A car company charges $42.50 per day to rent a car in 10 since for every mile driven Lydia wants to rent a car knowing that she plans to drive 225 miles and she has at most a $150 to spend
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Explanation:
d: number of days.
m: number of miles.
The total cost of the car 'y' is:
[tex]y=42.5d+0.1m[/tex]Lydia plans to drive 225 miles, this is m = 225. And she has $150 to spend, so we have to solve y ≤ 150.
The inequality to solve is:
[tex]\begin{gathered} y\le150 \\ 42.5d+0.1\cdot225\le150 \\ 42.5d+22.5\le150 \end{gathered}[/tex]Solving the inequality:
[tex]\begin{gathered} 42.5d+22.5\le150 \\ 42.5d\le150-22.5 \\ 42.5d\le127.5 \\ d\le\frac{127.5}{42.5} \\ d\le3 \end{gathered}[/tex]Lydia can afford 3 days at most.
Answers:
• Inequality: ,42.5d + 22.5 ≤ 150
,• d ≤ 3
Related Questions
Hi, can you help me answer this question please, thank you!
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We are asked to determine the test statistic for two populations. To do that we will use the following formula:
[tex]z=\frac{\bar{x_2}-\bar{x_1}}{\sqrt[]{\frac{SD^2_2}{n_2}+\frac{SD^2_1}{n_1}}}[/tex]Where:
[tex]\begin{gathered} \bar{x_1},\bar{x_2}=\text{ population means} \\ SD_1,SD_2=\text{ standard deviations} \\ n_1,n_2=\text{ population sizes} \end{gathered}[/tex]Substituting the values we get:
[tex]z=\frac{83.3_{}-75.4}{\sqrt[]{\frac{(17.8)^2_{}}{19}+\frac{(9.7)^2_{}}{12}}}[/tex]Solving the operations we get:
[tex]z=1.596[/tex]Therefore, the test statistic is 1.596.
To determine the P-value we will determine the probability that the test statistic is less than the value we determined. This is:
[tex]p-\text{value}=P(z<1.596)[/tex]The value of the probability we find it in the z-table using the value z = 1.596, we get:
[tex]p-\text{value}=0.9441[/tex]Therefore, the p-value is 0.9441.
✓ 2 ✓ 3 ✓4 Find the missing side length. Assume that all intersecting sides meet at right angles. Be sure to include the correct unit in your answer. 16 A 9 A 13 A Check PLUS EGEE E Com. Tune here to
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the given diagram:
We have to find the value of DE
Since all the lines intersect at right angle
[tex]\begin{gathered} \text{Here, AC + DE = BF} \\ 5\text{ + DE =13} \\ DE=13-5 \\ DE=8\text{ ft} \end{gathered}[/tex]Answer : DE = 8ft
Use the Pythagorean theorem to find the length of this segment.
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To find the length of the segment use the Pythagoras theorem below:
[tex]AC^2=AB^2+BC^2[/tex]From the graph, we have:
AB = 15 - 3 = 12
BC = 7 - 2 = 5
Let's solve for the length of the segment AC:
[tex]\begin{gathered} AC^2=12^2+5^2 \\ \\ AC^2=144+25 \\ \\ AC^2=169 \end{gathered}[/tex]Take the square-root of both sides:
[tex]\begin{gathered} \sqrt[]{AC^2}=\sqrt[]{169} \\ \\ AC\text{ = 13} \end{gathered}[/tex]Therefore, the length of this segment is 13
ANSWER:
13
Which values are solutions to the inequality below? Check all that apply. x^2
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Answer:
11, 7, and -8
I will show you a pic
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Answer
The red line represents y = x + 1
The blue line represents y = 2x - 7
We can see that the two lines and equation intersect at (8, 9)
Solution
x = 8
y = 9
Explanation
The two equations are
y = x + 1
y = 2x - 7
To solve this graphically, we will plot the two equations on the same graph and the solution will exist at the point where the two lines meet.
To plot the lines for each of these equations, we will use intercepts to obtain two points on each line and then connect the two points to get each of the lines.
y = x + 1
when x = 0
y = x + 1
y = 0 + 1
y = 1
First point on the line is (0, 1)
when y = 0
y = x + 1
0 = x + 1
x = -1
Second point on the line is (-1, 0)
The two points are (0, 1) and (-1, 0)
y = 2x - 7
when x = 0
y = 2x - 7
y = 2(0) - 7
y = 0 - 7
y = -7
First point on the line is (0, -7)
when y = 0
y = 2x - 7
0 = 2x - 7
2x = - 7
Divide both sides by 2
(2x/2) = (-7/2)
x = -3.5
Second point on the line is (-3.5, 0)
The two points are (0, -7) and (-3.5, 0)
The graph of this question will now be presented under answer and the point of intersection will bw evident.
Hope this Helps!!!
RIn this figure, sin ZQOP = cos 2and cos ZROQ = sin 2
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So, Sin QOP = QP/OQ = OPPOSITE/ Hyp
Then, cos PQO = QP/OQ = Adjacent/ Hyp
Using the same logic, Cos ROQ = RO/OQ = Adj/Hyp
Then, sin QPO = RO/OQ = Opp/Hyp
What E is. if E=2 when W=15, find E when W=10
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Since the elongation E varies directly with the weight W, they are related as follows
[tex]E=kW[/tex]where k is the constant of proportionality. In order to find k, we can substitute the given values, that is, when E=2, W=15, then we have
[tex]2=k\cdot15[/tex]Then, k is given as
[tex]k=\frac{2}{15}[/tex]Therefore, our formula for any E and W is
[tex]E=\frac{2}{15}W[/tex]Now, in order to find E in the second case, by replacing W=10, we get
[tex]E=\frac{2}{15}(10)[/tex]which yields
[tex]E=\frac{20}{15}=\frac{4}{3}[/tex]Therefore, the answer is
[tex]E=\frac{4}{3}[/tex]The expression(Toys =)equalsy"zsztwhere1, the exponent of y, is:s, the exponent of %, is:t, the exponent of X, is:Next Question
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The given expression can be simplified as follows:
[tex]\frac{(x^4y^5z^5)^6}{(x^5y^5z)^5}=\frac{x^{24}y^{30}z^{30}}{x^{25}y^{25}z^5}=x^{-1}y^5z^{25}=\frac{y^5z^{25}}{x}[/tex]Compare with the given term as follows:
[tex]\frac{y^5z^{25}}{x}=\frac{y^rz^s}{x^t}[/tex]Therefore r, the exponent of y is 5
s, the exponent of z is 25
t, the exponent of x is 1.
simplify: 9x^4-27x^6/3x^3
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simplify: 9x^4-27x^6/3x^3
we have the expression
[tex]\begin{gathered} 9x^4-\frac{27x^6}{3x^3} \\ \\ 9x^4-9x^{(6-3)} \\ 9x^4-9x^3 \end{gathered}[/tex]we have the expression
[tex]\begin{gathered} \frac{9x^4-27x^6}{3x^3} \\ \frac{9x^4}{3x^3}-\frac{27x^6}{3x^3} \\ 3x-9x^3 \end{gathered}[/tex]this is the answer
Target is having a sale on bath towels usually bath towels cost $15 today I paid only $12 what percent of the original price did I pay for the bath towels
Answers
80%
1) Since the prices on Target dropped from $15 to $12 to find the equivalent percentage of $12 in comparison to $15 we need to write down the following ratio:
[tex]\begin{gathered} 15----100\% \\ 12----x\% \\ \frac{15}{12}=\frac{100}{x} \\ 15x=12\cdot100 \\ \frac{15x}{15}=\frac{1200}{15} \\ x=80\% \end{gathered}[/tex]Note that we cross multiplied that ratio and then on the second step we have divided both sides by 15.
2) Hence, I paid 80% of the original price that day
Find the zeros of each function by using a graph and a table. f(x)=x^2+2x–24.
Answers
Explanation
Step 1
[tex]f(x)=x^2+2x-24[/tex]Zeros
A(-6,0) B(4,0)
because
[tex]\begin{gathered} f(x)=x^2+2x-24 \\ f(-6)=(-6)^2+2(-6)-24=36-12-24=0 \\ f(4)=4^2+2\cdot4-24=16+8-24=0 \end{gathered}[/tex]Step 2
table
[tex]\begin{gathered} (-6,0) \\ (4,0) \\ f(-1)=-1^2+2\cdot-1-24=1-2-24=-25 \\ (1,-25) \\ \end{gathered}[/tex]I hope this helps you
Paxton did the following multiplication problem. Where should he put the decimal point in his product? 9.18 % 73 2754 64260 67014 6.701.4 67.014 6.7014 670.14
Answers
We have the multiplication of 9.18 * 7.3 and want to know where the decimal point has to be.
In the multiplication we have to sum the decimal position of the numbers and taht is the position of the decimal point of the result.
In this case, the decimal point of 9.18 is 2 places and for 7.3 is one place, so the final result will have 2+1=3 places. The decimal point has to be 3 places.
The number is 67014 and the decimal point is three places from rigth, 67.014
At the end of the winter, coats are on sale for 75% off. Question: a. If a heavy coat was priced at $ 160, then how much money will you save since it is on sale?
Answers
$120
Explanation
remember
[tex]75\text{ \% =}\frac{75}{100}=0.75[/tex]it means you can find 75 % of any value, just by doing th product of the number and 0.75
[tex]\begin{gathered} 160\cdot0.75=120 \\ \end{gathered}[/tex]Hence, 75% is $120 ,
the discount is 120,the new price is 40
the money you save is the difference, it is 120
I hope this helps you
Kinsley measured a city park and made a scale drawing the scale of the drawing was 13 millimeters and 5 meters if the actual width of the soccer field is 65 meters how wide is the field in millimeters
Answers
Kinsley measured a city park
Scale of drawing : 13 milimeter and 5 meter
Actual width is 65 meter
Let the field in Milimeter is x
So,
[tex]\begin{gathered} 13\text{ milimeters and 5 meters=x milimeters and 65 meters} \\ \frac{13}{5}=\frac{x}{65} \\ \text{Apply cross multiplication} \\ x=\frac{13\times65}{5} \\ x=\frac{13\times13}{1} \\ x=169 \\ Park\text{in milimeters = 169} \\ So, \\ 13\text{ milimeters and 5 meters=169 milimeters and 65 meters} \end{gathered}[/tex]Answer: 13 milimeters and 5 meters = 169 milimeters and 65 meters
I have this question and I can’t figure it out
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Hello!
First, let's remember about the integers numbers.
These numbers can be positive or negative (and include the number 0). The main characteristic is that these numbers don't have a decimal part.
Knowing it, we can say that are integers:
• -1,
,• 0,
,• 2,
,• -2.
In the number line, we'll have:
Can you pls help me with this question thank you
Answers
The expression we have is:
[tex]2x+5[/tex]In this expression, we have two terms: 2x and 5.
The elements in these terms are variables, coefficients, and constants.
A variable is a letter that can take different values, a coefficient is a number that accompanies the variable, and a constant is a number that does not have any variable next to it (its value will not change).
In this case:
2 is the coefficient,
x is the variable,
and 5 is the constant.
Answer: a) 5
X-4-3-2-1012f(x)-6-4-1-2-5-8-16Which is a possible turning point for the continuousfunction f(x)?O(-3,-4)O (-2,-1)O (0,-5)O (1,-8)
Answers
Solution
[tex](-2,-1)[/tex]The final answer
Option B
When Ruby works out, she spends 2 minutes stretching for every 15 minutes of exercise. If Ruby spends 15 minutes stretching, how long did she spend exercising?
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The ratio of time spend for stretching to time spend for exercise remain same.
Equate the ratio of time spend for stretching to time spend for exercise in both cases.
[tex]\begin{gathered} \frac{2}{15}=\frac{15}{x} \\ x=\frac{15\cdot15}{2} \\ =112.5 \end{gathered}[/tex]So Ruby spend 112 and a half minute to spend 15 minutes in stretching.
So answer is 112.5 min or
[tex]112\frac{1}{2}[/tex]The pollution of Linton is 12 times as great as a pollution of Ellmore. The combine population of both sounds is 9,646 people. What is the population of Linton?
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Explanation
[tex]\begin{gathered} 12x+x=9646 \\ 13x=9646 \\ x=\frac{9646}{13} \\ x=742 \end{gathered}[/tex]The population of Linton is 742*12=8904
Answer
8904
3(t - 24) = 8t - 4(t + 15)
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We need to solve the equation:
[tex]3(t-24)=8t-4(t+15)[/tex]Then:
[tex]\begin{gathered} 3(t-24)=8t-4(t+15) \\ 3t-72=8t-4t-60 \\ 3t-72=4t-60 \\ 3t-4t=-60+72 \\ -t=12 \\ t=-12 \end{gathered}[/tex]Therefore, t=-12.
Select all of the ordered pairs that are solutions to the equation y= -6x + 7
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We must substitute each pair into our equation:
A. If we substitute point (1,1), we have
[tex]\begin{gathered} 1=-6(1)+7 \\ 1=-6+7 \\ 1=1 \end{gathered}[/tex]then, this point is a solution.
B. If we substitute point (-1,1), we have
[tex]\begin{gathered} 1=-6(-1)+7 \\ 1=6+7 \\ 1=13\text{ !!!} \end{gathered}[/tex]this means that this point is not a solution.
C. If we substitute point (-6,7), we have
[tex]\begin{gathered} 7=-6(-6)+7 \\ 7=36+7 \\ 7=43\text{ !!!} \end{gathered}[/tex]this means that this point is not a solution.
D. If we substitute point (3,-11), we have
[tex]\begin{gathered} -11=-6(3)+7 \\ -11=-18+7 \\ -11=-11 \end{gathered}[/tex]then, this point is a solution.
Therefore, the solutions are option A and D
For the rectangle shown below, which can be used to find the value of x?A. 3^2 + x^2 = 15^2B. (x + 3)^2 = 15^2C. 3^2 + 15^2 = x^2D. 3 + 15 = x^2
Answers
Since the rectangle is composed of 2 right triangles, we can apply the Pythagorean theorem to one of the triangles:
c^2 = a^2 +b^2
Where:
c= hypotenuse (longest side) = 15
a & b = the other two legs of the triangle ( x , 3 )
Replacing:
15^2 = 3^2 + x ^2
Solve for x
225 = 9 + x^2
225-9 = x^2
216 = x^2
√216 = x
x= 14.69
So, to find the value of x, we can use:
A. 3^2 + x^2 = 15^2
Volume of the box with the cone shape cut out of it. What are the side lengths of the box and what is the volume of the box/cube only?
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Answer:
Side length of the box: 6 cm
Volume of the box/cube: 216 cm³
Volume of the box without cone: 190.88 cm³
Explanation:
The sides of a cube are all equal, so in this case, the side length of the box is 6 cm.
Then, the volume can be calculated as
Volume = side x side x side
Volume = 6 cm x 6 cm x 6 cm
Volume = 216 cm³
To know the volume of the box with the cone shape cut of it, we need to calculate the volume of the cone with the following equation
[tex]Volume=\frac{1}{3}\pi r^2h[/tex]Where π = 3.14, r is the radius and h is the height. The diameter of the cone is 4 cm, so the radius is
r = 4 cm/2 = 2 cm
Then, replacing r = 2 cm and h = 6 cm, we get
[tex]\begin{gathered} Volume=\frac{1}{3}(3.14)(2\text{ cm\rparen}^2(6\text{ cm\rparen} \\ Volume=\frac{1}{3}(3.14)(4\text{ cm}^2)(6\text{ cm\rparen} \\ Volume=25.12\text{ cm}^3 \end{gathered}[/tex]Now, the volume of the box without the cone shape is
V = 216 cm³ - 25.12 cm³
V = 190.88 cm³
So, the answers are
Side length of the box: 6 cm
Volume of the box/cube: 216 cm³
Volume of the box without cone: 190.88 cm³
Photo attached. Total cost as function of x = Domain of total cost function =
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1. Identify the given from the statements.
• Let the side of the square base= s
• Height of the rectangular box be= h
,• Given that :
s^2h = 20ft^3
solving for sh: s^2h = 20
s^2=20/h
s*s/s = 20/h /s
s = 20/hs
• Therefore sh = 20/s
Expandingthe given statements , we learn that :
• Material cost for base per square foot = 20 cents
,• Material cost for sides per square foot = 18 cents
,• Material cost for the top per square foot = 14 cents
2. Calculate Total cost
Total cost = {(s^2*20) + 4(sh)*18 +s^2*14}cents
= 20s^2 + 72*20/s + 14s^2
=34s^2 + 1440/s
• Expressing Total cost in terms of x , let s = x .
• Then Total cost (x) = 34x^2 +1440/x
2. Calculating the domain of TC(x) =34x^2 +1440/x
x <0 or x> 0
Domain (-∞;0) U(0;∞)
7. Find the value of x in the figure below. Justify your answer. 4 pts 20° (x - 15)° X = Reason:
Answers
Answer
x = 85°
Explanation
The image of this question shows that the two angles 20° and (x - 15)° both sum up to give a right angle (90°). So,
20° + (x - 15)° = 90°
20° + x - 15° = 90°
x° + 20° - 15° = 90°
x + 5° = 90°
x = 90° - 5°
x = 85°
Hope this Helps!!!
Arlene buys a phone case and charging cord for 15% off. The original cost of the phone case is $18. Her total discount is $4.20.Write and solve an equation to find the original price of the charging cord
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The total discount in Arlene's purchase was 4.20, this includes the discount applied over both, the case and the charging cord. We already know the original price of the phone case, the discount applied and the value of the discount, use it to find the original price of the chargind cord:
[tex]\begin{gathered} 0.15(18+x)=4.20 \\ 18+x=\frac{4.20}{0.15} \\ 18+x=28 \\ x=28-18 \\ x=10 \end{gathered}[/tex]The original price of the charging cord is $10
Write an equation for a line perpendicular to y=-5x-3 and passing through the point (15,4)
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1) We need to consider the fact that perpendicular lines described by linear functions have opposite and reciprocal slopes when compared to the original linear function.
2) So we can state that the perpendicular line to y=-5x-3 has a slope of :
[tex]m=\frac{1}{5}[/tex]3) Now, the next step is to plug into the Slope-intercept form the following point (15,4) and then find the y-intercept:
[tex]\begin{gathered} y=mx+b \\ 4=\frac{1}{5}(15)+b \\ 4=3+b \\ 4-3=b \\ b=1 \end{gathered}[/tex]4) Thus, the equation of a perpendicular line to the line described by the linear function y=-5x-3 is:
[tex]y=\frac{1}{5}x+1[/tex]
1/4+3/8 in simplest term
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Rewrite barrel as a unit rate. hour O A. barrel/hour B. 10 barrels/hour O C. To barrel/hour OD. 1 barrels/hour
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Then the correct answer is A. 5/8 barrel/hour
A biologist records the number and types of fish caught in a local lake during a 2-yearperiod. The biologist reports that 796 of the fish caught during this period were trout,whereas 43% of the fish caught were bass. These reports of the number of trout andbass at this lake are examples ofcumulative frequencies.percentile ranks.relative frequencies.smooth curves.
Answers
Record for the number and types of fish caught in a local lake during a 2-year
period
Fish caught during this period were trout = 796
Fish caught during this period were bass = 43%
Total percentile = 100%
Fish caught during this period were trout = 100 - percentage of fish caught during this period were bass
fish caught during this period were trout = 100 - 43 = 57%
Hence the reports of the number of trout and
bass at this lake are examples of percentile ranks