Answer: $9,000 is invested in the 5% bond, and $24,000 is invested in the 6.5% bond.
Step-by-step explanation: x + y = 33,000 (the full sum contributed)
We too know that the yearly intrigued earned is $2,010.00, which is the whole of the intrigued earned from each bond. Utilizing the equation for basic intrigued:
intrigued = principal x rate x time
ready to calculate the intrigued earned from each bond:
0.05x (for the 5% bond)
0.065y (for the 6.5% bond)
So we have another condition:
0.05x + 0.065y = 2,010
Presently we have two conditions with two factors, which we are able illuminate utilizing substitution or end.
Let's utilize substitution:
x + y = 33,000 --> y = 33,000 - x
0.05x + 0.065y = 2,010 --> 0.05x + 0.065(33,000 - x) = 2,010
Streamlining:
0.05x + 2,145 - 0.065x = 2,010
-0.015x = -135
x = 9,000
So the sum contributed within the 5% bond is $9,000, and the sum contributed within the 6.5% bond is:
y = 33,000 - x = 33,000 - 9,000 = $24,000
Please help, I don't know how to do this !
The lengths of the legs, the circumference and the radius of the circles are;
First part; The diameter = 18 units
The second leg = 9 units
Second part; The circumference = 34·π units
Third part; The radius is r = 36 cm
What is a radius of a circle?The radius of a circle is the distance from the center to the circumference.
First part;
cos(30°) = 9·√3 ÷ The diameter = (√3)/2
The diameter = (9·√3) ÷ ((√3)/2) = 18
The length of the other leg, using the relationship between special triangles is 18/2 = 9
Second part;
The lengths of the legs of the right triangle indicates that we get;
The square of the diameter, D² = 30² + 16² = 1156
The length of the diameter, D = √(1156) = 34
The formula for finding the circumference of a circle, C, based on the diameter can be presented as follows;
C = π × D
Therefore;
The circumference of the circle, C = π × 34
The circumference of the circle = 34·π units
Third part
The length of the 20° arc = 4·π cm
Therefore;
4·π = 2·π·r × 20/360
Where;
r = The radius of the circle
Therefore;
r = (360/20) × 4·π ÷ (2·π) = 36
The radius of the circle, r = 36 cm
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sage team brought 224 cans they brought 3 times cans as asifs team joes team brought 4 times as asifs team how much cans did joes team bring in
The number of cans brought by Joes team is J = 299 cans
Given data ,
Let the number of cans brought by Joes team be J
Now , sage team brought 224 cans
And , sage team brought 3 times cans as asifs team
And , joes team brought 4 times as asifs team
On simplifying the equation , we get
Asif's team brought 224/3 = 74.67 (rounded to the nearest whole number) cans.
Joe's team brought 74.67 x 4 = 298.68
So , J = 299 cans
Hence , the equation is solved and J = 299 cans
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Find the 13th term of the geometric sequence 4, −8, 16, ...
Answer:
Step-by-step explanation:
6) The height h(t) of a projectile is given by
h(t) = −t² + 7t + 9.
Find the time (s) at which the rocket is
20 ft above the ground.
Answer:
2.38 s
4.62 s
Step-by-step explanation:
The height of a projectile is given by the function:
[tex]h(t) = -t^2 + 7t + 9[/tex]
where:
h(t) is the height of the rocket about the ground (in feet).t is the time (in seconds).To determine the time(s) at which the rocket is 20 ft above the ground, set h(t) = 20 and solve for t.
[tex]-t^2+7t+9=20[/tex]
To solve the equation for t, complete the square.
Subtract 9 from both sides:
[tex]-t^2+7t=11[/tex]
Divide both sides by -1:
[tex]t^2-7t=-11[/tex]
Add the square of half the coefficient of the term with the variable "t" to both sides of the equation:
[tex]t^2-7t+\left(\dfrac{-7}{2}\right)^2=-11+\left(\dfrac{-7}{2}\right)^2[/tex]
Simplify:
[tex]t^2-7t+\dfrac{49}{4}=\dfrac{5}{4}[/tex]
We have now created a perfect square trinomial on the left side of the equation. Factor the perfect square trinomial:
[tex]\left(t-\dfrac{7}{2}\right)^2=\dfrac{5}{4}[/tex]
To solve for t, square root both sides:
[tex]t-\dfrac{7}{2}=\pm \sqrt{\dfrac{5}{4}}[/tex]
[tex]t-\dfrac{7}{2}=\pm\dfrac{\sqrt{5}}{2}[/tex]
Add 7/2 to both sides of the equation:
[tex]t=\dfrac{7}{2}\pm\dfrac{\sqrt{5}}{2}[/tex]
[tex]t=\dfrac{7\pm\sqrt{5}}{2}[/tex]
Therefore, the times at which the rocket is 20 ft above the ground is:
[tex]t=\dfrac{7-\sqrt{5}}{2}=2.38\; \rm s[/tex]
[tex]t=\dfrac{7+\sqrt{5}}{2}=4.62\; \rm s[/tex]
If tan A= 28 /45 and cosB= 13 12 and angles A and B are in Quadrant I, find the value of tan(A+B).
The value of Tan A + B is 6.009.
How to solve for the tangent[tex]sin^2 A + cos^2 A = 1sin^2 A = 1 - cos^2 Asin A = sqrt(1 - cos^2 A)sin A = sqrt(1 - (28/45)^2)sin A = sqrt(1 - 784/2025)sin A = sqrt(1241/2025)cos A = 28/45\geq[/tex]
We have to solve for Tan B
[tex]sin^2 B + cos^2 B = 1sin^2 B = 1 - cos^2 Bsin B = sqrt(1 - cos^2 B)sin B = sqrt(1 - (13/12)^2)sin B = sqrt(55/144)Therefore, tanB = sinB / cosB = (sqrt(55/144)) / (13/12) = (12 sqrt(55)) / 143[/tex]
[tex]tan(A+B) = (3207 \sqrt{55}) + 1804 \sqrt{124})) / (3115 - 207 \sqrt{(341)}[/tex]
= 6.009.
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PLEASE HELP FAST I’M HAVING A LITTE TROUBLE PLEASE GIVE THE RIGHT ANSWER
Answer:
C) 1/2 and 8
Step-by-step explanation:
-2x + y = 7 Eq. 1
6x + y = 11 Eq. 2
From Eq. 1:
y = 7 + 2x Eq. 3
From Eq. 2:
y = 11 - 6x Eq. 4
Equalyzing Eq. 3 and Eq. 4:
7 + 2x = 11 - 6x
2x + 6x = 11 - 7
8x = 4
x = 4/8
x = 1/2
From Eq. 3:
y = 7 +2* 1/2
y = 7 + 1
y = 8
Check:
From Eq. 2
6x + y = 11
6*1/2 + 8 = 11
3 + 8 = 11
use the product to rewrite log16(256b)
Log₁₆(256b) in terms of the logarithm of b, which is the factor that was multiplied by 256 inside the logarithm is 2 + log₁₆(b)
We can use the product rule of logarithms, which states that the logarithm of a product is equal to the sum of the logarithms of the factors.
Therefore, we can write
log₁₆(256b) = log₁₆(256) + log₁₆(b)
We can simplify log₁₆(256) as follows:
log₁₆(256) = log₁₆(16^2) = 2
Therefore, we have:
log₁₆(256b) = log₁₆(256) + log₁₆(b)
= 2 + log₁₆(b)
So, we have rewritten log₁₆(256b) in terms of the logarithm of b, which is the factor that was multiplied by 256 inside the logarithm.
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I know the answer, but I don't understand how. The answer is 90 degrees
In the circle the measure of arc YP as per the given measurements is equal to option D. 166°
In the circle,
Measure of arc ST = 70°
Measure of arc RS = 96°
Measure of angle YZP = 128°
Arc formed by vertically opposite angles are equal
This implies,
Measure of arc YT = Measure of arc PR
Measure of arc ST + Measure of arc RS + Measure of arc YT = 180°
⇒70° + 96° + Measure of arc YT = 180°
⇒Measure of arc YT = 180° - 166°
⇒Measure of arc YT = 14°
⇒ Measure of arc PR = 14°
Measure of arc YP = 360° - ( Measure of arc (PR + RS + ST + TY ))
⇒ Measure of arc YP = 360° - ( 14° + 96° + 70° + 14° )
⇒Measure of arc YP = 360° -194°
⇒ Measure of arc YP = 166°
Therefore, the measure of arc YP is equal to option D. 166°
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10+x/2 evaluate the expression for givin values of the variable
The value of the expression when x = 4 is 12.
The value of the expression when x = -8 is 6.
We have,
To evaluate the expression 10 + x/2 for a given value of the variable x, we simply substitute the value of x into the expression and simplify.
For example:
If x = 4:
10 + x/2 = 10 + 4/2
= 10 + 2
= 12
So when x = 4, the value of the expression is 12.
If x = -8:
10 + x/2 = 10 + (-8)/2
= 10 - 4
= 6
So when x = -8, the value of the expression is 6.
Thus,
The value of the expression when x = 4 is 12.
The value of the expression when x = -8 is 6.
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In △ABC, AB=6 cm, AC=15 cm, and m∠A=48°.
What is the area of △ABC?
Enter your answer as a decimal in the box. Round only your final answer to the nearest hundredth.
The area of triangle ABC is : 33.44 cm²
What is the Area of a Triangle?The area of a triangle is the region enclosed within the sides of the triangle. The area of a triangle varies from one triangle to another depending on the length of the sides and the internal angles. The area of a triangle is expressed in square units, like, m2, cm2, in2, and so on.
We have the information from the question is:
In triangle ABC :
AB=6 cm,
AC=15 cm, and
m ∠A=48°.
We have to find the area of Triangle ABC
Area of triangle is : 1/2 × bc × (SinA)
Area of Triangle = 1/2 × 6 × 15 × (Sin 48° )
We know the value of Sin 48° is 0.7431
Area of Triangle = (1/2) × 66.88
Area of triangle = 33.44 cm²
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Area of a regular polygon work
Answer:
Step-by-step explanation:
[tex]\frac{x}{18} =cot~30\\x=18cot~30=18 \times \sqrt{3} =18\sqrt{3} \\base=36\sqrt{3} \\area~of~triangle=3\times \frac{1}{2} \times36\sqrt{3} \times 18=972\sqrt{3} \approx 1683.55~mm^2[/tex]
In the first group of 5, average was 6.5. In the second group of 15, the average was 4.5. What was the overall average of the two groups?
Step-by-step explanation:
First group of five total score = 5 * 6.5 = 32.5
Group of fifteen total score = 15 * 4.5 = 67.5
total score of 20 is 67.5 + 32.5 = 100
Average = 100 /20 = 5.0
In an election, the median number of votes a candidate received in 6 towns was 250. Which statement MUST be true about this election?
OA. The total number of votes the candidate received in the election was 1500.
OB. The candidate received at least 250 votes in half of the 6 towns.
OC. The candidate received exactly 250 votes in at least two of the towns.
O D. The total number of votes received by all the candidates in the election was 1500.
Answer:
B. The candidate received at least 250 votes in half of the 6 towns.
This is because the median number of votes is the middle value when all the vote counts are put in order. This means that at least three towns gave the candidate more than 250 votes, and at least three towns gave the candidate fewer than 250 votes. So, the candidate received at least 250 votes in half of the 6 towns. The total number of votes the candidate received in the election cannot be determined from this information. Similarly, the number of votes received in individual towns cannot be determined.
32 T W Proving Triangle Congruent; determine if AAS, SAS, SSS, HL, ASA
10: HL
11: AAS
12: SSS
13: SSS
14: HL
15: AAS
16: AAS
17: HL
18: AAS
19: SAS
20 AAS
1: SAS
2: ASA
3: HL
4: HL
5: ASA
6: ASA
7: HL
8: SSS
9: HL
I don’t understand how to solve and find the system here. Please help, I need it to be done in three hours.
The system of equations in the context of this problem is given as follows:
5h + 3c = 340.2h + 6c = 400.The solution is given as follows:
h = 35, c = 55.
How to model the system of equations?The variables used for the system of equations are given as follows:
Variable h: Amount earned with a haircut.Variable c: Amount earned with a color treatment.Considering the Monday's earnings, the equation is given as follows:
5h + 3c = 340.
Considering the Wednesday's earnings, the equation is given as follows:
2h + 6c = 400.
Hence the system is:
5h + 3c = 340.2h + 6c = 400.Simplifying the second equation by 2, we have that:
h + 3c = 200.
Subtracting the first equation by the second, we can obtain the value of h as follows:
4h = 140
h = 140/4
h = 35.
Hence the value of c is given as follows:
c = (200 - 35)/3
c = 55.
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Find the range of the following: 20, 3, 95, 54, 71, 66
Answer:
92.
Step-by-step explanation:
To find the range of the set of numbers 20, 3, 95, 54, 71, 66, we first need to find the difference between the largest and smallest numbers in the set.
Largest number = 95
Smallest number = 3
Range = Largest number - Smallest number
Range = 95 - 3
Range = 92
Therefore, the range of the set of numbers 20, 3, 95, 54, 71, 66 is 92.
Answer:92
Step-by-step explanation: highest-lowest
it helps to put the terms in order first
95-3=92
The time it takes to preform a task has a continuous uniform distribution between 43 min and 57 min. What is the the probability it takes between 48 and 49.9 min. Round to 4 decimal places. P(48 < X < 49.9) =
The probability of the task taking between 48 and 49.9 minutes is 0.1357 or 13.57% (rounded to four decimal places).
To calculate this probability, we first need to find the total probability of the entire range between 43 and 57 minutes. This can be found by subtracting the lower bound from the upper bound and dividing by the total range:
P(43 < X < 57) = (57 - 43) / (57 - 43) = 1
Since the probability of the entire range is 1, we can find the probability of any sub-interval by dividing the length of the sub-interval by the length of the total range. Therefore, the probability of the task taking between 48 and 49.9 minutes is:
P(48 < X < 49.9) = (49.9 - 48) / (57 - 43) = 1.9 / 14 = 0.1357 or 13.57%
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at+a+fair+Daniel+and+Claire+went+on+a+ride+that+has+two+separate+circular+tracks.+Daniel+rode+in+a+purple+car+that+travels+a+total+distance+of+265+feet+around+the+track+.+Ciara+rode+in+a+yellow+car+that+travels+a+total+distance+of+170+feet+around+the+track.+They+drew+drew+a+sketch+of+the+ride.+What+is+the+difference+ofthe+radii+of+the+two+circle+tracks
Note that the difference in the radii of the two circular tracks is about 13.68 ft
How did we get that?recall that the circumference of a circle is denoted by the expression
C = 2πr
In this case, C = Circumference
and r radius
Since we have two tracks
Let ra = radius of the purple track and
rb = radius of the yello track
So
265 = 2πra
170 = 2πrb
making ra and rb subject of the expression we have
ra = 256/(2π) =40.7436654315 ≈ 40.74
rb = 170 / (2π) = 27.0563403256 ≈ 27.06
Hence, the difference is
40.74 - 27.06 = 13.68ft
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Full Question:
at a fair Daniel and Claire went on a ride that has two separate circular tracks. Daniel rode in a purple car that travels a total distance of 265 feet around the track . Ciara rode in a yellow car that travels a total distance of 170 feet around the track. They drew drew a sketch of the ride. What is the difference ofthe radii of the two circle tracks
What transformation were used
Since ΔJKL was transformed to create ΔMNO, the sequence of transformations that were used include the following:
Reflection over the y-axis.
(x, y) → (x, y - 9).
What is a reflection over the y-axis?In Geometry, a reflection over or across the y-axis or line x = 0 is represented and modeled by this transformation rule (x, y) → (-x, y).
By applying a reflection over the y-axis to the coordinate of the given triangle JKL, we have the following coordinates:
Coordinate J = (3, 7) → Coordinate J' = (-(3), 7) = (-3, 7).
Next, we would apply a vertical translation by shifting triangle J'K'L' down by 9 units to produce ΔMNO;
(x, y) → (x, y - 9).
J' (-3, 7) → (-3, 7 - 9) = M' (-3, -2).
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Ellen bought a pedometer wristband to track how many steps she takes. Each day she recorded the number of steps in a table. At the end of the week, she found her mean number of steps was 10,000. She made a bar graph to show her daily steps for the week. Which graph could show Ellen’s steps?
THANKS SO MUCH!!!
Answer:
it is b got it right on my test
Step-by-step explanation:
The required graph that shows Ellen's steps where her mean steps for a week is 10000 is graph 2 where the data are as,
Day 1 = 11000 , Day 2 = 9000, Day 3 = 10000, Day 4 = 12000, Day 5 = 8000, Day 6 ≈ 10000, Day 7 ≈ 10000
Mean of ungrouped data is the average of all the values of the observation in the data set.
Mean = (Summation of all values in data set)/ (Number of observations)
Here there are 7 observations and the mean number of steps is 10000.
For graph 1 the values for each day of the week are as,
Day 1 = 8000 , Day 2 = 9000, Day 3 = 12000, Day 4 = 2000, Day 5 = 6000, Day 6 = 10000, Day 7 = 10000
Thus, the mean number of steps is = (8000+ 9000+ 12000+ 2000+ 6000+ 10000 + 10000) / 7
= 8143 (approximately) ≠ 10000
For graph 2 the values for each day of the week are as,
Day 1 = 11000 , Day 2 = 9000, Day 3 = 10000, Day 4 = 12000, Day 5 = 8000, Day 6 ≈ 10000, Day 7 ≈ 10000
Thus, the mean number of steps is = (11000+ 9000+ 10000+ 12000+ 8000+ 10000 + 10000) / 7
= 10000
For graph 3 the values for each day of the week are as,
Day 1 = 7000 , Day 2 = 14000, Day 3 = 11500, Day 4 = 10000, Day 5 = 4500, Day 6 = 6000, Day 7 = 11000
Thus, the mean number of steps is = (7000+ 14000+ 11500+ 10000+ 4500+ 6000 + 11000) / 7
= 9143 (approximately) ≠ 10000
For graph 4 the values for each day of the week are as,
Day 1 = 8000 , Day 2 = 9000, Day 3 = 9500, Day 4 = 10000, Day 5 = 9000, Day 6 = 7000, Day 7 = 11000
Thus, the mean number of steps is = (8000+ 9000+ 9500+ 10000+ 9000+ 7000 + 11000) / 7
= 9071 (approximately) ≠ 10000
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What is the solution to this system?
(1, 0)
(1, 6)
(8, 26)
(8, –22)
x = –2
Answer:
Step-by-step explanation:The system of equations represented by the given points is:1x + 0y = 1 (equation 1)
1x + 6y = 1 (equation 2)
8x + 26y = 1 (equation 3)
8x - 22y = 1 (equation 4)
We can use the first equation to solve for x in terms of y:1x + 0y = 1
1x = 1
x = 1
Then we can substitute x=1 into equations 2-4 to obtain three equations in terms of y:1(1) + 6y = 1 => 6y = 0 => y = 0
8(1) + 26y = 1 => 26y = -7/8 => y = -7/208
8(1) - 22y = 1 => -22y = -7/8 - 7 => y = 15/44
Therefore, the solution to the system is x = 1 and y = -7/208 or y = 0 or y = 15/44.
However, we are given that x = -2, which contradicts our solution of x = 1. Therefore, there is no solution to the system with the given values.
On the first day of school, the eighth graders get to select one of two dances to attend for no cost. They can attend either the Fall Fling or the Spring Social for free.
Of the 144 boys in eighth grade, 82 choose to attend the Fall Fling for free. Of the 156 girls in eighth grade, 75 choose to attend the Spring Social for free.
The president of the eighth grade class constructed the following two-way table to show the numbers of boys and girls that choose each dance. Some of the values and totals are incorrect.
If the following two-way table was constructed from the data, select all the values the president of the eighth grade class has in
the correct spots.
The two-way table constructed from the data showing the numbers of boys and girls that choose each dance category should look like this:
Boys Girls Total
Fall Fling 82 81 163
Spring Social 62 75 137
Total 144 156 300
What is a two-way table?A two-way table is a data display tool that displays the frequencies for two different categories collected from a single group.
A two-way table use rows to represent one category variable and the columns to represent a second category variable, while the cells display the proportions or frequencies of each row and column combination.
The total number of boys in eighth grade = 144
The number of boys who choose to attend the Fall Fling = 82
Therefore, the number of boys who attend the Spring Social = 62 (144 - 82)
The total number of girls in eighth grade = 156
The number of girls who choose to attend the Spring Social = 75
Therefore, the number of girls who attend the Fall Fling = 81 (156 - 75)
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Answer:
62, 82, 163, 300
Step-by-step explanation:
I got this on study island
 1.What did Ruskin write about Nocturne in Black and Gold: Falling Rocket?
2. How did Whistler defend his work in court? ( You need to mention the two paintings he brought with him to court.)
3. What was the outcome of the trial?
4. Why do you think the lawsuit was important in the history of art?
If a > O and y > 0, which expression is equivalent to v 768219 g372
• A.
162°y18 32g
• B.
-O c. 1604 ° V324,
O D.
8ay18 /12ry
The expression 16x⁹y¹⁸√3xy is equivalent to √769x¹⁹y³⁷
If a > O and y > 0, which expression is equivalent to √769x¹⁹y³⁷
Factor and rewrite the radicand in exponential form:
√16²×3x¹⁸×xy³⁶×y
Simplify the radical expression:
16x⁹y¹⁸ √2²×3xy
Factor and rewrite the radicand in exponential form:
8x⁹y¹⁸√2²×3xy
8x⁹y¹⁸.2√3xy
Multiply the monomials:
16x⁹y¹⁸√3xy
Hence, the expression 16x⁹y¹⁸√3xy is equivalent to √769x¹⁹y³⁷
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Anna took a job that paid $116 the first week. She was guaranteed a raise of 7% each week. How much money will she make in all over 10 weeks? Round the answer to the nearest cent, and number answer only.
Anna will make $1602.71 in all over 10 weeks.
Here, the first week salary of Anna = $116
i.e., the initial salary a = $116
She was guaranteed a raise of 7% each week.
so, her salary in the next week would be,
116 + 7% of 116
7 percent of 116 is:
116 × 7/100 = 8.12
so, her salary will be $124.12
So, the equation for the salary after 'm' weeks would be,
[tex]n = 116\times (1 + 0.07)^{m - 1}\\\\n = 116\times (1 .07)^{m - 1}[/tex]
Using this equation , the salary in the second week m = 2 would be,
n = 124.12
Salary in the third week m = 3 would be,
n = 116 × [tex](1.07)^{3-1}[/tex]
n = 132.81
Salary in the fourth week m = 4 would be,
n = 116 × [tex](1.07)^{4-1}[/tex]
n = 142.11
Salary in the fifth week m = 5 would be,
n = 116 × [tex](1.07)^{5-1}[/tex]
n = 152.05
Salary in the sixth week m = 6 would be,
n = 116 × (1.07)⁶⁻¹
n = 162.7
Salary in the seventh week m = 7 would be,
n = 116 × (1.07)⁷⁻¹
n = 174.08
Salary in the eighth week m = 8 would be,
n = 116 × (1.07)⁸⁻¹
n = 186.27
Salary in the nineth week m = 9 would be,
n = 116 × (1.07)⁹⁻¹
n = 199.31
And the salary after the tenth week m = 10 would be,
n = 116 × (1.07)¹⁰⁻¹
n = 213.26
The total money after 10 weeks would be,
T = 116 + 124.12 + 132.81+ 142.11 + 152.05 + 162.7 + 174.08 + 186.27 + 199.31 + 213.26
T = $1602.71
This is the required amount.
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$18,389
is invested, part at 10%
and the rest at 8%
. If the interest earned from the amount invested at 10%
exceeds the interest earned from the amount invested at 8%
by $621.02
, how much is invested at each rate? (Round to two decimal places if necessary.)
I need help! pls and thank u!
Evaluating the given function, we can see that the intensity at 35 cm is 0.816
How to find the intensity at 35 cm from the source?The intensity at a distance d in centimeters is given by the function:
I = 100*d⁻²
Here we want to find the intensity at 35 centimeters from the source, then we need to evaluate the function above at d = 35, doing that we will get:
I = 100*35⁻²
I = 100/(35²)
I = 0.816
The intensity at 35 centimeters from the source is 0.816.
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The parallel dotplots below display the number of absences for students in each of two classes.
2 dotplots titled class absences. The number lines go from 0 to 10 and are labeled number of absences. Class D, 0, 7; 1, 11; 2, 4; 3, 3. Class C, 0, 8; 1, 10, 2, 4; 3, 1; 5, 1; 10, 1.
Which of the following statements is true?
The range for the distribution of the number of absences is larger for class D.
The range for the distribution of the number of absences is larger for class C.
The IQR for the distribution of the number of absences is larger for class D.
The IQR for the distribution of the number of absences is larger for class C.
The range, difference of maximum and minimum values, is larger for Class C with a range of 10 compared to Class D with a range of 3. The data provided is not sufficient to calculate the IQR (Interquartile Range) for either class.
Explanation:To begin, let's understand what the terms 'range' and 'IQR' (Interquartile Range) refer to in statistics. The range is simply the difference between the largest and smallest data values. The IQR is the range of the middle 50% of the values when ordered from lowest to highest.
For Class D, the lowest absence number is 0 and the highest is 3, thus the range is 3-0 = 3. For Class C, the lowest absence number is 0 and the highest is 10, so the range is 10-0 = 10. Thus, the range for the distribution of absences is larger for Class C.
Calculating IQR requires determining the 1st Quartile (Q1) and 3rd Quartile (Q3) values and subtracting Q1 from Q3. The data given doesn't include enough information to directly calculate these values and hence, it is impossible to definitively say which class has a larger IQR based on the provided information.
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Help. I need game credits for GMM
The difference of p and z times the sum of p and z is p² - 2pz - z².
How to represent expression?An algebraic expression in mathematics is an expression which is made up of variables and constants, along with algebraic operations such as addition, subtraction, division and multiplication.
Hence, the difference of p and z times the sum of p and z can be represented as follows:
Therefore,
(p - z) × (p + z) = (p - z) (p + z)
(p - z) (p + z) = p² - pz - pz - z²
p² - pz - pz - z² = p² - 2pz - z²
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If the amount of food available is increased, then the cricket frog population will (increase/decrease/stay the same) over the span of five years.