Answer:
I think that the answer is - 8.-1=8
2) If licorice cost $6.59 a pound, how much would it cost to buy a quarter pound of licoric
(Hint: Convert the mixed fraction to an improper fraction or decimal and multiply by th
quantity required)
Answer:
$1.65
Step-by-step explanation:
[tex]6.59*.25=1.65[/tex]or
[tex]6.59*\frac{1}{4} =1.65[/tex]The square pyramid shown below has a base with sides of 10 units. The slant height of the pyramid is 8 units. What is the vertical height, h?
Round your answer to the nearest tenth.
Answer:
h = 6.2 units
Step-by-step explanation:
Given triangle ABC is a right triangle with the measures of the two sides,
BC = [tex]\frac{10}{2}[/tex] = 5 units
AC = 8 units
By applying Pythagoras theorem in the given triangle,
AC² = AB² + BC²
8² = AB² + 5²
AB² = 64 - 25
AB = √39
AB = 6.24 units
AB ≈ 6.2 units
Jeff has 20 coins. 2/5 of them are quarters. How many quarters does he have? How many coins are not quarters?
Test for symmetry and then graph the polar equation.
r=3−5sinθ
Answer:
Symmetric with respect to the x-axis
Symmetric with respect to the y-axis
Symmetric with respect to the origin
PLEASE HELPPP THIS IS DUE ASAPPPP!!!!!!!!!!!!!! WILL GIVE BRAINLIEST
Answer:
I think the answer is 5/6
Step-by-step explanation:
There are three even numbers and two uneven numbers less than four. Therefore, on a standard die, the.probability of Rollin a number that is even or less than for is 5/6.
F(4) =
If g(x) = 2, x=
An
Step-by-step explon:
Use the figure to find u.
Answer:
u = 2
Step-by-step explanation:
Since this is a right triangle, we can use trig functions
cos theta = adj side / hypotenuse
cos 45 = sqrt(2) / u
u cos 45 = sqrt(2)
u = sqrt(2) / cos 45
u = sqrt(2) / 1/ sqrt(2)
u = sqrt(2) * sqrt(2)
u =2
u=2
Answer:
Solution given:
Relationship between base and hypotenuse is given by cos angle.Cos 45°=base/hypotenuse
[tex]\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{u}[/tex]
doing crisscrossed multiplication
[tex]\sqrt{2}*\sqrt{2}=1*u[/tex]
u=2
Which sequence or sequences are correct and why?
Answer:
didn't get the question did u forget to put the sequence ???????
pls write the question fully so that I can help you
14. In a statistics class with 15 males and 13 females, five students are selected to put problems on the board. What is the probability that:
a. 3 females and 2 males are selected? b.all five students selected are males? c. all five students selected are females? d.at least one male is selected?
Answer:
a) 0.3056 = 30.56% probability that 3 females and 2 males are selected.
b) 0.0306 = 3.06% probability that all five students selected are males.
c) 0.0131 = 1.31% probability that all five students selected are females.
d) 0.9869 = 98.69% probability that at least one male is selected.
Step-by-step explanation:
The students are chosen from the sample without replacement, which means that the hypergeometric distribution is used to solve this question.
Hypergeometric distribution:
The probability of x successes is given by the following formula:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]
In which:
x is the number of successes.
N is the size of the population.
n is the size of the sample.
k is the total number of desired outcomes.
In this question:
15 + 13 = 28 students, which means that [tex]N = 28[/tex]
5 are selected, which means that [tex]n = 5[/tex]
13 females, which means that [tex]k = 13[/tex]
a. 3 females and 2 males are selected?
3 females, so this is P(X = 3).
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]P(X = 3) = h(3,28,5,13) = \frac{C_{13,3}*C_{15,2}}{C_{28,5}} = 0.3056[/tex]
0.3056 = 30.56% probability that 3 females and 2 males are selected.
b.all five students selected are males?
0 females, so this is P(X = 0).
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]P(X = 0) = h(0,28,5,13) = \frac{C_{13,0}*C_{15,5}}{C_{28,5}} = 0.0306[/tex]
0.0306 = 3.06% probability that all five students selected are males.
c. all five students selected are females?
This is P(X = 5). So
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]P(X = 5) = h(5,28,5,13) = \frac{C_{13,5}*C_{15,0}}{C_{28,5}} = 0.0131[/tex]
0.0131 = 1.31% probability that all five students selected are females.
d.at least one male is selected?
Less than five females, so:
[tex]P(X < 5) = 1 - P(X = 5) = 1 - 0.0131 = 0.9869[/tex]
0.9869 = 98.69% probability that at least one male is selected.
if ax^3+9x^2+4x-10 when divided by x-3 leaves the reminder 5,then a=
Let f(x) = ax³ + 9x² + 4x – 10
g(x) = 0⇒x - 3 = 0
⇒ x = 0 + 3
⇒ x = 3
On dividing f(x) by x - 3, it leaves a remainder 5.
Now keeping, f(3) = 5
⇒a(3)³ + 9(3)² + 4(3) - 10 = 5
⇒ a × 27 + 9 × 9 + 4 × 3 - 10 = 5
⇒ 27a + 81 + 12 - 10 = 5
⇒ 81 + 12 - 10 - 5 = 27a
⇒ 81 + 12 - 15 = 27a
⇒ 93 - 15 = 27a
⇒ 78 = 27a
⇒ a = 27/78
⇒ a = 0.3461
An alarm clock is slow. It falls behind 4 minutes every 24 hours. If the clock was showing the correct time at 6:00 this morning, how many seconds ahead was the clock at 10:00 last night?
Answer:
80 Seconds
I dont really want to type the whole thing out, just think about it again, or go to a tutor website, you should be able to get it, you have to use these, multiplication of three numbers, and multiplication and division by factorization of numbers.
A class contains 18 girls and 14 boys. For all parts of this question, each boy and girl are distinguishable from one another. Answer the following questions:a)In how many ways can a committee of one boy and one girl be chosen
Answer:
The total number of ways is 252.
Step-by-step explanation:
Number of girls = 18
number of boys = 14
Commitee of one girl and a boy
(18 C 1)(14 C 1)
= 252
Help
The table shows the results of a survey of students in two math classes.
Find P(more than 1 hour of TV | 6th period class). Round to the nearest thousandth. Be sure to show and explain your work.
Did You Watch More Than One Hour of TV Last Night?
Answer:
0.6
Step-by-step explanation:
In this question, the | symbol means "given", so we can phrase the question as "Find the probability that a student watches more than one hour of TV given that they are in the 6th period class"
Next, because there is a 100% chance that the student is in the 6th period class, we can disregard the results of the 3rd period class.
Given that the student is in the 6th period class, there are 15 total students (as 9+6=15 = the sum of the yes and no answers for 6th period) and 9 students that said yes. Therefore, there is a 9/15 = 3/5 = 0.6 probability that a student watches more than 1 hour of TV given that they are in the 6th period class
HELP PLEASE ASAP!!! So for this problem I got the scientific notation however I can not seem to figure out the standard notation. Can someone please help me out here please?
Answer:
0.000000073
Step-by-step explanation:
Given number is,
7.3E - 8
In scientific notation number will be,
7.3 × 10⁻⁸
In standard form the number will be,
0.000000073
what's the radius of 16x^2+16y^2=1 With a center of (0,0) ?
Answer:
The center is (0,0) and the radius is 1/4
Step-by-step explanation:
The formula for a circle is
(x-h)^2 + (y-k)^2 = r^2 where (h,k) is the center and r is the radius
16x^2+16y^2=1
Divide by 16
x^2+y^2=1/16
x^2 + y^2 = (1/4) ^2
(x-0)^2 + (y-0)^2 = (1/4) ^2
The center is (0,0) and the radius is 1/4
What is 23– 48? Need awnser now
Given: Line AC is parallel to DF, Line BE is perpendicular to DF, and angle AEB is congruent to angle CEB, prove angle BAE is congruent to angle BCE. Will give Brainliest if explained thoroughly.
9514 1404 393
Explanation:
There are several ways you can go at this. Here are a couple. All proofs will start with the given relations being repeated as part of the proof. Here are the next steps.
Angle Sum∠AED ≅ ∠BAE . . . . alternate interior angles are congruent
∠AED +∠AEB = 90° . . . . angle sum theorem
∠BAE +∠AEB = 90° . . . . substitution property of equality
∠CEF ≅ ∠BCE . . . . alternate interior angles are congruent
∠CEF +∠CEB = 90° . . . . angle sum theorem
∠BCE +∠CEB = 90° . . . . substitution property of equality
∠BAE +∠AEB = ∠BCE +∠CEB . . . . substitution property of equality
∠BAE +∠AEB = ∠BCE +∠AEB . . . . substitution property of equality
∠BAE = ∠BCE . . . . addition property of equality
Congruent Triangles∠ABE = ∠CBE = 90° . . . . BE ⊥ AC
BE ≅ BE . . . . reflexive property of congruence
ΔBEA ≅ ΔBEC . . . . ASA congruence theorem
∠BAE ≅ ∠BCE . . . . CPCTC
Solve the following equation by first writing the equation in the form a x squared = c:
3 a squared minus 21 = 27
A. a = 4
B. a = plus-or-minus 4
C. a = plus-or-minus 16
D. a = 16
9514 1404 393
Answer:
B. a = plus-or-minus 4
Step-by-step explanation:
3a² -21 = 27 . . . . . . . given
3a² = 48 . . . . . . . . . . add 21 to both sides (desired form)
a² = 16 . . . . . . . . . . . divide both sides by 3
a = ±4 . . . . . . . take the square root
60 units needed, 14 units per case. What is the number of cases and the number of additional units?
Answer:
5 cases
10 additional units
Step-by-step explanation:
Given that :
Total number of units needed = 60 units
Total number of units per case = 14
Hence, the total number of cases required will be :
Number of units needed / number of units per case
Number of cases required = 60 / 14 = 4.285 (this means that 5 cases are required as 4 cases won't be up to 60 units)
With 5 cases, we have exceeded the required units needed :
Additional units will be : (14 * 5) - 60
Additional units = 70 - 60 = 10 units
Each side of a square is increasing at a rate of 8 cm/s. At what rate (in cm2/s) is the area of the square increasing when the area of the square is 49 cm2
Answer:
Step-by-step explanation:
This is nice and simple. I'm going to walk through it like I do when teaching this concept to my class for the first time. This is a good problem for that.
We are given a square and we are looking for the rate at which the area is increasing when a certain set of specifics are given. That means that the main equation for this problem is the area of a square, which is:
[tex]A=s^2[/tex] where s is a side.
Since we are looking for the rate at which the area is changing, [tex]\frac{dA}{dt}[/tex], we need to take the derivative of area formula implicitly:
[tex]\frac{dA}{dt}=2s\frac{ds}{dt}[/tex] that means that if [tex]\frac{dA}{dt}[/tex] is our unknown, we need values for everything else. We are given that the initial area for the square is 49. That will help us determine what the "s" in our derivative is. We plug in 49 for A and solve:
[tex]49=s^2[/tex] so
s = 7
We are also given at the start that the sides of this square are increasing at a rate of 8cm/s. That is [tex]\frac{ds}{dt}[/tex]. Filling it all in:
[tex]\frac{dA}{dt}=2(7)(8)[/tex] and
[tex]\frac{dA}{dt}=112\frac{cm^2}{s}[/tex]
The surface area of a square of side L is given by
[tex]A = L^2[/tex]
The rate of change of the area per unit time is
[tex]\dfrac{dA}{dt} = 2L\dfrac{dL}{dt}[/tex]
We can express the length L on the right hand side in terms of the area A [tex](L = \sqrt{A})[/tex]:
[tex]\dfrac{dA}{dt} = 2\sqrt{A}\dfrac{dL}{dt}[/tex]
[tex]\:\:\:\:\:\:\:=2(\sqrt{49\:\text{cm}^2})(8\:\text{cm/s})[/tex]
[tex]\:\:\:\:\:\:\:=112\:\text{cm}^2\text{/s}[/tex]
A human resources office is working to implement an increase in starting salaries for new
administrative secretaries and faculty at a community college. An administrative secretary
starts at $28,000 and new faculty receive $40,000. The college would like to determine the
percentage increase to allocate to each group, given that the college will be hiring 8
secretaries and 7 faculty in the upcoming academic year. The college has at most $5,000 to
put towards raises. What should the percentage increase be for each group?
Answer:
Step-by-step explanation :
Let % increase in administrative secretary be = x
Let % increase in new faculty receive be = y
Administrative secretary salary = 28,000
New faculty receive Salary = 40,000
(8)*(x/100)* (28000) + (7)*(y/100)*(40000) = 5,000
2240x +2800 y = 5,000
224x +280 y = 500
56x +70y = 125
Therefore, x and y should be chosen such that it satisfy the above equation.
Simplify the following expression: (4x2)2 • (3x3)3
Answer:
432x^13
Step-by-step explanation:
(4x^2)^2 • (3x^3)^3
We know that a^b^c = a^(b*c)
4^2 x^2^2 * 3^3 x^3^3
16 x^4 * 27 x^9
We know that a^b ^ a^c = a^(b+c)
16*27 x^(4+9)
432x^13
Answer:
432x¹³
Step-by-step explanation:
( 4x² ) ² • ( 3x³ ) ²
( 16x²)² • ( 27x³)²
[tex]16 x{}^{2 \times 2} \times 6 {}^{3 \times 3 } \\ 16x {}^{4} \times27 {}^{9} [/tex]
[tex](16 \times 27)x {}^{4 + 9} [/tex]
432x¹³
Remember the dataset of alligators which was about the length and weight of several aligators in Florida. The variable X is the length of aligator and the Y variable is the weight of them. A researcher decided to use decision tree and designed two steps: X<4, X>4. What is the name of this method of splitting?A. Multi-way splitting.B. Entropy classification.C. Binary splitting.D. Gini index.
Answer:
A. multi-way split.
Step-by-step explanation:
Multi way split consists of internal at decision tree branches. Gini index measures probability of impurity in the random variables chosen. Entropy is measure of uncertainty in the sample selected. Binary splitting is used to speed up numerical evaluation.
Find the exact values of the six trigonometric functions at “a” given cos(2a) = - 4/5 and a is
in the 2nd quadrant.
If a is in the second quadrant, then cos(a) < 0 and sin(a) > 0.
Recall the double angle identity for cosine:
cos(2a) = 2 cos²(a) - 1 = 1 - 2 sin²(a)
It follows that
2 cos²(a) - 1 = -4/5 ==> cos²(a) = 1/10 ==> cos(a) = -1/√10
1 - 2 sin²(a) = -4/5 ==> sin²(a) = 9/10 ==> sin(a) = 3/√10
Then we find
1/cos(a) = sec(a) = -√10
1/sin(a) = csc(a) = √10/3
sin(a)/cos(a) = tan(a) = -3
1/tan(a) = cot(a) = -1/3
In a poll, adults in a region were asked about their online vs. in-store clothes shopping. One finding was that % of respondents never clothes-shop online. Find and interpret a % confidence interval for the proportion of all adults in the region who never clothes-shop online.
The question is incomplete. The complete question is :
In a poll, 1100 adults in a region were asked about their online vs. in-store clothes shopping. One finding was that 43% of respondents never clothes-shop online. Find and interpret a 95% confidence interval for the proportion of all adults in the region who never clothes-shop online.
Solution :
95% confidence interval for p is :
[tex]$\hat p - Z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}} < p < \hat p + Z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}$[/tex]
[tex]$0.43 - 1.96\times \sqrt{\frac{0.43(1-0.43)}{1100}} < p < 0.43 + 1.96\times \sqrt{\frac{0.43(1-0.43)}{1100}}$[/tex]
0.401 < p < 0.459
Therefore, 95% confidence interval is from 0.401 to 0.459
A paper factory makes cardboard sheets like the one shown below. If the area of each sheet is given by the expression 6x ^ 2 + 7x + 2, what are the dimensions of each sheet of cardboard?
Answer:
(3x+2) by (2x+1)
Step-by-step explanation:
A cardboard is a rectangle, and has two dimensions. Given a quadratic equation, you should find a way to split it in two.
The easiest way to do so is through factoring. (There are many ways to do this, take a look at the plethora of sources offered on the internet.)
When the expression 6x^2 + 7x + 2 is factored, it is (3x+2)(2x+1). Hence, these are your dimensions.
What is the difference of the two polynomials? (NineX squared plus 8X) minus (twoX squared plus 3X)
Answer:
[tex]7x {}^{2} + 5x[/tex]
Step-by-step explanation:
[tex]9x {}^{2} + 8x - (2x {}^{2} + 3x) \\ \\ = 9x {}^{2} + 8x - 2x {}^{2} - 3x (remove \: brackets) \\\ \\ = 7x {}^{2} - 5x [/tex]
If the total income generated from Gasoline for AER was $408 millions, how much would be the cost for a barrel of gasoline
5(6t-3)=5t+35
find the value of t
Answer:
t = 2
Step-by-step explanation:
5(6t-3)=5t+35
Step 1 distribute the 5 by multiplying 5 to what is inside of the parenthesis
5(6t) - 5(3) = 5t + 35
Outcome: 30t - 15 = 5t + 35
Step 2 add 15 to both sides
30t - 15 + 15 = 5t + 35 + 15
Outcome: 30t = 5t + 50
Step 3 subtract 5t from both sides
30t - 5t = 5t - 5t + 50
Outcome: 25t = 50
Step 4 divide both sides by 25
25t/25 = 50 we're left with t = 2
Help with solving this Functions problem
Answer:
See answers below
Step-by-step explanation:
Given the following functions:
r(x) = x - 6
s(x) = 2x²
r(s(x)) = r(2x²)
Replacing x with 2x² in r(x) will give;
r(2x²) = 2x² - 6
r(s(x)) = 2x² - 6
(r-s)(x) = r(x) - s(x)
(r-s)(x) = x - 6 - 2x²
Rearrange
(r-s)(x) = - 2x²+x-6
(r+s)(x) = r(x) + s(x)
(r-s)(x) = x - 6 + 2x²
Rearrange
(r-s)(x) = 2x²+x-6
