From the diagram attached below;
We consider;
[tex]\overline {BC}[/tex] to be the height of the tree and [tex]\overline {DE}[/tex] to be the height of how tall you could be and still be completely in the shadow of the tree.
∠D = ∠B = 90°
Also;
ΔEAD = ΔBAC (similar triangles)
Therefore, their sides will also be proportional
i.e
[tex]\dfrac{\overline {DE}}{ \overline {BC}}= \dfrac{\overline{AD}}{ \overline{AC}}[/tex]
[tex]\dfrac{x}{ 45}= \dfrac{225-220}{225}[/tex]
[tex]\dfrac{x}{ 45}= \dfrac{25}{225}[/tex]
By cross multiply
225x = 45 × 25
[tex]x = \dfrac{45 \times 25}{225}[/tex]
[tex]x = \dfrac{1125}{225}[/tex]
x = 5.0 ft
Therefore, you could stand at 5.0 ft and still be completely in the shadow of the tree
Zhi and her friends moved on to the card tables at the casino. Zhi wanted to figure out the probability of drawing a king of clubs or an ace of clubs
Answer:
There is not enough information to the problem.
The things we know are:
Zhi wants to draw a king of clubs or an ace of clubs.
There is a 52 card deck, but we do not know which cards are in play.
Now, the probability of drawing two specific cards out of a deck of 52 cards is equal to:
P = 2/52 = 0.038.
Now, suppose that there are X cards in game, and Zhi knows the cards (and he can see that the king of clubs and the ace of clubs are not in the table, so the must be on the card pile), now the probability is bigger, because now we want to draw two specific cards out of 52 - X cards, so the probability now is:
p = 2/(52 - X)
and 52 - X is smaller than 52, so the denominator is smaller, which means that the probability is actually larger.
Answer:
31%
Step-by-step explanation:
Since the two events, drawing a face card and drawing an ace card, are non-overlapping, we can use the following formula:
P left parenthesis A c e space o r space F a c e right parenthesis equals P left parenthesis A c e right parenthesis plus P left parenthesis F a c e right parenthesis equals 4 over 52 plus 12 over 52 equals 16 over 52 equals 4 over 13 equals space 0.308 space o r space 31 percent sign
solve 27 to the power of (2/3)
Answer:
9Step-by-step explanation:
[tex]27^{\frac{2}{3}}\\\mathrm{Factor\:the\:number:\:}\:27=3^3\\=\left(3^3\right)^{\frac{2}{3}}\\\mathrm{Apply\:exponent\:rule}:\\\\\quad \left(a^b\right)^c=a^{bc},\:\quad \:a\ge 0\\\\\left(3^3\right)^{\frac{2}{3}}=3^{3}\times \frac{2}{3}}\\\\3\=times \frac{2}{3}=2\\\\=3^2 \\\\=9[/tex]
[tex]27^{2/3}=(3^3)^{2/3}=3^2=9[/tex]
A soccer ball is made of 32 pieces of leather: white hexagons and black pentagons. Each black piece borders only with white pieces, each white piece borders with three black pieces and three white pieces. How many black pieces are needed to manufacture the ball?
Answer:
24
Step-by-step explanation:
1+3=4
4 divided by 32 is 8
8 white ones
then 8-32 is 24
24 black ones
When is it easier to use the addition method rather than the substitution method to solve a system of equations?
Answer: When the addition of two or more equations leads to the elimination of one of the variables.
Step-by-step explanation:
When we have a system of equations, the addition method seems to be useful only when adding the equations will lead to the elimination of one of the variables:
An example of this can be, for the variables x and y:
3*x + x*y - 2*y = 3
x^2 + x*y - 2y = 42
now we can "add" (actually subtract) the equations and get (eq2 minus eq1)
(x^2 + x*y - 2y) - (3*x + x*y - 2*y ) = 42 - 3
x^2 - 3*x = 39
x^2 - 3*x - 39 = 0
And now we can solve it for x, and then find the value of y.
Simplify your answer as much as possible
You said - 1/3 - 3/5 x = 1/2
Multiply each side by 3 :
- 1 - 9/5 x = 3/2
Multiply each side by 5 :
- 5 - 9x = 15/2
Multiply each side by 2 :
- 10 - 18x = 15
Add 10 to each side :
- 18x = 25
Divide each side by -18 :
x = - 25/18
or x = - 1 and 7/18 (same thing)
Given that
[tex]\sqrt{2p-7}=3[/tex]
and
[tex]7\sqrt{3q-1}=2[/tex]
Evaluate
[tex]p + {q}^{2} [/tex]
Answer:
Below
Step-by-step explanation:
The two given expressions are:
● √(2p-7) = 3
● 7√(3q-1) = 2
We are told to evaluate p+q^2
To do that let's find the values of p and q^2
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Let's start with p.
● √(2p-7) = 3
Square both sides
● (2p-7) = 3^2
● 2p-7 = 9
Add 7 to both sides
● 2p-7+7 = 9+7
● 2p = 16
Divide both sides by 2
● 2p/2 = 16/2
● p = 8
So the value of p is 8
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Let's find the value of q^2
● 7√(3q-1) = 2
Square both sides
● 7^2 × (3q-1) = 2^2
● 49 × (3q-1) = 4
● 49 × 3q - 49 × 1 = 4
● 147q - 49 = 4
Add 49 to both sides
● 147q -49 +49 = 4+49
● 147q = 53
Divide both sides by 147
● 147q/147 = 53/147
● q = 53/ 147
Square both sides
● q^2 = 53^2 / 147^2
● q^2 = 2809/21609
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● p+q^2 = 8 +(2809/21609)
● p+q^2 = (2809 + 8×21609)/21609
● p+q^2 = 175681 / 21609
● p + q^2 = 8.129
Round it to the nearest unit
● p+ q^2 = 8
