Data:
The city received 12 (1/4) inches of rain in 25 (1/2) hours.
Procedure:
Rewriting the numbers as decimals.
[tex]12\cdot\frac{1}{4}=12.25[/tex][tex]25\cdot\frac{1}{2}=25.5[/tex]To find the approximate rainfall rate in inches per hour, we have to do as follows:
[tex]\frac{12.25}{25.5}\approx0.48\frac{in}{h}[/tex]Rounding the result, we get...
[tex]0.48\approx0.5\approx\frac{1}{2}[/tex]Answer: D. about 1/2 inch per hour
Finish the other half of the graph if it was even and odd.
To solve this problem, first, let's remember the definitions of even and odd functions.
• A function f is ,even, if the graph of f is ,symmetric about the y-axis,.
,• A function f is ,odd, if the graph of f is ,symmetric about the origin.
a) To make the function even, we must complete the graph such the graph result is symmetric about the y-axis (the vertical axis). Doing that we get:
b) To make the function odd, we must complete the graph such the graph result is symmetric about the origin (the horizontal axis). Doing that we get:
Which of the following measurements form a right triangle? Select all that apply.
We are asked to find which of the measurements form a right triangle.
A right triangle is a triangle that has an angle of 90°, and also we can use the Pythagorean theorem in them.
The Pythagorean theorem tells us that the sum of the two legs of the triangle squared is equal to the hypotenuse squared:
[tex]a^2+b^2=c^2[/tex]Where a and b are the legs of the triangle and c is the Hypotenuse. Also, in the right triangle, the hypotenuse is the longest side of the triangle.
We will use the Pythagorean theorem formula on all of the options using the first two given measures as a and b, and check that we the third measure as the value of c.
Option A. 7in, 24in, and 25 in.
We define:
[tex]\begin{gathered} a=7 \\ b=24 \end{gathered}[/tex]And apply the Pythagorean theorem:
[tex]7^2+24^2=c^2[/tex]And we solve for c. If the result for x is 25, the triangle will be a right triangle, if not, this will not be an answer.
-Solving for c:
[tex]\begin{gathered} 49+576=c^2 \\ 625=c^2 \end{gathered}[/tex]Taking the square root of both sides we find c:
[tex]\begin{gathered} \sqrt[]{625}=c \\ 25=c \end{gathered}[/tex]Since we get the third measure as the value of c option A is a right triangle.
Option B. 18ft, 23ft, and 29 ft.
we do the same as did with option A. First, define a and b:
[tex]\begin{gathered} a=18 \\ b=23 \end{gathered}[/tex]Apply the Pythagorean theorem:
[tex]18^2+23^2=c^2[/tex]And solve for c:
[tex]\begin{gathered} 324+529=c^2 \\ 853=c^2 \\ \sqrt[]{853}=c \\ 29.2=c \end{gathered}[/tex]We get 29.2 instead of just 29, thus option B is NOT a right triangle.
Option C. 10in, 24in, and 26 in.
Define a and b:
[tex]\begin{gathered} a=10 \\ b=24 \end{gathered}[/tex]Apply the Pythagorean theorem:
[tex]10^2+24^2=c^2[/tex]Solve for c:
[tex]\begin{gathered} 100+576=c^2 \\ 676=c^2 \\ \sqrt[]{676}=c \\ 26=c \end{gathered}[/tex]We get 26 which is the third measure given, thus, option C is a right triangle.
Option D. 10yd, 15yd, and 20yd.
Define a and b:
[tex]\begin{gathered} a=10 \\ b=15 \end{gathered}[/tex]Apply the Pythagorean theorem:
[tex]\begin{gathered} 10^2+15^2=c^2 \\ 100+225=c^2 \\ 325=c^2 \\ \sqrt[]{325}=c \\ 18.03=c \end{gathered}[/tex]We don't get 20yd as the value of c, thus, option D is NOT a right triangle.
Option E. 15mm, 18mm, and 24 mm
Define a and b:
[tex]\begin{gathered} a=15 \\ b=18 \end{gathered}[/tex]Apply the Pythagorean theorem
[tex]\begin{gathered} 15^2+18^2=c^2 \\ 225+324=c^2 \\ 549=c^2 \\ \sqrt[]{549}=c \\ 23.43=c \end{gathered}[/tex]We don't get 24 as the value of c, thus, option E is Not a right triangle.
Answer:
Option A and Option C are right triangles.
ill take a pic of it
The line passes through the points given.
Select any two points from the table, (-4,2) and (-3,5).
The slope is given by:
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]Hence the slope is:
[tex]\begin{gathered} m=\frac{5-2}{-3-(-4)} \\ m=\frac{3}{1} \\ m=3 \end{gathered}[/tex]The slope is 3.
Inequality-x less than or equal to 18
Answer: x[tex]\leq[/tex]18
What is the area of a triangle that has a height of 10 feet and a base of 6 feet?A. 16 feet squaredO B. 30 feet squaredO C.35 feet squaredD. 60 feet squared
Th formula for the area of the triangle is as follows.
[tex]A=\frac{bh}{2}[/tex]where b is the base of the triangle and h is its height.
Substitute the given values into the formula and then simplify.
[tex]\begin{gathered} A=\frac{(6)(10)}{2} \\ =\frac{60}{2} \\ =30 \end{gathered}[/tex]Therefore, the area of the triangle is 30 square feet, which is option B.
Lena eats an apple every otherday. Suppose today is Monday,October 1. Lena eats an appletoday.When will Lena eat an appleon a Monday again?AnsLe
Find to the nearest degree the measure of the angle of elevation of the sun when a woman 150 cm tall casts a shadow 40 cm long.
The triangle formed is shown in the diagram below
The angle of elevation of the sun is represented by x. To determine x, we would apply the tangent trigonometric ratio which is expressed as
Tan# = opposite side/adjacent side
opposite side = 150
adjacent side = 40
Tan x = 150/40 = 3.75
x = Tan^-1(3.75)
x = 75.069
To the nearest degree, the measure of the angle of elevation of the sun is 75 degrees
Simplify this fraction: 30/36
To simplify this fraction, we will have to find the common factors of both the numerator and denominator, then divide.
Common factors of 30 and 36 are: 2, 3, and 6
Now both numerator and denominator by the highest common factor which is 6:
[tex]\frac{30}{36}\text{ = }\frac{5}{6}[/tex]
After simplifying the fraction, we have:
[tex]\frac{5}{6}[/tex]compute the value of the discriminant and give the number of real solutions of the quadratic equation. -2x²+3x+5=0
Given a quadratic equation in standard form
[tex]y=ax^2+bx+c[/tex]The discriminant D
[tex]D=b^2-4ac[/tex]tells the types of roots the equation has.
In this case, we have
[tex]\begin{gathered} -2x^2+3x+5=0 \\ a=-2 \\ b=3 \\ c=5 \end{gathered}[/tex]Then, the discriminant of this quadratic equation will be
[tex]\begin{gathered} D=b^2-4ac \\ D=(3)^2-4(-2)(5) \\ D=9+40 \\ \mathbf{D=49} \end{gathered}[/tex]Finally, the value of discriminat is 49 and as he discriminant is greater than zero then this quadratic equation has 2 different real solutions.
4Select the correct equations.Gracie, Mary, and Nancy each have a small collection of seashells. Gracie has 5 more than  times the number of shells Mary has. Nancy has 1 more than  times the number of shells Mary has. Gracie and Nancy have the same number of shells. If x is the number of shells Mary has, identify the equation that represents this situation and identify its solution.
Given data:
Gracie has 5 more than times mary have G=5+a(x).
Nancy has 1 ore than ties mary have N=1+b(x)
Given that G=N
5+ax=1+bx
4=x(b-a)
graph and label each figure and it's image under the given reflection. give the new coordinates. you don't have to graph it for me, could you just helps with the coordinates
Explanation
Step 1
Let the vertices
[tex]\begin{gathered} C(-4,7) \\ D(-2,4) \\ E(-4,1) \\ F(-6,4) \end{gathered}[/tex]When you reflect a point across the y-axis, the y-coordinate remains the same, but the x-coordinate is transformed into its opposite (its sign is changed)
[tex]P(x,y)\rightarrow reflect\text{ across y a }\xi s\rightarrow P^{\prime}(-x,y)[/tex]then, apply the rule to find the new coordinates
[tex]\begin{gathered} C(-4,7)\rightarrow C^{\prime}(4,7) \\ D(-2,4)\rightarrow D^{\prime}(2,4) \\ E(-4,1)\rightarrow E^{\prime}(4,1) \\ F(-6,4)\rightarrow F^{\prime}(6,4) \end{gathered}[/tex]I hope this helps you
Find all the powers of four in the range of 4 and 1000
a plant is already 44 cm tall, and will grow one cm every month. let H be height in cm and M months. write and equation relating H to M . then use equation to find plants height after 32 months
H = height in cm
M = months
The plant is already 44 cm tall
GRowth every month = 1 cm
Equation:
H (m) = 44 + m
The height after m months, will be equal to the initial height (44) plus the number of months.
For 32 months, replace m by 32 and solve:
H (32) = 44+32
H (32) = 76 cm
After 32 months, the plant will be 76 cm tall
Factor the common factor1) -36m + 16
Given:
-36m + 16
To factor out the common factor, let's find the Greatest Common Factor (GCF) of both values.
GCF of -36 and 16 = -4
Factor out -4 out of -36 and 16:
[tex]-4(9m)-4(-4)[/tex]Factor out -4 out of [-4(9m) - 4(-4)] :
[tex]-4(9m\text{ - 4)}[/tex]ANSWER:
[tex]-4(9m-4)[/tex]1.The histogram (next page) summarizes the data on the body lengths of 143 wild bears. Write a fewsentences describing the distribution of body lengths.403020103035404570 7580 8550 55 60 65length in inchesBe sure to comment on the shape, center, and spread of the distribution.
The shape of the distribution is bell-shaped. This is because the distribution presents a normal distribution
The distribution is almost symmetrically skewed with no outlier
The center is about 60 inches(about 59 wild bears before the center and about 84 wild bears beyond the center)
The distribution is widely spread: The data range is the highest inches minus the lowest inches
Therefore, the spread of the distribution is 85 inches - 35 inches, which equals 50 inches.
Please tell me if these are correct if theyre not please help and tell me which ones are the right answers
Answer:
They're correct
Step-by-step explanation:
solve the following d. be sure to take into account whether a letter is capitalized or not .3y^3 ×m=5Qd
Step 1: Write out the equation.
[tex]3g^3+m=5Qd[/tex]Step 2: Divide both sides of the equation by 5Q, we have
[tex]\frac{3g^3+m}{5Q}=\frac{5Qd}{5Q}[/tex]this implies that
[tex]d=\frac{3g^3+m}{5Q}[/tex]Gordin has 4.1 GB of data in one 4-week billing cycle on his cell phone. Any unused data will roll over to the next cycle. The table shows how much data Gordon uses each week before his next billing cycle. WEEK. DATA USAGE (IN GB) 1. 0.8 2. 1.3 3. 0.9 4. 1.1How much data does Gordon have left to roll over to the next billing cycle at the end of the four weeks?A. 0 GB of dataB. 0.8 GB of datac. 3 GB of data
Gordan have left 0 GB to roll over to the next billing cycle at the end of the four weeks.
What is Addition?
The sum of all the numbers are called the Addition of numbers.
Given that;
Gordin has 4.1 GB of data in one 4-week billing cycle on his cell phone.
And, The table shows how much data Gordon uses each week before his next billing cycle as;
WEEK = 1 2 3 4
DATA USAGE = 0.8 1.3 0.9 1.1
Now,
Since, Gordin has 4.1 GB of data in one 4-week billing cycle on his cell phone.
And, The table shows how much data Gordon uses each week before his next billing cycle.
So, We can add all the given data usage in 1 to 4 week as;
Data usage = 0.8 + 1.3 + 0.9 + 1.1
= 4.1 GB
Thus, Gordan have left 0 GB to roll over to the next billing cycle at the end of the four weeks.
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What is the sequence that has a recursive formula A(n)= A(n-1)+4 where A(1)=3
1) Considering that, let's find each term:
[tex]\begin{gathered} a_1=3 \\ a_n=a_{n-1}+4 \\ a_2=a_1+4\Rightarrow a_2=3+4=7 \\ a_3=a_2+4\Rightarrow a_3=7+4\text{ =11} \\ a_4=11+4\text{ }\Rightarrow a_4=15 \end{gathered}[/tex]2) So the sequence is
[tex](3,7,11,15,\ldots)[/tex]As each term, from the 2nd one is 4 units more that's why we can make it using a recursive formula
A sole trader operates his business from
a warehouse, which has been damaged
by a fire, which occurred at the end of
the financial year. After the fire, the
remaining inventory that is undamaged
amounts to GHC 2,000 (cost). The
accountant establishes the following
information: I Inventory at the
beginning of the year was GHC 16,000 II
Purchases during the year were GHC
115,000 III Sales during the year were
GHC 140,000 IV The trader sells his
goods at a mark-up of 25% of cost What
is the value of the inventory lost in fire?
Beginning inventory = 16,000 Purchase = 115,000 Sales = 140,000 Mark up = 25% on cost Undamanged inventory = 2,000
The value of the inventory lost in fire is 2,000
How do you take inventory loss due to fire into account?
The cost of products available for sale is then subtracted from the cost of goods sold. The quantity will reflect how much inventory the fire has actually destroyed. As an illustration, $275,000 minus $80,000 = $195,000, which represents the amount of inventory destroyed in the fire.
Calculate the quantity of inventory destroyed by deducting the cost of sold goods from the cost of goods that are still in stock. In this scenario, the amount of merchandise lost in the fire was $275,000 less $70,000, or $205,000.
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8 Solve: 2 3= 4x + 2 - O- O 1 2 AP 4 ( ) 1 4. O 1 2
We will have the following:
[tex]3=4x+2\Rightarrow1=4x[/tex][tex]\Rightarrow x=\frac{1}{4}[/tex][Third option]
the prompt is in the photo
By using the given box and whisker plot, the number of students that earned a score from 77 and 90 is: N. 13.
What is a box and whisker plot?A box and whisker plot is also referred to as boxplot and it can be defined as a type of chart that can be used to graphically represent the five-number summary of a data set with respect to locality, skewness, and spread.
In Mathematics, the five-number summary of any box and whisker plot include the following:
MinimumFirst quartileMedianThird quartileMaximumWhat is an interquartile range?IQR is an abbreviation for interquartile range and it can be defined as a measure of the middle 50% of data values when they are ordered from lowest to highest.
Mathematically, interquartile range (IQR) is the difference between first quartile (Q₁) and third quartile (Q₃):
IQR = Q₃ - Q₁
Based on the given box and whisker plot, we can logically deduce the following quartile ranges:
Third quartile, Q₃ = 90
First quartile, Q₁ = 77
Now, we can calculate the interquartile range (IQR) is given by:
Interquartile range, IQR = Q₃ - Q₁
Interquartile range, IQR = 90 - 17
Interquartile range, IQR = 13
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Examine the graph and write a statement about the data. Use specific information from the graph.
This bar graph represents the percentage of the public's trust in the Federal Government from year 1960 to 2015. The public trust was highest in year 1960 at 74% and it was at its lowest in 2015 at 18%.
Find the probability of drawing a red ace and then a spade when two cards are dranw (without replacement) from a standard deck of cards.a. 1/102b. 31/102c. 1/2d. 31/64
a. probability of drawing a red ace (first draw)
In a standard deck, there are 52 cards. Out of these 52 cards, two are red aces. Hence, the probability of drawing a red ace is 2/52 or 1/26.
b. probability of drawing a spade (second draw)
On the second draw, 51 cards are left. Assuming that a red ace was taken on the first draw, 13 spades are left on the deck. Hence, the probability of drawing a spade is 13/51.
So, to get the probability of drawing a red ace AND a spade, simply multiply the two probabilities above.
[tex]\frac{1}{26}\times\frac{13}{51}=\frac{13}{1326}[/tex]Then, reduce 13/1326 into its simplest form by dividing both numerator and denominator by 13.
[tex]\frac{13\div13}{1326\div13}=\frac{1}{102}[/tex]Hence, the probability of drawing a red ace AND a spade is 1/102. (Option A)
Given circle O with diameter AC, tangent AD, and the measure of arc BC is 74 degrees, find the measures of all other indicated angles.
We want to find the measure of the angles 1 to 8, given that the diameter is AC and the measure of the Arc BC is 74°.
The angle 5, ∡BOC is central and it is equal to the measure of the arc it intercepts, the arc BC. Thus the angle 5 is 74°.
The angle 4, ∡AOB also is central, and it is equal to the measure of the arc AB. As the line AC is the diameter of the circle O, the arc AC is equal to 180°, and thus, the sum of the angles 4 and 5 will be 180°:
[tex]\begin{gathered} \measuredangle4+\measuredangle5=180^{\circ} \\ \measuredangle4=180^{\circ}-\measuredangle5=180^{\circ}-74^{\circ}=106^{\circ} \end{gathered}[/tex]Thus, the angle 4 is 106°.
The angle 6 is an inscribed angle, and thus it is half of the arc it intersects, the arc AB. This means that the angle 6 is 106°/2=54°.
The angle 2 also is an inscribed angle, half of the arc BC, and thus, the angle 2 is 74°/2=37°.
Now, the triangle BOC has the angles 5, 6 and 7, and the sum of those angles is 180°. This means that:
[tex]\begin{gathered} \measuredangle5+\measuredangle6+\measuredangle7=180^{\circ} \\ 74^{\circ}+54^{\circ}+\measuredangle7=180^{\circ} \\ 128^{\circ}+\measuredangle7=180^{\circ} \\ \measuredangle7=180^{\circ}-128^{\circ}=52^{\circ} \end{gathered}[/tex]Thus, the angle 7 is 52°.
Following a same argument, we can get the angle 8, as being part of the triangle AOB.
[tex]\begin{gathered} \measuredangle2+\measuredangle4+\measuredangle8=180^{\circ} \\ \measuredangle8=180^{\circ}-37^{\circ}-106^{\circ}=37^{\circ} \end{gathered}[/tex]This means that the angle 8 is 37°.
As the line AD is tangent to the circle O, this means that the lines AC and AD are perpendicular, and thus, the angle 1 is 90°.
Lastly, as the angles 1, 2 and 3 are coplanar, their sum is 180°. This is:
[tex]\begin{gathered} \measuredangle1+\measuredangle2+\measuredangle3=180^{\circ} \\ \measuredangle3=180^{\circ}-\measuredangle1-\measuredangle2 \\ \measuredangle3=180^{\circ}-90^{\circ}-37^{\circ}=180^{\circ}-127^{\circ}=53^{\circ} \end{gathered}[/tex]Thus, the angle 3 is 53°.
the volume of a sphere is 2304pi in^3 the radius of the sphere is ___ inches.
Answer:
The radius = 12 inches.
Explanation:
Given a sphere with radius, r units:
[tex]\text{Volume}=\frac{4}{3}\pi r^3[/tex]If the volume of a sphere is 2304π in³, then:
[tex]\frac{4}{3}\pi r^3=2304\pi[/tex]We solve the equation for r:
[tex]\begin{gathered} \frac{4\pi r^3}{3}=2304\pi \\ 4\pi r^3=2304\pi\times3 \\ r^3=\frac{2304\pi\times3}{4\pi} \\ r^3=1728 \end{gathered}[/tex]Next. take cube roots of both sides.
[tex]\begin{gathered} r=\sqrt[3]{1728} \\ r=12\text{ inches} \end{gathered}[/tex]The radius of the sphere is 12 inches.
Solve the inequality |3x+3| + 3 > 15Write the answer in interval notation
Solution:
Given the inequality:
[tex]|3x+3|+3>15[/tex]To solve the inequality,
step 1: Add -3 to both sides of the inequality.
Thus,
[tex]\begin{gathered} |3x+3|+3-3>-3+15 \\ \Rightarrow|3x+3|>12 \end{gathered}[/tex]Step 2: Apply the absolute rule.
According to the absolute rule:
[tex]\mathrm{If}\:|u|\:>\:a,\:a>0\:\mathrm{then}\:u\:<\:-a\:\quad \mathrm{or}\quad \:u\:>\:a[/tex]Thus, from step 1, we have
[tex]\begin{gathered} 3x+3<-12\text{ or 3x+3>12} \\ \end{gathered}[/tex][tex]\begin{gathered} when \\ 3x+3<-12 \\ add\text{ -3 to both sides of the inequality} \\ 3x-3+3<-3-12 \\ \Rightarrow3x<-15 \\ divide\text{ both sides by the coefficient of x, which is 3} \\ \frac{3x}{3}<-\frac{15}{3} \\ \Rightarrow x<-5 \end{gathered}[/tex][tex]\begin{gathered} when \\ 3x+3>12 \\ add\text{ -3 to both sides of the inequality} \\ 3x+3-3>12-3 \\ \Rightarrow3x>9 \\ divide\text{ both sides by the coefficient of x, which is 3} \\ \frac{3x}{3}>\frac{9}{3} \\ \Rightarrow x>3 \end{gathered}[/tex]This implies that
[tex]x<-5\quad \mathrm{or}\quad \:x>3[/tex]Hence, in interval notation, we have:
[tex]\left(-\infty\:,\:-5\right)\cup\left(3,\:\infty\:\right)[/tex]shows four types of polygons which type of polygon shown has no pairs of parallel sides
Accoring to the given figures, the pentagon is the one without parallel sides.
Hence, the answer is Pentagon.Question 17. 4 pts
In 98 years of football, Loudon has averaged 296 points per season and the standard deviation is 14. What percent of the years has Loudon scored between 254 and 338 points per season?
Answer:
Over 98 years, London scored 75.14% per season between 254 and 338 points.
Step-by-step explanation:
Write an equation that represents a vertical shrink by a factor of 1/4 of the graph of g(x)=|x|.
h(x)=?
an equation that represents a vertical shrink by a factor of 1/4 of the graph of g(x)=|x| is y = |x|/4
What is vertical stretch/vertical compression ?
• A vertical stretch is derived if the constant is greater than one while the vertical compression is derived if the constant is between 0 and 1.
• Vertical stretch means that the function is taller as a result of it being stretched while vertical compress is shorter due to it being compressed and is therefore the most appropriate answer.
The y-values are multiplied by a value between 0 and 1, which causes them to travel in the direction of the x-axis. This is known as a vertical shrink and tends to flatten the graph. A point (a,b) on the graph of y=f(x) y = f (x) shifts to a point (a,kb) (a, k b) on the graph of y=kf(x) y = k f (x) in both scenarios.
The function g(x) is defined as |x|.
To vertically shrink the graph of g(x) by a factor of 1/4, divide the function by 4.
g(x) = f(x)/3
f(x) is equal to (|x|)/4.
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