The mean age will increase, and the median age will remain the same.
Option B is the correct answer.
What is a mean?It is the average value of the set given.
It is calculated as:
Mean = Sum of all the values of the set given / Number of values in the set
We have,
Ages:
14, 15, 15, 16, 16, 16, 16, 17, 17, 17, 17, 17, 18
The mean of the ages.
= (14 + 15 + 15 + 16 + 16 + 16 + 16 + 17 + 17 + 17 + 17 + 17 + 18) / 13
= 211/13
= 16.23
The median of the ages.
= 16
Now,
18 becomes 19.
14, 15, 15, 16, 16, 16, 16, 17, 17, 17, 17, 17, 19.
Mean = 212/13 = 16.3
Median = 16
Thus,
The mean age is increased.
The median is the same.
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HELP NEEDED PLEASE :(
Identify the center and the radius of a circle that has a diameter with endpoints at (5, 8)
and (7,6).
Answer:
Centre is,
((5+7)/2,(8+6)/2)
or, (12/2,14/2)
or, (6,7)
radius is,
[√{(5-7)²+(8-6)²}]/2
= [√(2²+2²)]/2
= [√(4+4)]/2
= [√8 ]= [2√2]/2 = √2
The center of a circle = (6, 7)
the radius of a circle = [tex]\sqrt{2}[/tex] units
What is the distance formula?"The distance between two points [tex](x_1,y_1),(x_2,y_2)[/tex] is given by [tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_2)^2}[/tex]"
What is midpoint formula?"The coordinates of the midpoint of the line segment having endpoints [tex](x_1,y_1),(x_2,y_2)[/tex] is given by [tex]m=(\frac{x_1+x_2}{2} ,\frac{y_1+y_2}{2} )[/tex] "
What is diameter of circle?"It is the line segment through the center and touching two points on its edge. "
For given question,
The diameter of circle has endpoints at (5, 8) and (7,6).
Let [tex](x_1,y_1)=(5,8),(x_2,y_2)=(7,6)[/tex]
First we find the length of the diameter of the circle.
Using the distance formula,
[tex]\Rightarrow d=\sqrt{(x_2-x_1)^2+(y_2-y_2)^2} \\\\\Rightarrow d=\sqrt{(7-5)^2+(6-8)^2} \\\\\Rightarrow d=\sqrt{2^2+(-2)^2}\\\\\Rightarrow d=\sqrt{4+4}\\\\\Rightarrow d=2\sqrt{2}~units[/tex]
We know that the diameter = 2 × radius
⇒ radius (r) = [tex]\frac{2\sqrt{2} }{2}[/tex]
⇒ r = [tex]\sqrt{2}[/tex] units
We know that the midpoint of the diameter is the center of the circle.
Using midpoint formula the center of the circle would be,
[tex]\Rightarrow O=(\frac{x_1+x_2}{2} ,\frac{y_1+y_2}{2} )\\\\\Rightarrow O=(\frac{5+7}{2} ,\frac{8+6}{2} )\\\\\Rightarrow O=(\frac{12}{2} ,\frac{14}{2} )\\\\\Rightarrow O=(6,7)[/tex]
Therefore, the center of the circle is (6,7)
Hence, the center of a circle = (6, 7)
the radius of a circle = [tex]\sqrt{2}[/tex]
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(1,-19),(-2,-7) finding slope
Answer:
The slope is -4.
Step-by-step explanation:
Slope(m)=(y2-y1)/(x2-x1)
y2=-7, y1=-19, x2=-2, x1=1
(-7+19)/(-2-1)
=12/-3
=-4
Answer: -4
Step-by-step explanation:
The slope formula is: [tex]y_{2} -y_{1}/x_{2}-x_{1} \\[/tex]
So it is: (-7+19)/(-2-1) = 12/-3 = -4
I hope this helped!
take away 4/5 from 6 1/2
Answer:
3-4/5=2.2
hope it helps
Hi there!
»»————- ★ ————-««
I believe your answer is:
[tex]5\frac{7}{10}[/tex]
»»————- ★ ————-««
Here’s why:
⸻⸻⸻⸻
[tex]\boxed{\text{Calculating the answer...}}\\\\6\frac{1}{2}-\frac{4}{5} \\-------------\\6\frac{1}{2} = \frac{13}{2} \\\\\frac{13}{2} - \frac{4}{5}\\\\LCM(2,5): 10\\\\\frac{13}{2} =\frac{13*5}{2*5} =\frac{65}{10}\\\\\frac{4}{5}=\frac{4*2}{5*2}=\frac{8}{10}\\\\\frac{65}{10}-\frac{8}{10}\\\\ \frac{57}{10}\\\\\frac{57}{10}\rightarrow\boxed{5\frac{7}{10}}[/tex]
»»————- ★ ————-««
Hope this helps you. I apologize if it’s incorrect.
Triangle ABC has vertices of A(-6, 7), B(4, -1), and C(-2, -9). Find the length of the median from ZB in triangle ABC
A. 4
B. 18
C. 8
D. 768
Please select the best answer from the choices provided
Ο Α
D
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Answer:
C. 8
Step-by-step explanation:
The median from vertex B is the line segment between there and the midpoint of side AC. That midpoint is ...
D = (A +C)/2 = ((-6, 7) +(-2, -9))/2 = (-8, -2)/2 = (-4, -1)
So, we want the length of the line between (-4, -1) and (4, -1). These points are on the same horizontal line (y=-1), so the length is the difference of the x=coordinates:
median AD = 4 -(-4) = 8 . . . . units in length
2 6 + 3 * 4 2 + 7 * - 2 /
Answer:
26 + 3 x 42 + 7 x -2 = 138
Step-by-step explanation:
Ok bud, first step we must convert our symbols (Makes it easier to solve)
26 + 3 x 42 + 7 x -2
* subsitutes for multiplication.
I recommend using PEMDAS at times:
1 - Parentheses
2 - Exponents and Roots
3 - Multiplication
4 - Division
5 - Addition
6 - Subtraction
Yet again your numbers were spaced out could they be exponents? if so:
3x^{42}+7x+24
Our answer would round to 24 but he equation was not put in a valid or straight forward way.
An urn contains 12 balls, five of which are red. Selection of a red ball is desired and is therefore considered to be a success. If three balls are selected, what is the expected value of the distribution of the number of selected red balls
The expected value of the distribution of the number of selected red balls is 0.795.
What is the expected value?The expected value of the distribution is the mean or average of the possible outcomes.
There are 12 balls in an urn, five of which are crimson. The selection of a red ball is desired and hence considered a success.
In this case, the possible outcomes are 0, 1, 2, or 3 red balls.
To calculate the expected value, we need to find the probability of each outcome and multiply it by the value of the outcome.
The probability of selecting 0 red balls is :
[tex]$\frac{7}{12} \cdot \frac{6}{11} \cdot \frac{5}{10} = \frac{105}{660}$[/tex].
The probability of selecting 1 red ball is :
[tex]$3 \cdot \frac{5}{12} \cdot \frac{7}{11} \cdot \frac{6}{10} + 3 \cdot \frac{7}{12} \cdot \frac{5}{11} \cdot \frac{6}{10} + 3 \cdot \frac{7}{12} \cdot \frac{6}{11} \cdot \frac{5}{10} = \frac{315}{660}$[/tex].
The probability of selecting 2 red balls is
[tex]:$\dfrac{5}{12} \cdot \frac{7}{11} \cdot \frac{6}{10} + \frac{7}{12} \cdot \frac{5}{11} \cdot \frac{6}{10} + \frac{7}{12} \cdot \frac{6}{11} \cdot \frac{5}{10} = \frac{105}{660}$.[/tex]
The probability of selecting 3 red balls is
[tex]$\dfrac{5}{12} \cdot \frac{4}{11} \cdot \frac{3}{10} = \frac{15}{660}$[/tex]
The expected value is then :
[tex]$0 \cdot \frac{105}{660} + 1 \cdot \frac{315}{660} + 2 \cdot \frac{105}{660} + 3 \cdot \frac{15}{660} = \frac{525}{660} = \frac{175}{220} \approx \boxed{0.795}$[/tex]
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Walnut High Schools Enrollment is exactly five times as large as the enrollment at walmut junior high the total enrollment for the two schools is 852 what is the enrollment at each school
Answer:
142
Step-by-step explanation:
If the enrollment is five times as big, then that is six different things. So, take 852 and divide it by 6 to get 142. Or in other words:
852÷6=142
142x6=856
I hope this helps. Cheers^^
Find the length of AB
Since this is a right triangle, we can use one of the three main trigonometric functions: sine, cosine, or tangent.
Remember: SOH-CAH-TOA
Looking from the given angle, we know the opposite side and want to know the hypotenuse. Therefore, we should use the sine function.
sin(54) = 16/AB
AB = 16/sin(54)
AB = 19.78 units
Hope this helps!
Factorize (256⁴-1).
Use appropriate identity.
(256⁴-1)
= (256-1)⁴
Using identity (a-b)⁴ = a⁴−4a³b+6a²b²−4ab³+b⁴
Let a be 256 and b be 1
Then
256⁴−4(256)³(1)+6(256)²(1)²−4(256)(1)³+(1)⁴
After solving
(256²-1)²
(a-b)² = a²-2ab+b²
256²-2×256×1+1²
= (256²-1)(256²+1)
Must click thanks and mark brainliest
Answer:
Use identity:
a² - b² = (a + b)(a - b)Consider that:
256 = 2⁸Now factorize:
256⁴ - 1 = (2⁸)⁴ - 1 = 2³² - 1 = (2¹⁶ - 1)(2¹⁶ + 1) = (2⁸ - 1)(2⁸ + 1)(2¹⁶ + 1) = (2⁴ - 1)(2⁴ + 1)(2⁸ + 1)(2¹⁶ + 1) = (2² - 1)(2² + 1)(2⁴ + 1)(2⁸ + 1)(2¹⁶ + 1) = (2 - 1)(2 + 1)(2² + 1)(2⁴ + 1)(2⁸ + 1)(2¹⁶ + 1)a figure skating school offers introductory lessons at $25 per session. There is also a registration fee of $30
Answer:
Part A: y = 25x + 30
Part B: $180
Step-by-step explanation:
for part A, 25 is the constant so that goes with the x, and you just add 30 because it is the extra fee. slope intercept form is y = mx + b
for part B you just plug 6 into the slope intercept equation
y = 25(6) + 30
y = 150 + 30
y = 180
$180
What is the distance from point N to LM in the figure below?
N
8.4
8.1
7.8
O
O A. 3.11
B. 0.8
C. 8.1
D. 2.18
E. 7.8
F. 8.4
Answer:
the answer to your question is 7.8 (E)
The distance from point N to LM is 7.8, 8.1 and 8.4 unit.
What is perpendicular?Perpendicular lines are those that cross at a straight angle to one another. Examples include the opposite sides of a rectangle and the steps of a straight staircase. the icon used to represent two parallel lines.
Perpendicular lines are two separate lines that cross one other at a right angle, or a 90° angle.
Given:
In ΔNOM
The Perpendicular distance is 7.8 unit
and, Hypotenuse distance id 8.1 unit
Now, In ΔNOL
The Perpendicular distance is 7.8 unit
and, Hypotenuse distance id 8.4 unit
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X
1
2
3
4
P
0,2
0,3
?
0,1
Answer:
(0,4) will be point (P) at 3 because,
Step-by-step explanation:
by using newton interpolation method we can find P(0,4) at 3 .
Which function describes this graph
Answer:
C.
Step-by-step explanation:
A P E X
Algebra word problem plz help me
Step-by-step explanation:
here's the answer to your question
A single die is rolled twice. The 36 equally-likely outcomes are shown to the right. Find the probability of getting two numbers whose sum is 10 .
Answer:
The probability of getting two numbers whose sum is 10 is 25%.
Step-by-step explanation:
Given that a single die is rolled twice, and there are 36 equally-likely outcomes, to find the probability of getting two numbers whose sum is 10 the following calculation must be performed:
1 = +9
2 = +8
3 = +7
4 = +6
5 = +5
6 = +4
7 = +3
8 = +2
9 = +1
9/36 = 0.25
Therefore, the probability of getting two numbers whose sum is 10 is 25%.
Solve the following equation for
d
d. Be sure to take into account whether a letter is capitalized or not.
Current
How many years will it take for an initial investment of $60,000 to grow to $90,000? Assume a rate of interest of
4% compounded continuously.
>It will take about _years for the investment to grow to $90,000.
(Round to two decimal places as needed.)
Answer:
I think i don't know the answer i am so sorry!!!
maybe someone else can Answer
forty-five percent of the students in a dorm are business majors and fifty-five percent are non-business majors. business majors are twice as likely to do their studying at the library as non-business majors are. half of the business majors study at the library. if a randomly slected student from the dorm studies at the library, what is the probability the student is a business major
Solution :
Defining the following events as :
B : Being a Business major
α : Studying at the library
∴ Given that :
[tex]$P(B) = \frac{45}{100}$[/tex]
= 0.45
Again, P [ Studying at the library | Being a Business major ] = 2 P [ Studying at the library | Being a non business major ]
[tex]$P[ \alpha | B] = 2 P[\alpha | B^C]$[/tex] .......(1)
Again,
[tex]$P[\text{Studying at the library } | \text{ Being a business major}] = \frac{1}{2} = 0.50$[/tex]
[tex]$P(\alpha | B) = 0.50$[/tex]
From (1), we get
[tex]$P(\alpha | B^C) = \frac{1}{2} . P(\alpha | B)$[/tex]
[tex]$=\frac{1}{2} \times 0.50$[/tex]
= 0.25
Therefore, we need,
= P[ The students is a Business major | The student studies at the library ]
[tex]$=P(B | \alpha)$[/tex]
By Bayes theorem
[tex]$=\frac{P(B). P(\alpha | B)}{P(B).P(\alpha | B) + P(B^C). P(\alpha | B^C)}$[/tex]
[tex]$=\frac{0.45 \times 0.50}{0.45 \times 0.50 + 0.55 \times 0.25}$[/tex]
= 0.6207
Solve this equation for x. Round your answer to the nearest hundredth.
1 = In(x + 7)
Answer:
[tex]\displaystyle x \approx -4.28[/tex]
General Formulas and Concepts:
Pre-Algebra
Equality PropertiesAlgebra II
Natural logarithms ln and Euler's number eStep-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle 1 = ln(x + 7)[/tex]
Step 2: Solve for x
[Equality Property] e both sides: [tex]\displaystyle e^1 = e^{ln(x + 7)}[/tex]Simplify: [tex]\displaystyle x + 7 = e[/tex][Equality Property] Isolate x: [tex]\displaystyle x = e - 7[/tex]Evaluate: [tex]\displaystyle x = -4.28172[/tex]e^1 = x+7
e - 7 = x
x = -4.28
In the figure, find the measure of TU⎯⎯⎯⎯⎯⎯⎯⎯
Answer:
TU = 27
Step-by-step explanation:
We are given two secant segments that are drawn from a circle to meet at an exterior point of the circle. Thus, according to the secant secant theorem, the product of the measure of one secant segment and its external secant segment equals that of the product of the other and its external secant segment.
Thus:
VU*TU = VW*BW
Substitute
7(x + 4) = 9(-2 + x)
7x + 28 = -18 + 9x
Collect like terms
7x - 9x = -18 - 28
-2x = -46
Divide both sides by -2
x = -46/-2
x = 23
✔️TU = x + 4
Plug in the value of x
TU = 23 + 4
TU = 27
Jean can swim 100 meters in 1.86 minutes. Sean can swim the same distance in 2.12 minutes.
please where is the question
please give me correct answer
by picture
Answer:
hello,
a) answer: 600
b) answer: 840
Step-by-step explanation:
a)
20=2²*5,25=5²,30=2*3*5,40=2³*5
l.c.m(20,25,30,40)=2³*3*5²=8*3*25=800
b)
24=2³*3, 42=2*3*7,35=5*7
l.c.m=(24,42,35)=2³*3*5*7=840
What are the domain and range of f(x) = |x + 6|?
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Answer:
domain: all real numbersrange: y ≥ 0Step-by-step explanation:
The function is defined for all values of x, so its domain is all real numbers.
The function can produce values of f(x) that are 0 or greater, so its range is ...
y ≥ 0
Simplify tan(arcsec 1)
Answer:
0
Step-by-step explanation:
Arc sec(1)=0, tan(0)=0
Find the mean for the amounts: $17.482: $14.987: $13.587$14.500, $18.580. $14.993
Answer:
The mean of these numbers is 15.68816 with a repeating 6.
Step-by-step explanation:
Two coins are tossed. Assume that each event is equally likely to occur. a) Use the counting principle to determine the number of sample points in the sample space. b) Construct a tree diagram and list the sample space. c) Determine the probability that no tails are tossed. d) Determine the probability that exactly one tail is tossed. e) Determine the probability that two tails are tossed. f) Determine the probability that at least one tail is tossed.
Answer:
(a) 4 sample points
(b) See attachment for tree diagram
(c) The probability that no tail is appeared is 1/4
(d) The probability that exactly 1 tail is appeared is 1/2
(e) The probability that 2 tails are appeared is 1/4
(f) The probability that at least 1 tail appeared is 3/4
Step-by-step explanation:
Given
[tex]Coins = 2[/tex]
Solving (a): Counting principle to determine the number of sample points
We have:
[tex]Coin\ 1 = \{H,T\}[/tex]
[tex]Coin\ 2 = \{H,T\}[/tex]
To determine the sample space using counting principle, we simply pick one outcome in each coin. So, the sample space (S) is:
[tex]S = \{HH,HT,TH,TT\}[/tex]
The number of sample points is:
[tex]n(S) = 4[/tex]
Solving (b): The tree diagram
See attachment for tree diagram
From the tree diagram, the sample space is:
[tex]S = \{HH,HT,TH,TT\}[/tex]
Solving (c): Probability that no tail is appeared
This implies that:
[tex]P(T = 0)[/tex]
From the sample points, we have:
[tex]n(T = 0) = 1[/tex] --- i.e. 1 occurrence where no tail is appeared
So, the probability is:
[tex]P(T = 0) = \frac{n(T = 0)}{n(S)}[/tex]
This gives:
[tex]P(T = 0) = \frac{1}{4}[/tex]
Solving (d): Probability that exactly 1 tail is appeared
This implies that:
[tex]P(T = 1)[/tex]
From the sample points, we have:
[tex]n(T = 1) = 2[/tex] --- i.e. 2 occurrences where exactly 1 tail appeared
So, the probability is:
[tex]P(T = 1) = \frac{n(T = 1)}{n(S)}[/tex]
This gives:
[tex]P(T = 1) = \frac{2}{4}[/tex]
[tex]P(T = 1) = \frac{1}{2}[/tex]
Solving (e): Probability that 2 tails appeared
This implies that:
[tex]P(T = 2)[/tex]
From the sample points, we have:
[tex]n(T = 2) = 1[/tex] --- i.e. 1 occurrences where 2 tails appeared
So, the probability is:
[tex]P(T = 2) = \frac{n(T = 2)}{n(S)}[/tex]
This gives:
[tex]P(T = 2) = \frac{1}{4}[/tex]
Solving (f): Probability that at least 1 tail appeared
This implies that:
[tex]P(T \ge 1)[/tex]
In (c), we have:
[tex]P(T = 0) = \frac{1}{4}[/tex]
Using the complement rule, we have:
[tex]P(T \ge 1) + P(T = 0) = 1[/tex]
Rewrite as:
[tex]P(T \ge 1) = 1-P(T = 0)[/tex]
Substitute known value
[tex]P(T \ge 1) = 1-\frac{1}{4}[/tex]
Take LCM
[tex]P(T \ge 1) = \frac{4-1}{4}[/tex]
[tex]P(T \ge 1) = \frac{3}{4}[/tex]
construct the truth table (p ∧ q) =⇒ [(q ∧ ¬p) =⇒ (r ∧ q)]
[tex]\begin{array}{c|c|c|c|c|c} p & q & r & p\land q & q\land \neg p & r \land q \\&&&&\\ T & T & T & T & F & T \\ T & T & F & T & F & F \\ T & F & T & F & F & F \\ T & F & F & F & F & F \\ F & T & T & F & T & T \\ F & T & F & F & T & F \\ F & F & T & F & F & F \\ F & F & F & F & F & F\end{array}[/tex]
An implication A => B is true if either A is false, or both A and B are true. So
[tex]\begin{array}{c|c|c}p\land q & (q\land\neg p) \implies (r\land q) & (p\land q) \implies \big[(q\land\neg p) \implies (r\land q)\big] \\&&\\T & T & \mathbf T\\T & T & \mathbf T\\F & T & \mathbf T\\F & T & \mathbf T\\F & T & \mathbf T\\F & F & \mathbf T\\F & T & \mathbf T\\F & T & \mathbf T\end{array}[/tex]
and the given statement is a tautology.
find b for (b-1)/4=(7b+2)/12
Answer:
-1.25
Step-by-step explanation:
you first have to cross multiply
12(b-1)=4(7b+2)
12b-12=28b+8
group the like terms
12b-28b=8+12
-16b/-16=20/-16
b= -1.25
I hope it helps
Step-by-step explanation:
Answer is in the picture..
hope it helps
What is the equation of a circle with a center at (4, -9) and a radius of 5?
Answer:
(x - 4)² + (y + 9)² = 25
Step-by-step explanation:
The equation of a circle is written as seen below.
(x – h)² + (y – k)² = r²
Where (h,k) represents the center of the circle and r represents the radius
We want to find the equation of a circle that has a center at (4,-9) and a radius of 5.
We know that (h,k) represents the center so h = 4 and k = -9
We also know that r represents the radius so r = 5
Now to find the equation of this specific circle we simply plug in these values into the equation of a circle formula
Equation: (x – h)² + (y – k)² = r²
h = 4, k = -9 and r = 5
Plug in values
(x - 4)² + (y - (-9))² = 5²
5² = 25
The two negative signs in front of the 9 cancel out and it changes to + 9
The equation of a circle with a center at (4,-9) and a radius of 5 is
(x - 4)² + (y + 9)² = 25
The job Andrew has this summer paid 7.25 an hour and the job he had last Summer paid 6.50 an hour. how much more does Andrew earn in a 40 hour week this summer than he did in a 40 hour week last summer
Answer:
30 Dollars more
Step-by-step explanation:
This summer he earned = 7.25 X 40 = 290
Last summer he earned = 6.5 X 40 = 260
How much more he earned in 40 hours = 290-260= 30 dollars more
Answer from Gauthmath