The average height of the Viking's starting lineup is 1.4 inches greater than the average height of the Panther's starting lineup.
The Panthers and the Vikings are competing for the state basketball championship. The data shows the height in inches of the starting lineup for each team.
The heights of the players on the Panther's team are 72, 74, 71, 73, and 75. The heights of the players on the Viking's team are 71, 77, 76, 74, and 74.
Let the average height of the Panther's starting lineup be denoted by the variable "A1".
A1 = (72 + 74 + 71 + 73 + 75)/5
A1 = 365/5
A1 = 73
So, the average height of the Panther's starting lineup is 73 inches.
Let the average height of the Viking's starting lineup be denoted by the variable "A2".
A2 = (71 + 77 + 76 + 74 + 74)/5
A2 = 372/5
A2 = 74.4
So, the average height of the Viking's starting lineup is 74.4 inches.
The difference in the average height is calculated below.
d = A2 - A1
d = 74.4 - 73
d = 1.4
Hence, the average height of the Viking's starting lineup is 1.4 inches greater than the average height of the Panther's starting lineup.
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Suppose that only two factories make Playstation machines. Factory 1 produces 70% of the machines and Factory 2 produces the remaining 30%. Of the machines produced in Factory 1, 2% are defective. Of the machines produced in Factory 2, 5% are defective. What proportion of Playstation machines produced by these two factories are defective? Suppose that you purchase a playstation machine and it is defective. What is the probability that it was produced by Factory 1?
Given:
Factory 1 produces 70%
Factor 2 produces 30%
Defective machines in factory 1 = 2%
Defective machines in factory 2 = 5%
Find-:
What is the probability that it was produced by Factory 1?
Explanation-:
Probability of machines produced by factory1
[tex]\begin{gathered} P(F_1)=70\% \\ \\ P(F_1)=\frac{70}{100} \\ \\ P(F_1)=\frac{7}{10} \\ \end{gathered}[/tex]Probability of machines produced by factory 2
[tex]\begin{gathered} P(F_2)=30\% \\ \\ P(F_2)=\frac{30}{100} \\ \\ P(F_2)=\frac{3}{10} \end{gathered}[/tex]Probability of factory 1 produced defective item,
[tex]\begin{gathered} P(\frac{x}{F_1})=2\% \\ \\ P(\frac{x}{F_1})=\frac{2}{100} \\ \\ P(\frac{x}{F_1})=\frac{1}{50} \end{gathered}[/tex]Probability of factory 2 produced defective item,
[tex]\begin{gathered} P(\frac{x}{F_2})=5\% \\ \\ P(\frac{x}{F_2})=\frac{5}{100} \\ \\ P(\frac{x}{F_2})=\frac{1}{20} \end{gathered}[/tex]So, the probability that randomly selected items was form factor 1.
[tex]P(\frac{F_1}{x})\text{ is}[/tex]Now, apply Bayes theorem is:
[tex]P(\frac{F_1}{x})=\frac{P(F_1)P(\frac{x}{F_1})}{P(F_1)P(\frac{x}{F_1})+P(F_2)P(\frac{x}{F_2})}[/tex]So, the value is:
[tex]\begin{gathered} =\frac{\frac{7}{10}\times\frac{1}{50}}{\frac{7}{10}\times\frac{1}{50}+\frac{3}{10}\times\frac{1}{20}} \\ \\ =\frac{\frac{7}{500}}{\frac{7}{500}+\frac{3}{200}} \\ \\ =\frac{\frac{7}{5}}{\frac{7}{5}+\frac{3}{2}} \\ \\ =\frac{\frac{7}{5}}{\frac{14}{10}+\frac{15}{10}} \\ \\ =\frac{\frac{7}{5}}{\frac{14+15}{10}} \\ \\ =\frac{7}{5}\times\frac{10}{29} \\ \\ =\frac{14}{29} \end{gathered}[/tex]So, the probability is 14/29.
Solve: 6 · x=42What dose x=?
We have to solve this expression.
We can solve it dividing both sides by 6:
[tex]\begin{gathered} 6x=42 \\ \frac{6x}{6}=\frac{42}{6} \\ x=7 \end{gathered}[/tex]Answer: x = 7
Is an irrational number?
In this case a rational number is a number that could be represented as p/q where q is different to 0 in this case pi can't be represete
12What mistake did the student make when solvingtheir two-step equation?(a)b) If correctly solved what should the value of be?
Given the equation:
[tex]\frac{x}{6}+3=-18[/tex](a) You can identify that the student applied the Subtraction Property of Equality by subtraction 3 from both sides of the equation:
[tex]\frac{x}{6}+3-(3)=-18-(3)[/tex]However, the student made a mistake when adding the numbers on the right side.
Since you have two numbers with the same sign on the right side of the equation, you must add them, not subtract them and use the same sign in the result. Then, the steps to add them are:
- Add their Absolute values (their values without the negative sign).
- Write the sum with the negative sign.
Then:
[tex]\frac{x}{6}=-21[/tex](b) The correct procedure is:
1. Apply the Subtraction Property of Equality by subtracting 3 from both sides (as you did in the previous part):
[tex]\begin{gathered} \frac{x}{6}+3-(3)=-18-(3) \\ \\ \frac{x}{6}=-21 \end{gathered}[/tex]2. Apply the Multiplication Property of Equality by multiplying both sides of the equation by 6:
[tex]\begin{gathered} (6)(\frac{x}{6})=(-21)(6) \\ \\ x=-126 \end{gathered}[/tex]Hence, the answers are:
(a) The student made a mistake by adding the numbers -18 and -3:
[tex]-18-3=-15\text{ (False)}[/tex](b) The value of "x" should be:
[tex]x=-126[/tex]Solve the following equation on the interval [0°, 360º). Round answers to the nearest tenth. If there is no solution, indicate "No Solution."2sec^2(x) - 13tan(x) = -13
Given
[tex]2\sec ^2(x)-13\tan (x)=-13[/tex]Add 13 to both sides
[tex]\begin{gathered} 2\sec ^2(x)-13\tan (x)+13=-13+13 \\ 2\sec ^2(x)-13\tan (x)+13=0 \end{gathered}[/tex]We have that
[tex]\sec ^2(x)=1+\tan ^2(x)[/tex]So, substitute in the above equation
[tex]2(1+\tan ^2(x))-13\tan (x)+13=0[/tex]Simplify
[tex]\begin{gathered} 2+2\tan ^2(x)-13\tan (x)+13=0 \\ 15+2\tan ^2(x)-13\tan (x)=0 \end{gathered}[/tex]Reordering the equation
[tex]2\tan ^2(x)-13\tan (x)+15=0[/tex]We get a quadratic equation, then solve by factoring
[tex](2\tan (x)-3)(\tan (x)-5)=0[/tex]Separate the solutions
[tex]\begin{gathered} 2\tan (x)-3=0 \\ 2\tan (x)-3+3=0+3 \\ 2\tan (x)=3 \\ \frac{2\tan (x)}{2}=\frac{3}{2} \\ \tan (x)=\frac{3}{2} \end{gathered}[/tex]And
[tex]\begin{gathered} \tan (x)-5=0 \\ \tan (x)-5+5=0+5 \\ \tan (x)=5 \end{gathered}[/tex]Next, solve for x for each solution
[tex]\begin{gathered} \tan (x)=\frac{3}{2} \\ x=\tan ^{-1}(\frac{3}{2}) \\ x=56.3 \end{gathered}[/tex]And
[tex]\begin{gathered} \tan (x)=5 \\ x=\tan ^{-1}(5) \\ x=78.7 \end{gathered}[/tex]Answer:
x = 56.3° and x = 78.7°
Write down 2 fractions where the denominator of one is a multiple of the denominator of other
The two fractions are 1/3 and 1/6.
What is a fraction?A fraction has two parts: Numerator and Denominator.
It is in the form of a Numerator / Denominator. A fraction is a numerator divided by the denominator.
We need to write 2 fractions where the denominator of one is a multiple of the denominator of the other.
Let's consider the one fraction as;
1/3
Then another one must be multiple of the denominator of the other.
So, 1/6
We see that "the denominator of one is a multiple of the denominator of other".
Thus the two fractions are 1/3 and 1/6.
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find the value of the investment at the end of 5 years
Given: Following details for an amount compounded annually-
[tex]\begin{gathered} P=34900 \\ R=8\% \\ t=5\text{ years} \end{gathered}[/tex]Required: To determine the amount after 5 years.
Explanation: The formula for compound interest is as follows-
[tex]A=P(1+\frac{r}{n})^{\frac{t}{n}}[/tex]Here, A is the amount accrued.
P is the principal amount.
r is the annual rate as a decimal.
t is the time.
n is the number of times interest is compounded in a year.
In this case, the value of n=1 as we are calculating for annual compounding if the interest is compounded semiannually, n=2. For monthly, n=12. Finally, for daily n=365.
Now substituting the values in the formula as-
[tex]\begin{gathered} A=34900(1+0.08)^5 \\ =34900(1.08)^5 \\ =\text{\$}51279.55 \end{gathered}[/tex]Final Answer: Investment after 5 years compounded annually is $51279.55
2. The water level in a reservoir is now 52 meters. Which equation can be used to find the initial depth, d, if this is the water level after a 23% increase? * O 0.23. d = 52 O d = 52 · 0.23 O 1.23. d = 52 O d = 52. 1.23
Answer:
1.23d = 52
Explanation:
If 52 meters is the water level after a 23% increase, then we can say that the initial depth d added to the 23% of d is equal to 52 meters. So:
d + 23%d = 52 meters
Since 23% is equivalent to 0.23, we get:
d + 0.23d = 52
Finally, adding the like terms, we get:
(1 + 0.23)d = 52
1.23d = 52
So, the equation is:
1.23d = 52
7) The point spreads on 12 football games for a season are:1, 3, 14,9,7,3,6, 27, 3, 13, 8, 17.a (3pts) Make a histogram for the data.1-511-1516-2021-2526-30Symmetric6-10b. (2 pts) Describe the distribution of the data, (Circle One)FrequencySkewed RightSkewed Left(2 pts)Which measure of center would be most accurate? (circle one)MeanMedianModeC.d. (2 pts) Which measure of spread would be most accurate? (circle one)RangeInterquartile rangeStandard Deviation
Table of frequencies.
Interval Frequency
1-5 4
6-10 4
11-15 2
16-20 1
21-25 0
26 - 30 1
Therefore the graph would be
B. As we can see from the graph it is skewed left.
C. Since the graph is skewed, the better option would be the median.
D. Since the graph is skewed, the better option would be the interquartile range.
A tank has a capacity of 13 gallons. When it is full, it contains 20% alcohol. How many gallons must be replaced with an 70% alcohol solution to give 13 gallons of 30% solution? Round your final answer to 1 decimal place if necessary.
Given:
A tank has a capacity of 13 gallons. When it is full.
The tank contains 20% alcohol.
We will find the number of gallons that must be replaced with a 70% alcohol solution to give 13 gallons of 30% solution
Let the number of gallons that must be replaced = x
so, there are x gallons with a 70% alcohol and (13 -x) with a 20% alcohol.
So, we can write the following equation:
[tex]70x+20(13-x)=30*13[/tex]Solve the equation to find (x):
[tex]\begin{gathered} 70x+20*13-20x=30*13 \\ 50x+260=390 \\ 50x=390-260 \\ 50x=130 \\ x=\frac{130}{50}=2.6\text{ gallons} \end{gathered}[/tex]So, the answer will be 2.6 gallons
Hi, can you help me answer this question please, thank you!
The t-statistic of the hypothesis is -2.1075 and the P value is 0.04 .
Given that
Sample Size п, = 80 proportion of mean P₁ = 45%
P₁ = 0·45
Sample size п₂ = 40
proportion of mean P₂ = 55%
P₂=0·55
q₁ = 1- P₁=1-0·45 = 0.55
q₂= 1 - P₂ =1-0.55 = 0.45
V₁ = 0.65
Mean= P₁- P₂ = 0.35 -0.55 =-0.20
standard deviation
SE (P₁ P₂) = 0.0949
Test statistic = 0.0949 = P₁- P₂ / SE( P₁- P₂) = -2.1075
t = -2-1075
DF = (N-1)+(N2-1)
Significance level=0.05
CS = 79+39
df = 118
This is a two tailed test for this hypothesis
P = 0.037236
P = 0.037
Hence the t-statistic of the hypothesis is -2.1075 and the P value is 0.037
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In the expansion of (3a + 4b)^8, which of the following are possible variable terms?
Remember the Binomial Theorem:
[tex](a+b)^n\text{ =}\sum_{i\mathop{=}0}^n\begin{bmatrix}{n} & \\ {i} & {}\end{bmatrix}a^{(n\text{ - i})}b^i[/tex]Now, consider the following polynomial:
[tex]\left(3a+4b\right)^8[/tex]Applying the Binomial Theorem, where:
a = 3a
b= 4b
we get:
[tex](3a+4b)^8\text{ =}\sum_{i\mathop{=}0}^8\begin{bmatrix}{8} & \\ {i} & {}\end{bmatrix}3a^{(8\text{ - i})}4b^i[/tex]thus, expanding the sum, we get:
[tex]\begin{gathered} \frac{8!}{0!(8\text{ -0})!}(3a)^8(4b)^0+\frac{8!}{1!(8\text{ -1})!}(3a)^7(4b)^1+\frac{8!}{2!(8\text{-2})!}(3a)^6(4b)^2 \\ +\frac{8!}{3!(8\text{ - 3})!}(3a)^5(4b)^3\text{ + ........+}\frac{8!}{8!(8\text{ -8})!}(3a)^0(4b)^8 \end{gathered}[/tex]Now, simplifying we get:
[tex]\begin{gathered} 6561a^8\text{ + 6998a}^7b\text{ + 326592a}^6b^2+870912a^5b^3+1451520a^4b^4 \\ +1548288a^3b^5+1032192a^2b^6+393216ab^7+65536b^8 \end{gathered}[/tex]then, we can conclude that the correct answer is:
Answer:The variable terms are:
[tex]\begin{gathered} a^8\text{ ,a}^7b\text{ , a}^6b^2,\text{ }a^5b^3,\text{ }a^4b^4 \\ ,\text{ }a^3b^5,\text{ }a^2b^6,\text{ }ab^7\text{ and }b^8 \end{gathered}[/tex]Write a cosine function that Has a midline of 2 an amplitude of 3 and a period of 7pi/4
Given:
Amplitude of cosine function, A=3.
Period, T=7π/4.
Midline, D=2.
The time period can be expressed as:
[tex]T=\frac{2\pi}{B}[/tex]Put T=7π/4 to find the value of B.
[tex]\begin{gathered} \frac{7\pi}{4}=\frac{2\pi}{B} \\ B=\frac{4\times2}{7} \\ =\frac{8}{7} \end{gathered}[/tex]The general cosine function can be expressed as,
[tex]f(x)=A\cos (Bx)+D[/tex]Substitute B=8/7, A=3 and D=2 in above equation.
[tex]f(x)=3\cos (\frac{8}{7}x)+2[/tex]Therefore, the cosine function is,
[tex]f(x)=3\cos (\frac{8}{7}x)+2[/tex]For a certain kind of plaster work, 1.5 cu yd of sand are needed for every 100 sq yd of surface. How much sand will be needed for 350 sq yd of surface?
We are told that we need 1.5 cu yd of sand for every 100 sq yd of surface, then we can express the ratio of sand to surface like this:
[tex]\text{ratio}=\frac{1.5}{100}[/tex]In order to find how much sand we need for 350 sq yd of surface, we just have to multiply 350 by this ratio, then we get:
[tex]350\times\frac{1.5}{100}=5.25[/tex]Then, we need 5.25 cubic yards of sand.
Find the sum of (3x2 + 18x – 7) and (-13x2 + 7x – 11)A –13x3 + 3x2 + 25x – 18B –13x3 + 10x2 + 7x – 7C-13x3 + 10x2 + 18x – 18D -10x2 + 25x – 18
Answer:
The correct option is D, the sum of the given polynomials is
[tex]-10x^2+25x-18[/tex]Explanation:
To find the sum of:
[tex]3x^2+18x-7[/tex]and
[tex]-13x^2+7x-11[/tex]We write:
[tex]\begin{gathered} (3x^2+18x-7)+(-13x^2+7x-11) \\ =3x^2+18x-7-13x^2+7x-11 \end{gathered}[/tex]Collect like terms:
[tex]\begin{gathered} 3x^2-13x^2+18x+7x-7-11 \\ =-10x^2+25x-18 \end{gathered}[/tex]how much money will be in Devon's retirement account if she continues to make the same monthly investment for 40 years
Annuities
It refers to a special form to accumulate interest over a regular payment or cash flow (C) per period.
Devon decides to save money for her retirement by depositing C=$524 each month in an account that is expected to earn interest with an APR of r=5.25% compounded monthly.
We will calculate the future value (FV) of her investment over a period of n=40 years.
The future value can be calculated with the formula:
[tex]FV=C\cdot\frac{(1+i)^n-1}{i}[/tex]Where i is the interest rate adjusted for the compounding period. Since there are 12 months in one year:
[tex]i=\frac{r}{12}=\frac{0.0525}{12}=0.004375[/tex]The number of periods is also adjusted for monthly compounding:
n = 40*12 = 480
Now apply the formula:
[tex]FV=524\cdot\frac{(1+0.004375)^{480}-1}{0.004375}[/tex]Calculating:
[tex]\begin{gathered} FV=524\cdot1,629.45 \\ FV=853,832.69 \end{gathered}[/tex]There will be $853,832.69 in Devon's retirement account in 40 years
Flex Gym charges a membership fee of $150.00 plus $41.00 per month to join the gym. Able gym charges a membership fee of $120.00 plus $46.00 per month. Find the number of months for which you would pay the same total fee to both gyms.
We have to write an equation for each gym of the cost as a function of the months, so:
[tex]\begin{gathered} We\text{ call c=the total cost and m=months.} \\ \text{For Flex Gym:} \\ c_F=41\cdot m+150 \\ \text{For Able Gym}\colon \\ c_A=46\cdot m+120 \end{gathered}[/tex]Now, we want to find the number of months at which the both gym have the same cost, so:
[tex]\begin{gathered} c_F=c_A \\ 41\cdot m+150=46\cdot m+120 \\ 150-120=46\cdot m-41\cdot m \\ 30=5\cdot m \\ m=\frac{30}{5}=6 \end{gathered}[/tex]At 6 months the cost of the both gyms is the same.
The plot below represents the function f ( x ) : 1 2 3 4 5 -1 -2 -3 -4 -5 1 2 3 4 5 -1 -2 -3 -4 -5 Evaluate f ( 3 ) : f ( 3 ) =
Solution
The function represented by the graph is
The root of the equation are -0.5 , 1.5
[tex]\begin{gathered} x=-0.5,x=2.5 \\ (x+0.5)(x-2.5) \\ x^2-2.5x+0.5x-1.25 \\ x^2-2x-1.25 \end{gathered}[/tex]Therefore the function of x =
[tex]\begin{gathered} f(x)=x^2-2x-1.25 \\ f(3)=3^2-2(3)-1.25 \\ f(3)=9-6-1.25 \\ f(3)=1.75 \end{gathered}[/tex]Hence the correct value of f(3) = 1.75
995
× 55 ?? What’s the partial product of this?
Find the solutions to the following quadric equation 2Xsquared -1x-2=0
Given the quadratic equation:
[tex]2x{}^2-1x-2=0[/tex]We can use the general solution for the quadratic equation ax² + bx + c = 0:
[tex]x=\frac{-b\pm\sqrt{b{}^2-4ac}}{2a}[/tex]From the problem, we identify:
[tex]\begin{gathered} a=2 \\ b=-1 \\ c=-2 \end{gathered}[/tex]Finally, using the general solution:
[tex]\begin{gathered} x=\frac{-(-1)\pm\sqrt{(-1)^2-4(2)(-2)}}{2\cdot2}=\frac{1\pm\sqrt{1+16}}{4} \\ \\ \therefore x=\frac{1\pm\sqrt{17}}{4} \end{gathered}[/tex]Solve the following equation for "b".b/3 = M
In order to solve an equation for a variable we need to isolate it on the left side. In this case we want to find the value of "b", therefore we must perform operations in such a way that it will be the only thing on the left side of the equation. To do so we need to switch the operation of each term we don't want to be on the left side, this means that if a term is adding it should go to the right side subtracting and if it is multiplying it should go dividing. In this case there is only one term that is dividing "b", so it should go to the right side by multiplying. With this in mind lets solve the problem:
[tex]\begin{gathered} \frac{b}{3}\text{ = M} \\ b\text{ = 3}\cdot M \end{gathered}[/tex]graph the system of linear inequalities.x + 2y ≥ 2-x + y ≤ 0
INFORMATION:
We have the next system of equations
[tex]\begin{gathered} x+2y\ge2 \\ -x+y\leq0 \end{gathered}[/tex]And we must graph it
STEP BY STEP EXPLANATION:
To graph the system, we need to graph first the two inequalities as equations. So, we would have
[tex]\begin{gathered} x+2y=2 \\ -x+y=0 \end{gathered}[/tex]- x + 2y = 2:
To graph it, we can find the x and y intercepts.
x intercept:
To find it, we need to replace y = 0, and solve for x
[tex]\begin{gathered} x+2(0)=2 \\ x=2 \end{gathered}[/tex]y intercept:
To find it, we need to replace x = 0, and solve for y
[tex]\begin{gathered} 0+2y=2 \\ y=1 \end{gathered}[/tex]So, the graph would be a line that passes through the points (2, 0) and (0, 1).
Since the symbol of this inequality is ≥, the graph would be the values that are on the line and above it.
- -x + y = 0:
To graph it, we can rewrite the equation as
[tex]y=x[/tex]And this is the identity line.
So, since the symbol of this inequality is ≤, the graph would be the identity line and the values below it.
Finally, the graph of the system would be the common part of the graph of each inequality
So, the graph of the system is the part colored in red and blue at the same time
ANSWER:
Paisley is going to invest in an account paying an interestrate of 34% compounded daily. How much would Paisleyneed to invest, to the nearest dollar, for the value of theaccount to reach $400 in 16 years?
Answer:
$2
Explanation:
To solve the given problem, we'll use the below compound interest formula;
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]where A = future amount = $400
P = the initial amount( principal)
r = annual interest rate in decimal form = 34/100 = 0.34
n = number of compounding periods in a year = 365
t = time in years = 16
Let's go ahead and substitute the above values into our formula and solve for P;
[tex]\begin{gathered} 400=P(1+\frac{0.34}{365})^{365\times16} \\ 400=P(1.0009)^{5840} \\ 400=229.86P \\ P=\frac{400}{229.86} \\ \therefore P=2\text{ dollars} \end{gathered}[/tex]Margo borrows $1400, agreeing to pay it back with 6% annual interest after 16 months. How much interest will she pay?
Given,
The principal amount is $1400.
The rate of interest is 6%.
The time period is 16 months.
Required
The interest paid by Morrow.
The simple interest is calculated as,
[tex]Simple\text{ interest=}\frac{P\times R\times T}{100}[/tex]Substituting the values then,
[tex]\begin{gathered} S.I=\frac{1400\times6\times16}{100\times12} \\ S.I=14\times2\times4 \\ S.I=56\times2 \\ S.I=112 \end{gathered}[/tex]Hence, the interest she will pay is $112.
Jody invested $4400 less in account paying 4% simple interest than she did in an account paying 3 percent simple interest. At the end of the first year, the total interest from both accounts was $592. find the amount invested in each account
The rule of the simple interest is
[tex]I=P\times R\times T[/tex]P is the initial amount
R is the rate in decimal
T is the time
Assume that she invested $x in the account that paid 3% simple interest
then she invested x - 4400 dollars in the account that paid 4% simple interest
Then let us find each interest, then add them, equate the sum by 592
[tex]\begin{gathered} P1=x-4400 \\ R1=\frac{4}{100}=0.04 \\ T1=1 \\ I1=(x-4400)\times0.04\times1 \end{gathered}[/tex]Let us simplify it
[tex]\begin{gathered} I1=0.04(x)-0.04(4400) \\ I1=0.04x-176 \end{gathered}[/tex][tex]\begin{gathered} P2=x \\ R2=\frac{3}{100}=0.03 \\ T2=1 \\ I2=x\times0.03\times1 \\ I2=0.03x \end{gathered}[/tex]Since the total interest is $592, then
[tex]\begin{gathered} I1+I2=592 \\ 0.04x-176+0.03x=592 \end{gathered}[/tex]Add the like terms on the left side
[tex]\begin{gathered} (0.04x+0.03x)-176=592 \\ 0.07x-176=592 \end{gathered}[/tex]Add 176 to both sides
[tex]\begin{gathered} 0.07x-176+176=592+176 \\ 0.07x=768 \end{gathered}[/tex]Divide both sides by 0.07 to find x
[tex]\begin{gathered} \frac{0.07x}{0.07}=\frac{768}{0.07} \\ x=10971.42857 \end{gathered}[/tex]Then She invested about 10971 dollars in the account of 3%
Since 10971 - 4400 = 6571
Then she invested about
if FE measures 20 centimeters, the approximate area of circle B is what
If FE measures 20 cm, then the area is 314 cm², if BE measure 3.5 cm, then the area is 38.5 cm², if AB measures 11 cm, then the area is 380 cm² and is EF measures 12 cm, then the area is 113 cm².
Area of a circle:
A = π r²
r = 1 /2 of diameter.
FE is the diameter
r = 20 / 2
r = 10 cm
Area of circle using FE:
A = π ( 10 )² = π × 100 = 314 cm²
BE is a radius:
Area = π × 3.5² = π × 12.25 = 38.465
A = 38.5 cm²
AB is a radius:
Area = π × 11²
A = π × 121
A = 379.94 = 380 cm²
EF is a diameter:
r = 12 / 2 = 6 cm
Area = π × 6²
A = π × 36
A = 113.04 = 113 cm²
Therefore, if FE measures 20 cm, then the area is 314 cm², if BE measure 3.5 cm, then the area is 38.5 cm², if AB measures 11 cm, then the area is 380 cm² and is EF measures 12 cm, then the area is 113 cm².
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Your question was incomplete, Please refer the content below:
1) if FE measures 20 centimeters, the approximate area of circle B is what?
2) if BE measures 3.5 centimeters, the approximate area of circle B is what?
3) if AB measures eleven centimeters, the approximate area of circle B is what
4) If EF measures twelve centimeters, the approximate area of circle B is what?
Suppose an auto racer won a 400 mile race with a time of 1:48:51. At one point the racer was 50 miles closer to the finish than the start. How far had the racer gone at that point?How far from the start was the racer? __ miles
To solve this problem, let's use the variables x and y to represent the distance of the racer to the start and to the finish, respectively.
If the total distance of the race is 400 miles, we have that the distance traveled by the racer until now (x) plus the distance he needs to travel to finish the race (y) is 400:
[tex]x+y=400[/tex]Also, at one point the racer was 50 miles closer to the finish than the start, so at that point we have that he is farther away from the beginning (that is, x is 50 units bigger than y):
[tex]x=y+50[/tex]Now, using this value of x in the first equation, we have:
[tex]\begin{gathered} (y+50)+y=400 \\ 2y+50=400 \\ 2y=350 \\ y=\frac{350}{2}=175 \end{gathered}[/tex]Now, finding the value of x, we have:
[tex]x=y+50\to x=175+50\to x=225[/tex]So the racer was 225 miles far from the start.
Chuck's age is five years less than twice Larry's age. If Chuck's age is 150% of Larry's age, then what is Larry's age, in years?A. 6B. 8C. 10D. 15
Answer:
Larry's age is 10 years
Explanation:
Let Chuck's age be c
Let Larry's age be L
Chuck's age is five years less than twice Larry's age
Mathematically:
[tex]c\text{ = 2l-5}[/tex]Chuck's age is 150% of Larry's age
What this mean is that Chuck's age is 1.5 times multiplied by Larry's age
Mathematically, we have this as:
[tex]c\text{ = 1.5l}[/tex]Now, we can proceed to equate the two equations as follows:
[tex]\begin{gathered} 2l-5\text{ = 1.5l} \\ 2l-1.5l\text{ = 5} \\ 0.5l\text{ = 5} \\ l\text{ = }\frac{5}{0.5} \\ l\text{ = 10 } \end{gathered}[/tex]Round the number. Write the result as a product of a single digit and a power of 10 0.00063718
EXPLANATION
Given the number 0.00063718, rounding and writting as a product of a single digit and a power of 10 give us:
6x10^-4
At the Dollar Spot, Carl bought pencils for $3.75, sharpies for $5 69, and glue sticks for ? 1. In the box below type which operation you would use: Division Addition Subtraction Multiplication 2. Why did you pick this operation?
Given that Carl bought pencils for $3.75, sharpies for $5 69, and glue sticks for
Although the question didn't give the value for glue sticks, the operation you would use here is Addition.
Addition symbol: +
2. I picked addition because to find the total amount Carl spent at the Dollar spot, you will need to add the amount he spent on pencils, sharpies and glue together.