Answer:
Welcome to Brainly! The answer is C. 56
Step-by-step explanation:
I don't believe I was shown how to complete this
At a school, there are 120 athletes. The ratio of boy athletes to giri athletes is 3:5. How many of the athletes are girls?
А
75
B
45
с
24
D 5
Answer:
a
Step-by-step explanation:
because there are over half girls and half of 120 is 60 and 75 is over half
Please help with this easy fraction problem <3
18 + 7/9
14 + 7/8
First, rewrite each fraction in terms of a common denominator. 8 and 9 don't share a common multiple until 8 * 9 = 72, so we have
7/9 = (7/9) • (8/8) = (7 • 8)/(9 • 8) = 56/72
7/8 = (7/8) • (9/9) = (7 • 9)/(8 • 9) = 63/72
Next, write each whole number in terms of fractions with the same denominator. We have
18 • 72 = 1296 ==> 18 = 1296/72
14 • 72 = 1008 ==> 14 = 1008/72
Write each mixed number as an improper fraction:
18 + 7/9 = 1296/72 + 56/72 = (1296 + 56)/72 = 1352/72
14 + 7/8 = 1008/72 + 63/72 = (1008 + 63)/72 = 1071/72
If the backpack and book together weigh 18 + 7/9 pounds, and the backpack without the book weighs 14 + 7/8 pounds, then the book alone weighs the difference, call it b :
b = (18 + 7/9) - (14 + 7/8)
b = 1352/72 - 1071/72
b = (1352 - 1071)/72
b = 281/72
Convert this to a mixed number. 72 • 4 = 288, so 72 • 4 - 7 = 281:
b = (72 • 4 - 7)/72
b = (72 • 4)/72 - 7/72
b = 4 - 7/72
b = (3 + 1) - 7/72
b = 3 + (1 - 7/72)
b = 3 + 65/72
I need help with a math problem comment if u want i really need help with this
The temperature at 6 a.m. was −12°F. At 11 a.m. the temperature was 0°F. Which of the following shows the temperature change from 6 a.m. to 11 am?
Answer:
+12 degrees
Step-by-step explanation:
You never showed a picture, but im pretty sure this is correct
I hope this helped, please mark Brainliest, thank you!
he pulse rates of 179 randomly selected adult males vary from a low of 43 bpm to a high of 115 bpm. Find the minimum sample size required to estimate the mean pulse rate of adult males. Assume that we want 90% confidence that the sample mean is within 3 bpm of the population mean. Complete parts (a) through (c) belo
Answer:
97
Step-by-step explanation:
Given that:
Low = 43bpm
High = 115 bpm
Error = 3
α = 90%
To find the range estimate :
Standard destviation = (high - low) / 4
Standard deviation = (115 - 43) / 4
Standard deviation = 72/4 = 18
The sample size,(n) :
n = [(Zα/2 * sd)/E] ²
2 - tail = (Zα/2) = 1.645
[(Zα/2 * sd)/E] ² = [(1.645 * 18) / 3]² = (29.61/3)²
= 97.41
Sample size = 97
What value of a will make the following equation true?
Answer:
a=16
Step-by-step explanation:
Answer:
A=4
Step-by-step explanation:
(3)/(5)x = (2)/(5)x + 8
Answer: x=-19
Step-by-step explanation:
Problem 9.4 Given the system of equations -3x2 + 7x3 = 4 x1 + 2x2 - x3 = 0 5x1 -2 x2 = 3 (a) Compute the determinant. (b) Use Cramer’s rule to solve for the x’s. (c) Use Gauss elimination with partial pivoting to solve for the x’s. As part of the computation, calculate the determinant in order to verify the value computed in (a) (d) Substitute your results back into the original equations to check your solution.
- 3 x₂ + 7 x₃ = 3
x₁ + 2 x₂ - x₃ = 0
5 x₁ - 2 x₂ = 3
(a) I suppose this is asking for the determinant of the coefficient matrix.
[tex]\begin{vmatrix}0&-3&7\\1&2&-1\\5&-2&0\end{vmatrix}[/tex]
Using a cofactor expansion along the first row, this reduces to
[tex]0\begin{vmatrix}2&-1\\-2&0\end{vmatrix}-(-3)\begin{vmatrix}1&-1\\5&0\end{vmatrix}+7\begin{vmatrix}1&2\\5&-2\end{vmatrix}= 3(0 - (-5))+7(-2-10)=\boxed{-69}[/tex]
(b) Using Cramer's rule, we have
[tex]x_1=\dfrac{\begin{vmatrix}\mathbf 3&-3&7\\\mathbf 0&2&-1\\\mathbf 3&-2&0\end{vmatrix}}{\begin{vmatrix}0&-3&7\\1&2&-1\\5&-2&0\end{vmatrix}}[/tex]
[tex]x_2=\dfrac{\begin{vmatrix}0&\mathbf 3&7\\1&\mathbf 0&-1\\5&\mathbf 3&0\end{vmatrix}}{\begin{vmatrix}0&-3&7\\1&2&-1\\5&-2&0\end{vmatrix}}[/tex]
[tex]x_3=\dfrac{\begin{vmatrix}0&-3&\mathbf 3\\1&2&\mathbf 0\\5&-2&\mathbf 3\end{vmatrix}}{\begin{vmatrix}0&-3&7\\1&2&-1\\5&-2&0\end{vmatrix}}[/tex]
That is, solving for n-th variable consists of dividing [the determinant of the coefficient matrix with its n-th column replaced with the right side of the system, the numbers in boldface,] by [the determinant of the coefficient matrix].
Compute each determinant:
[tex]\begin{vmatrix}\mathbf 3&-3&7\\\mathbf 0&2&-1\\\mathbf 3&-2&0\end{vmatrix}=3\begin{vmatrix}2&-1\\-2&0\end{vmatrix}+3\begin{vmatrix}-3&7\\2&-1\end{vmatrix}=3(0-2)+3(3-14)=-39[/tex]
(expanding along the first column)
[tex]\begin{vmatrix}0&\mathbf 3&7\\1&\mathbf 0&-1\\5&\mathbf 3&0\end{vmatrix}=-1\begin{vmatrix}3&7\\3&0\end{vmatrix}+5\begin{vmatrix}3&7\\0&-1\end{vmatrix}=-1(0-21)+5(-3-0)=6[/tex]
(again, along the first column)
[tex]\begin{vmatrix}0&-3&\mathbf 3\\1&2&\mathbf 0\\5&-2&\mathbf 3\end{vmatrix}=-1\begin{vmatrix}-3&3\\-2&3\end{vmatrix}+5\begin{vmatrix}-3&3\\2&0\end{vmatrix}=-1(-9-(-6))+5(0-6)=-27[/tex]
(first column)
So, we get the solution
[tex]x_1=\dfrac{-39}{-69}=\boxed{\dfrac{13}{23}},x_2=\dfrac6{-69}=\boxed{-\dfrac2{23}},x_3=\dfrac{-27}{-69}=\boxed{\dfrac9{23}}[/tex]
(c) Using elimination:
- 3 x₂ + 7 x₃ = 3
x₁ + 2 x₂ - x₃ = 0
5 x₁ - 2 x₂ = 3
Swap the first two equations:
x₁ + 2 x₂ - x₃ = 0
- 3 x₂ + 7 x₃ = 3
5 x₁ - 2 x₂ = 3
Add -5(equation 1) to equation 3:
x₁ + 2 x₂ - x₃ = 0
- 3 x₂ + 7 x₃ = 3
- 12 x₂ + 5 x₃ = 3
Add -4(equation 2) to equation 3:
x₁ + 2 x₂ - x₃ = 0
- 3 x₂ + 7 x₃ = 3
- 23 x₃ = -9
Multiply through equation 3 by -1/23:
x₁ + 2 x₂ - x₃ = 0
- 3 x₂ + 7 x₃ = 3
x₃ = 9/23
Add -7(equation 3) to equation 2:
x₁ + 2 x₂ - x₃ = 0
- 3 x₂ = 6/23
x₃ = 9/23
Multiply through equation 2 by -1/3:
x₁ + 2 x₂ - x₃ = 0
x₂ = -2/23
x₃ = 9/23
Add -2(equation 2) and equation 3 to equation 1:
x₁ = 13/23
x₂ = -2/23
x₃ = 9/23
Gwendolyn was physically present in the United States for 96 days in 2019, 198 days in 2018, and 66 days in 2017. Under the substantial presence test formula, how many days is Gwendolyn deemed physically present in the United States in 2019 g
Answer:
173 days
Step-by-step explanation:
The formula for substantial presence test is;
SBT = (Total of number of days present in the current tax year) + (1/3)(number of days in the year that was before the tax year) + (1/6)(number of days in the year that was two years before the tax year)
From the question, present tax year is 2019 and number of days is 96 days.
Year before tax year is 2018 and number of days is 198 days
2 years before tax year is 2017 and number of days is 66 days.
Thus;
SBT = 96 + ((1/3)198) + ((1/6)66)
SBT = 173 days
1. Let the test statistics Z have a standard normal distribution when H0 is true. Find the p-value for each of the following situations:
a) H1:μ>μ0,z=1.88
b) H1:μ<μ0,z=−2.75
c) H1:μ≠μ0,z=2.88
2. Let the test statistics T have t distribution when H0 is true. Find the p-value for each of the following situations (provide an interval if the exact one cannot be found using a table):
a) H1:μ>μ0,n=16,t=3.733
b) H1:μ<μ0,df=23,t=−2.500
c) H1:μ≠μ0,n=7,t=−2.250
Answer:
1 a [tex]p -value = 0.030054[/tex]
1b [tex]p -value = 0.0029798[/tex]
1c [tex]p -value = 0.0039768[/tex]
2a [tex]p-value = 0.00099966[/tex]
2b [tex]p-value = 0.00999706[/tex]
2c [tex]p-value = 0.0654412[/tex]
Step-by-step explanation:
Considering question a
The alternative hypothesis is H1:μ>μ0
The test statistics is z =1.88
Generally from the z-table the probability of z =1.88 for a right tailed test is
[tex]p -value = P(Z > 1.88) = 0.030054[/tex]
Considering question b
The alternative hypothesis is H1:μ<μ0
The test statistics is z=−2.75
Generally from the z-table the probability of z=−2.75 for a left tailed test is
[tex]p -value = P(Z < -2.75) = 0.0029798[/tex]
Considering question c
The alternative hypothesis is H1:μ≠μ0
The test statistics is z=2.88
Generally from the z-table the probability of z=2.88 for a right tailed test is
[tex]p -value = P(Z >2.88) = 0.0019884[/tex]
Generally the p-value for the two-tailed test is
[tex]p -value = 2 * P(Z >2.88) = 2 * 0.0019884[/tex]
=> [tex]p -value = 0.0039768[/tex]
Considering question 2a
The alternative hypothesis is H1:μ>μ0
The sample size is n=16
The test statistic is t = 3.733
Generally the degree of freedom is mathematically represented as
[tex]df = n - 1[/tex]
=> [tex]df = 16 - 1[/tex]
=> [tex]df = 15[/tex]
Generally from the t distribution table the probability of t = 3.733 at a degree of freedom of [tex]df = 15[/tex] for a right tailed test is
[tex]p-value = t_{3.733 , 15} = 0.00099966[/tex]
Considering question 2b
The alternative hypothesis is H1:μ<μ0
The degree of freedom is df=23
The test statistic is ,t= −2.500
Generally from the t distribution table the probability of t= −2.500 at a degree of freedom of df=23 for a left tailed test is
[tex]p-value = t_{-2.500 , 23} = 0.00999706[/tex]
Considering question 2c
The alternative hypothesis is H1:μ≠μ0
The sample size is n= 7
The test statistic is ,t= −2.2500
Generally the degree of freedom is mathematically represented as
[tex]df = n - 1[/tex]
=> [tex]df = 7 - 1[/tex]
=> [tex]df = 6[/tex]
Generally from the t distribution table the probability of t= −2.2500 at a degree of freedom of [tex]df = 6[/tex] for a left tailed test is
[tex]t_{-2.2500 , 6} = 0.03272060[/tex]
Generally the p-value for t= −2.2500 for a two tailed test is
[tex]p-value = 2 * 0.03272060 = 0.0654412[/tex]
I want to decorate my room with leaves and turkeys for Thanksgiving. The number of leaves is twenty more than twice the number of turkeys. The number of leaves is ten less than 4 times the number of turkeys. How many of each type of decoration should I use?
Answer:
50 leaves, 15 turkeys
Step-by-step explanation:
set up an equation
l = leaves
t = turkeys
l = 2t+20
l = 4t-10
2t+20=4t-10
30=2t
15=t (so that means that there would be 15 turkey decorations, but we're not done yet)
substitute (either equation):
l = 2(15)+20
l = 30+20
l = 50 (so this means there would be 50 leaves)
hope this helps!
The zeroes of the function are:
A) -2 and 0
B) 0 and 4
C) -2, 0, and 4
NEED HELP PLEASE ANSWER
Answer:
I think its 1430 miles
Step-by-step explanation:
Why do we study?give me reason
Answer:
Simple to help us improve our lives
Answer:
To do good in school
Step-by-step explanation:
the thing is teachers want us to study so we remember whatever is on the test for an example and that is good but not alt of people always study.
Question Help When purchasing bulk orders of batteries, a toy manufacturer uses this acceptance sampling plan: Randomly select and test 36 batteries and determine whether each is within specifications. The entire shipment is accepted if at most 2 batteries do not meet specifications. A shipment contains 3000 batteries, and 3% of them do not meet specifications. What is the probability that this whole shipment will be accepted? Will almost all such shipments be accepted, or will many be rejected?
Answer:
The probability is [tex]P(X \le 2 ) = 0.9072[/tex]
The company will accept 90.72% of the shipment and will reject [tex](100 -90.72) = 9.2\%[/tex] of the shipment , so many of the shipment are rejected
Step-by-step explanation:
From the question we are told that
The sample size is n = 36
The proportion that did not meet the requirement is [tex]p = 0.03[/tex]
Generally the probability that the whole shipment is accepted is equivalent to the probability that there is at most 2 batteries that do not meet the requirement , this is mathematically represented as
[tex]P(X \le 2 ) = [ P(X = 0 ) + P(X = 1 ) + P(X = 0)][/tex]
=> [tex]P(X \le 2 ) = [ [^{n}C_0 * (p)^{0} *(1-p)^{n-0} ] + [^{n}C_1 * (p)^{1} *(1-p)^{n-1} ] + [^{n}C_2 * (p)^{2} *(1-p)^{n-2} ]][/tex]
Here C stands for Combination (so we will be making the combination function in our calculators )
So
=> [tex]P(X \le 2 ) = [ [^{36}C_0 * (0.03)^{0} *(1-0.03)^{36-0} ] + [^{36}C_1 * (0.03)^{1} *(1-0.03)^{36-1} ] + [^{36}C_2 * (0.03)^{2} *(1-0.03)^{36-2} ]][/tex]
=> [tex]P(X \le 2 ) = [ [1 * 1 * 0.3340 ] + [36* 0.03 *0.3444 ] + [630 * 0.0009 *(0.355 ]][/tex]
=>[tex]P(X \le 2 ) = 0.9072[/tex]
The company will accept 90.72% of the shipment and will reject [tex](100 -90.72) = 9.2\%[/tex] of the shipment , so many of the shipment are rejected
Can someone who knows Pre Calculus help me on these questions.
Answer:
the answer should be B:12/5
Will give brainliest if correct what is the discriminant of the quadratic equation: (multiple choice)
Plot the x- and y-intercepts to graph the equation.
y=−x−5
Answer:
See the graph
Step-by-step explanation:
Graph the line using the slope and y-intercept, or two points.
Slope: − 1
x-intercept: (-5,0)
y-intercept: ( 0 , − 5 )
What does "a' control in a graph of * f(x) = ax + bx+ c?
Answer:
the stretch or compression of a quadratic, and if it's opening up to down.
Step-by-step explanation:
Statistical significance at the 0.01 level is __________ than significance at the 0.05 level .
a. less informative
b. more difficult to achieve
c. easier to achieve
d. less costly
Answer:
d. less costly
Step-by-step explanation:
Statistical significance level is the probability of rejecting the null hypothesis. Statistical significance at the 0.05 level is the standard level for rejecting the null hypothesis.
A significance of 0.01 shows that there is strong evidence to conclude that the null hypothesis is not true. A 0.05 significance is the most used level while 0.01 is used when multiple tests are conducted or when the researcher is not so sure if the null hypothesis should be rejected. It increases the amount of evidence required to prove the null hypothesis and it also reduces the possibility of obtaining a false positive.
What is the opposite of the opposite of -1.4?
Answer:
-1.4
Step-by-step explanation:
opposite of -1.4 = 1.4
opposite of the opposite of -1.4 = the opposite of 1.4 = -1.4
Solve: x^2=10
A) x= +5
B) x= +100
C) x= +3.5
D) x= 10 squared
Answer:
D) x=√10
Step-by-step explanation:
√(10)=3.16(approximately )
Tell me what quadrant this angle is in.
Answer:
Second quadrant
Step-by-step explanation:
[tex] \because \: \frac{ \pi}{2} < \theta < \pi \: bounds \: second \: quadrant \\ \\ \therefore \: \sin \theta = \frac{4}{5 } \: lies \: in \: second \: quadrant.[/tex]
I REALLY NEED THE HELP ASAP PLEASE
Solve - 4(x + 10) – 6 = -3(x - 2). (1 point)
-40
-46
-52
52
Answer:
X=-52
Step-by-step explanation:
-4(X+10)-6=-3(x-2)
-4x-40-6=-3x+6
-4x+3x=6+46
-x=52
X=-52
15 tens 7 ones what is the answer?
Answer:
157
Step-by-step explanation:
Answer: 15 tens and 7 ones in standard form would be 157. If you have 15 tens this means that you are adding 10, 15 times or multiplying 10 by 15, which gives you 150.
Step-by-step explanation: Hopefully this helped!
Manuel can walk 5 miles In one hour. At this rate, how many miles can he walk in 8 minutes? Round to the nearest tenth
Answer:0.6666664
Step-by-step explanation:
You take 5 and divide by 60. Then multiply by 8.
2/3 of a mile.
If he can walk 5 MILES a hour, lets use 5/60.
Multiply 5/60 by 8 since each minute he travels 5/60 of a MILE.
You get 0.66(repeating) OR 2/3
Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-life of the substance? The half-life is:_______.
(A) Less than 600 years
(B) Between 600 and 700 years
(C) Between 700 and 800 years
(D) Between 800 and 900 years
(E) More than 900 years
Answer:
The correct answer is:
Between 600 and 700 years (B)
Step-by-step explanation:
At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:
[tex]A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years[/tex]
First, let us calculate the decay constant (k)
[tex]75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036[/tex]
Next, let us calculate the half-life as follows:
[tex]\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2} \approx 669\ years[/tex]
Therefore the half-life is between 600 and 700 years
The mean and standard deviation of a random sample of n measurements are equal to and , respectively. a. Find a % confidence interval for if n. b. Find a % confidence interval for if n. c. Find the widths of the confidence intervals found in parts a and b. What is the effect on the width of a confidence interval of quadrupling the sample size while holding the confidence coefficient fixed?
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
[tex]33.55 < \mu < 35.5[/tex]
b
[tex]34.03 < \mu < 34.969 [/tex]
c
Generally the width at n = 49 is mathematically represented as
[tex]w = 2 * E[/tex]
[tex]w = 2 * 0.952 [/tex]
[tex]w = 1.904 [/tex]
Generally the width at n = 196 is mathematically represented as
[tex]w = 2 * E[/tex]
[tex]w = 2 * 0.4687 [/tex]
[tex]w = 0.9374 [/tex]
d
The correct option is E
Step-by-step explanation:
From the question we are told that
The sample mean is [tex]\= x = 34.5[/tex]
The standard deviation is [tex]s = 3.4[/tex]
Generally given that the confidence level is 95% then the level of significance is
[tex]\alpha = (100 - 95)\%[/tex]
=> [tex]\alpha = 0.05 [/tex]
Generally from the normal distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] is
[tex]Z_{\frac{\alpha }{2} } = 1.96[/tex]
Considering question a
From the question n = 49
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * \frac{s }{\sqrt{n} }[/tex]
=> [tex]E = 1.96* \frac{ 3.4 }{\sqrt{49} }[/tex]
=> [tex]E = 0.952 [/tex]
Generally 95% confidence interval is mathematically represented as
[tex]\= x -E < p < \=x +E[/tex]
[tex]34.5 -0.952 < p < 34.5 + 0.952[/tex]
=> [tex]33.55 < \mu < 35.5[/tex]
Considering question b
From the question n = 196
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * \frac{s }{\sqrt{n} }[/tex]
=> [tex]E = 1.96* \frac{ 3.4 }{\sqrt{196} }[/tex]
=> [tex]E = 0.4687 [/tex]
Generally 95% confidence interval is mathematically represented as
[tex]\= x -E < p < \=x +E[/tex]
[tex]34.5 -0.4687 < p < 34.5 +0.4687[/tex]
=> [tex]34.03 < \mu < 34.969 [/tex]
Considering question c
Generally the width at n = 49 is mathematically represented as
[tex]w = 2 * E[/tex]
[tex]w = 2 * 0.952 [/tex]
[tex]w = 1.904 [/tex]
Generally the width at n = 196 is mathematically represented as
[tex]w = 2 * E[/tex]
[tex]w = 2 * 0.4687 [/tex]
[tex]w = 0.9374 [/tex]
Now when the sample size is quadrupled i.e from n = 49 to n = 196
The width of the confidence interval decrease by 2 from 1.904 to 0.9374
Caleb and Emily are standing 100 yards from each other. Caleb looks up at a 45° angle to see a hot air balloon. Emily looks up at a 60° angle to see the same hot air balloon. Approximately how far is the hot air balloon off the ground?
a) 44.3 yd.
b) 63.4 yd.
c) 73.2 yd.
d) 89.7 yd.
Answer:
B. 63.4 yards.
Step-by-step explanation: