Answer:
n= (z)22E2
n=10× 99%÷ 0.07
There are two rectangles Jared is examining. He knows the width of the first rectangle measures
2.48 cm, and the length is twice its width.
Jared also knows that the width of the second rectangle is equal to the length of the first rectangle,
and that the area of the second rectangle is 9.92. Given this information, find the length of the
second rectangle for Jared.
1st rectangle:
width: 2.48cm
length: 4.96
2nd rectangle:
width: 4.96 (equals to the length of the 1st rectangle)
area: 9.92
length: 9.92/4.96 = 2
2/8 of a rope is 28 meters.What is the length of the rope? A.32 B.42 C.4 D.21
let length be x
ATQ
[tex]\\ \sf\longmapsto \dfrac{2}{8}\times x=28[/tex]
[tex]\\ \sf\longmapsto \dfrac{2x}{8}=28[/tex]
[tex]\\ \sf\longmapsto \dfrac{x}{4}=28[/tex]
[tex]\\ \sf\longmapsto x=4(28)[/tex]
[tex]\\ \sf\longmapsto x=112[/tex]
Step-by-step explanation:
there is something wrong with your problem description.
the offered answer options do not fit to the solution as it is described.
2/8th of a rope is 28 meters long. how long is the whole rope ?
as the other answer said : 2/8 = 1/4
and 1/4th of the rope x = 28 m
1/4 × x = 28
x (the length of the whole rope) = 4×28 = 112 meters
but - maybe the original problem said that 7/8th (and not 2/8th) of a rope is 28 m.
7/8 × x = 28
1/8 × x = 4
x = 32 m
then A (32) would be the right answer !
power sharing helps the ruling party to retain power for a long time. tick or wrong
[infinity]
Substitute y(x)= Σ 2 anx^n and the Maclaurin series for 6 sin3x into y' - 2xy = 6 sin 3x and equate the coefficients of like powers of x on both sides of the equation to n= 0. Find the first four nonzero terms in a power series expansion about x = 0 of a general
n=0
solution to the differential equation.
У(Ñ)= ___________
Recall that
[tex]\sin(x)=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}[/tex]
Differentiating the power series series for y(x) gives the series for y'(x) :
[tex]y(x)=\displaystyle\sum_{n=0}^\infty a_nx^n \implies y'(x)=\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty (n+1)a_{n+1}x^n[/tex]
Now, replace everything in the DE with the corresponding power series:
[tex]y'-2xy = 6\sin(3x) \implies[/tex]
[tex]\displaystyle\sum_{n=0}^\infty (n+1)a_{n+1}x^n - 2\sum_{n=0}^\infty a_nx^{n+1} = 6\sum_{n=0}^\infty(-1)^n\frac{(3x)^{2n+1}}{(2n+1)!}[/tex]
The series on the right side has no even-degree terms, so if we split up the even- and odd-indexed terms on the left side, the even-indexed [tex](n=2k)[/tex] series should vanish and only the odd-indexed [tex](n=2k+1)[/tex] terms would remain.
Split up both series on the left into even- and odd-indexed series:
[tex]y'(x) = \displaystyle \sum_{k=0}^\infty (2k+1)a_{2k+1}x^{2k} + \sum_{k=0}^\infty (2k+2)a_{2k+2}x^{2k+1}[/tex]
[tex]-2xy(x) = \displaystyle -2\left(\sum_{k=0}^\infty a_{2k}x^{2k+1} + \sum_{k=0}^\infty a_{2k+1}x^{2k+2}\right)[/tex]
Next, we want to condense the even and odd series:
• Even:
[tex]\displaystyle \sum_{k=0}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=0}^\infty a_{2k+1}x^{2k+2}[/tex]
[tex]=\displaystyle \sum_{k=0}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=0}^\infty a_{2k+1}x^{2(k+1)}[/tex]
[tex]=\displaystyle a_1 + \sum_{k=1}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=0}^\infty a_{2k+1}x^{2(k+1)}[/tex]
[tex]=\displaystyle a_1 + \sum_{k=1}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=1}^\infty a_{2(k-1)+1}x^{2k}[/tex]
[tex]=\displaystyle a_1 + \sum_{k=1}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=1}^\infty a_{2k-1}x^{2k}[/tex]
[tex]=\displaystyle a_1 + \sum_{k=1}^\infty \bigg((2k+1)a_{2k+1} - 2a_{2k-1}\bigg)x^{2k}[/tex]
• Odd:
[tex]\displaystyle \sum_{k=0}^\infty 2(k+1)a_{2(k+1)}x^{2k+1} - 2\sum_{k=0}^\infty a_{2k}x^{2k+1}[/tex]
[tex]=\displaystyle \sum_{k=0}^\infty \bigg(2(k+1)a_{2(k+1)}-2a_{2k}\bigg)x^{2k+1}[/tex]
[tex]=\displaystyle \sum_{k=0}^\infty \bigg(2(k+1)a_{2k+2}-2a_{2k}\bigg)x^{2k+1}[/tex]
Notice that the right side of the DE is odd, so there is no 0-degree term, i.e. no constant term, so it follows that [tex]a_1=0[/tex].
The even series vanishes, so that
[tex](2k+1)a_{2k+1} - 2a_{2k-1} = 0[/tex]
for all integers k ≥ 1. But since [tex]a_1=0[/tex], we find
[tex]k=1 \implies 3a_3 - 2a_1 = 0 \implies a_3 = 0[/tex]
[tex]k=2 \implies 5a_5 - 2a_3 = 0 \implies a_5 = 0[/tex]
and so on, which means the odd-indexed coefficients all vanish, [tex]a_{2k+1}=0[/tex].
This leaves us with the odd series,
[tex]\displaystyle \sum_{k=0}^\infty \bigg(2(k+1)a_{2k+2}-2a_{2k}\bigg)x^{2k+1} = 6\sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}[/tex]
[tex]\implies 2(k+1)a_{2k+2} - 2a_{2k} = \dfrac{6(-1)^k}{(2k+1)!}[/tex]
We have
[tex]k=0 \implies 2a_2 - 2a_0 = 6[/tex]
[tex]k=1 \implies 4a_4-2a_2 = -1[/tex]
[tex]k=2 \implies 6a_6-2a_4 = \dfrac1{20}[/tex]
[tex]k=3 \implies 8a_8-2a_6 = -\dfrac1{840}[/tex]
So long as you're given an initial condition [tex]y(0)\neq0[/tex] (which corresponds to [tex]a_0[/tex]), you will have a non-zero series solution. Let [tex]a=a_0[/tex] with [tex]a_0\neq0[/tex]. Then
[tex]2a_2-2a_0=6 \implies a_2 = a+3[/tex]
[tex]4a_4-2a_2=-1 \implies a_4 = \dfrac{2a+5}4[/tex]
[tex]6a_6-2a_4=\dfrac1{20} \implies a_6 = \dfrac{20a+51}{120}[/tex]
and so the first four terms of series solution to the DE would be
[tex]\boxed{a + (a+3)x^2 + \dfrac{2a+5}4x^4 + \dfrac{20a+51}{120}x^6}[/tex]
0.003 is 1/10 of
Please help I need this for homework !!!!!!!!!!!!
Answer:
0.03
Step-by-step explanation:
It is known that the variance of a population equals 1,936. A random sample of 121 has been selected from the population. There is a .95 probability that the sample mean will provide a margin of error of _____. Group of answer choices 31.36 or less 1,936 or less 344.96 or less 7.84 or less
Answer:
Option d (7.84 or less) is the right alternative.
Step-by-step explanation:
Given:
[tex]\sigma^2=1936[/tex]
[tex]\sigma = \sqrt{1936}[/tex]
[tex]=44[/tex]
Random sample,
[tex]n = 121[/tex]
The level of significance,
= 0.95
or,
[tex](1-\alpha) = 0.95[/tex]
[tex]\alpha = 1-0.95[/tex]
[tex]Z_{\frac{\alpha}{2} } = 1.96[/tex]
hence,
The margin of error will be:
⇒ [tex]E = Z_{\frac{\alpha}{2} }(\frac{\sigma}{\sqrt{n} } )[/tex]
By putting the values, we get
[tex]=1.96(\frac{44}{\sqrt{121} } )[/tex]
[tex]=1.96(\frac{44}{11} )[/tex]
[tex]=1.96\times 4[/tex]
[tex]=7.84[/tex]
Beginning in January, a person plans to deposit $1 at the end of each month into an account earning
15% compounded monthly. Each year taxes must be paid on the interest earned during that year. Find
the interest earned during each year for the first 3 years.
Answer:
hi I am a Nepal
[tex] {233333}^{2332} [/tex]
Previous studies suggest that use of nicotine-replacement therapies and antidepressants can help people stop smoking. The New England Journal of Medicine published the results of a double-blind, placebo-controlled experiment to study the effect of nicotine patches and the antidepressant bupropion on quitting smoking. The target for quitting smoking was the 8th day of the experiment.
In this experiment researchers randomly assigned smokers to treatments. Of the 162 smokers taking a placebo, 28 stopped smoking by the 8th day. Of the 272 smokers taking only the antidepressant buproprion, 82 stopped smoking by the 8th day.
Calculate the 99% confidence interval to estimate the treatment effect of buproprion (placebo-treatment). (The standard error is about 0.0407. Use critical value z = 2.576.)
( ), ( )
Round your answer to three decimal places. Put lower bound in the first box and upper bound in the second box.
Using the z-distribution, it is found that the 99% confidence interval to estimate the treatment effect of buproprion (placebo-treatment) is (-0.234, -0.024).
What is a t-distribution confidence interval?The confidence interval is:
[tex]\overline{x} \pm zs[/tex]
In which:
[tex]\overline{x}[/tex] is the sample mean.z is the critical value.s is the standard error.In this problem, we are given that z = 2.576, s = 0.0407. The sample mean is the difference of the proportions, hence:
[tex]\overline{x} = \frac{28}{162} - \frac{82}{272} = -0.129[/tex]
Then, the bounds of the interval are given by:
[tex]\overline{x} - zs = -0.129 - 2.576(0.0407) = -0.234[/tex]
[tex]\overline{x} + zs = -0.129 + 2.576(0.0407) = -0.024[/tex]
The 99% confidence interval to estimate the treatment effect of buproprion (placebo-treatment) is (-0.234, -0.024).
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Which number would be rounded UP to the nearest ten but DOWN to the nearest hundred?
A. 232
B. 238
C. 262
D. 268
Answer:
B
Step-by-step explanation:
Michelle would like to know how much of her loan payments will go toward interest. She has a $124,500 loan with a 5.9% interest rate that is compounded monthly. The loan has a term of 10 years. Calculate the total amount of interest that Michelle will pay over the course of the loan.
9514 1404 393
Answer:
$40,615.20
Step-by-step explanation:
The amortization formula will tell you Michelle's monthly payment.
A = P(r/12)/(1 -(1 +r/12)^(-12t)) . . . . loan value P at interest rate r for t years
A = $124,500(0.059/12)/(1 -(1 +0.059/12)^(-12·10)) ≈ $1375.96
__
The total of Michelle's 120 monthly payments is ...
12 × $1375.96 = $165,115.20
This amount pays both principal and interest, so the amount of interest she pays is ...
$165,115.20 -124,500 = $40,615.20
Michelle will pay $40,615.20 in interest over the course of the loan.
__
A calculator or spreadsheet can figure this quickly.
To calculate the volume of a chemical produced in a day a chemical manufacturing company uses the following formula below:
[tex]V(x)=[C_1(x)+C_2(x)](H(x))[/tex]
where represents the number of units produced. This means two chemicals are added together to make a new chemical and the resulting chemical is multiplied by the expression for the holding container with respect to the number of units produced. The equations for the two chemicals added together with respect to the number of unit produced are given below:
[tex]C_1(x)=\frac{x}{x+1} , C_2(x)=\frac{2}{x-3}[/tex]
The equation for the holding container with respect to the number of unit produced is given below:
[tex]H(x)=\frac{x^3-9x}{x}[/tex]
a. What rational expression do you get when you combine the two chemicals?
b. What is the simplified equation of ?
c. What would the volume be if 50, 100, or 1000 units are produced in a day?
d. The company needs a volume of 3000 How many units would need to be produced in a day?
Answer:
[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]
[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]
[tex]V(50) = 2548.17[/tex] [tex]V(100) = 10098.10[/tex] [tex]V(1000) = 999201.78[/tex]
[tex]x = 54.78[/tex]
Step-by-step explanation:
Given
[tex]V(x) = [C_1(x) + C_2(x)](H(x))[/tex]
[tex]C_1(x) = \frac{x}{x+1}[/tex]
[tex]C_1(x) = \frac{2}{x-3}[/tex]
[tex]H(x) = \frac{x^3 - 9x}{x}[/tex]
Solving (a): Expression for V(x)
We have:
[tex]V(x) = [C_1(x) + C_2(x)](H(x))[/tex]
Substitute known values
[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]
Solving (b): Simplify V(x)
We have:
[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]
Solve the expression in bracket
[tex]V(x) = [\frac{x*(x-3) + 2*(x+1)}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]
[tex]V(x) = [\frac{x^2-3x + 2x+2}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]
[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]
Factor out x
[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * \frac{x(x^2 - 9)}{x}[/tex]
[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * (x^2 - 9)[/tex]
Express as difference of two squares
[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * (x- 3)(x + 3)[/tex]
Cancel out x - 3
[tex]V(x) = [\frac{x^2-x+2}{(x + 1)}] *(x + 3)[/tex]
[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]
Solving (c): V(50), V(100), V(1000)
[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]
Substitute 50 for x
[tex]V(50) = [\frac{(50^2-50+2)(50 + 3)}{(50 + 1)}][/tex]
[tex]V(50) = \frac{(2452)(53)}{(51)}][/tex]
[tex]V(50) = 2548.17[/tex]
Substitute 100 for x
[tex]V(100) = [\frac{(100^2-100+2)(100 + 3)}{(100 + 1)}][/tex]
[tex]V(100) = \frac{9902)(103)}{(101)}[/tex]
[tex]V(100) = 10098.10[/tex]
Substitute 1000 for x
[tex]V(1000) = [\frac{(1000^2-1000+2)(1000 + 3)}{(1000 + 1)}][/tex]
[tex]V(1000) = [\frac{(999002)(10003)}{(10001)}][/tex]
[tex]V(1000) = 999201.78[/tex]
Solving (d): V(x) = 3000, find x
[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]
[tex]3000 = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]
Cross multiply
[tex]3000(x + 1) = (x^2-x+2)(x + 3)[/tex]
Equate to 0
[tex](x^2-x+2)(x + 3)-3000(x + 1)=0[/tex]
Open brackets
[tex]x^3 - x^2 + 2x + 3x^2 - 3x + 6 - 3000x - 3000 = 0[/tex]
Collect like terms
[tex]x^3 + 3x^2- x^2 + 2x - 3x - 3000x + 6 - 3000 = 0[/tex]
[tex]x^3 + x^2 -3001x -2994 = 0[/tex]
Solve using graphs (see attachment)
[tex]x = -54.783[/tex] or
[tex]x = -0.998[/tex] or
[tex]x = 54.78[/tex]
x can't be negative. So:
[tex]x = 54.78[/tex]
A cell phone company charges a monthly fee of $18 plus five cents for each call. A
customer's total cell phone bill this month is $50.50. Use n to represent the number of
calls.
Answer:
650 calls
Step-by-step explanation:
so since you have 18$ per month plus 5 cents per call you would do
18+0.5n(n represent the number of calls)= the total fee of $50.50 cents.
thus,now you need to figure out how much the phone calls were without the monthly fee so you would do:
50.50-18=32.50
so 32.50 is the price of all the phone calls
then you divide 32.50 by 0.05 which equals to 650
meaning that n=650
hope I helped!
In a certain animal species, the probability that a healthy adult female will have no offspring in a given year is 0.24, while the probabilities of 1, 2, 3, or 4 offspring are respectively 0.25, 0.19, 0.17, and 0.15. Find the expected number of offspring.
Answer:
The expected number of offspring is 2
Step-by-step explanation:
The given parameters can be represented as:
[tex]\begin{array}{cccccc}x & {0} & {1} & {2} & {3} & {4} \ \\ P(x) & {0.24} & {0.25} & {0.19} & {0.17} & {0.15} \ \end{array}[/tex]
Required
The expected number of offspring
This implies that we calculate the expected value of the function.
So, we have:
[tex]E(x) = \sum x * P(x)[/tex]
Substitute known values
[tex]E(x) = 0 * 0.24 + 1 * 0.25 + 2 * 0.19+ 3 * 0.17 + 4 * 0.15[/tex]
Using a calculator, we have:
[tex]E(x) = 1.74[/tex]
[tex]E(x) = 2[/tex] --- approximated
Help please guys thanks
Answer:
D
Step-by-step explanation:
sqrt_{4}(81)^5=(81^(5))^(1/4)=81^(5/4)
Answer:
D
Step-by-step explanation:
if it was properly typed, it would have been All of the above but the most correct option is D.
A ball is dropped from a height of 14 ft. The elasticity of the ball is such that it always bounces up one-third the distance it has fallen. (a) Find the total distance the ball has traveled at the instant it hits the ground the fourth time. (Enter an exact number.)
Answer:
Hello,
742/27 (ft)
Step-by-step explanation:
[tex]h_1=14\\\\h_2=\dfrac{14}{3} \\\\h_3=\dfrac{14}{9} \\\\h_4=\dfrac{14}{27} \\\\[/tex]
[tex]d=14+2*\dfrac{14}{3} +2*\dfrac{14}{9} +2*\dfrac{14}{27} \\=14*(1+\dfrac{1}{3}+\dfrac{2}{9} +\dfrac{2}{27} )\\=14*\dfrac{53}{27} \\=\dfrac{742}{27} \\[/tex]
The total distance the ball has traveled at the instant it hits the ground the fourth time [tex]28ft.[/tex]
What is the total distance?
Distance is a numerical measurement of how far apart objects or points are. It is the actual length of the path travelled from one point to another.
Here given that,
A ball is dropped from a height of [tex]14[/tex] ft. The elasticity of the ball is such that it always bounces up one-third the distance it has fallen.
So, after striking with the ground it covers the distance [tex]14[/tex] ft. so it rebounds to the height is [tex]\frac{1}{3}(14)[/tex].
Then again it hits the ground and covers the distance [tex]\frac{1}{3}(14)[/tex] and again after rebounding it goes to the height is
[tex]\frac{1(1)}{3(3)}.(14)=\frac{(1)^2}{(3)^2}(14)[/tex]
Then it falls the same distance and goes back to the height
[tex]\frac{1}{3}[/tex] ×[tex](\frac{(1)^2}{(3)^2})[/tex] ×[tex]14[/tex] = [tex]\frac{(1)^3}{(3)^3}(14)[/tex]
So, the total distance travelled is
[tex]14+2[\frac{1}{3}(14)+(\frac{1}{3})^2(14)+(\frac{1}{3})^3(14)+...][/tex]
We take the sum is twice because it goes back to the particular height and falls to the same distance.
[tex]S=14+2(\frac{\frac{1}{3}(14)}{1-\frac{1}{3}})\\\\\\S=\frac{a}{1-r}\\\\\\S=14+2(\frac{\frac{14}{3}}{\frac{2}{3}})\\\\S=14+2(\frac{14}{2})\\\\S=14+2(7)\\\\S=14+14\\\\S=28ft[/tex]
Hence, the total distance the ball has traveled at the instant it hits the ground the fourth time [tex]28ft.[/tex]
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Here is a number sequence. The rule for finding the next term is to add
a, where a is an integer. ! ! 8 ........! ! ........! ! 29 Work out the two
missing terms.
Answer:
8,15,22,29
Step-by-step explanation:
the interger a is 7,so to find the next term you have to add 7 plus the 8,
8+7=15
15+7=22
22+7=29
8,15,22,29
I hope this helps
What is the approximate length of arc s on the circle below? Use 3.14 for Pi. Round your answer to the nearest tenth.
-5.8 ft
-6.3 ft
-27.5 ft
-69.1 ft
9514 1404 393
Answer:
69.1 ft
Step-by-step explanation:
The diameter of the circle is 24 ft. The length of the arc is more than twice the diameter, so cannot be less than about 50 ft. The only reasonable choice is ...
69.1 ft
__
The circumference of the circle is ...
C = 2πr = 2(3.14)(12 ft) = 75.36 ft
The arc length of interest is 330° of the 360° circle, so is 330/360 = 11/12 times the circumference.
s = (11/12)(75.36 ft) = 69.08 ft ≈ 69.1 ft
Answer:D
Step-by-step explanation:
y varies directly as the cube of x. When x = 3, then y = 7. Find y when x = 4.
Answer:
[tex]y \: \alpha \: {x}^{3} \ \\ y \: = k {x}^{3} \\ where \: y = 7 \: and \:x = 3 \\ y = k {x}^{3} \\ 7 = k ( {3)}^{3} \\ 7 = 27k \\ k = \frac{7}{27} \\ \\ so \: \: y = \frac{7}{27} {x}^{3} \\ \\ y = \frac{7}{27} {4}^{3} \\ y = \frac{448}{27} [/tex]
the required value of y at x = 4 is 16.64.
Given that,
y varies directly as the cube of x. When x = 3, then y = 7.To determine the y when x = 4.
Proportionality is defined as between two or more sets of values, and how these values are related to each other in the sense that are they directly proportional or inversely proportional to each other.
Here,
y is directly proportional to the cube of x i.e
y ∝ x³
y = kx³ - - - - - (1)
where k is proportionality constant,
At x = 3 y = 7
7 = k (3)³
7 / 27 = k
k = 0.26
Put k in equation 1
y = 0.26 x³
Now at x = 4
y = 0.26 * 4³
y = 0.26 * 64
y = 16.64
Thus, the required value of y at x = 4 is 16.64.
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Question 4 (2 marks)
Justin works 14 hours at a normal pay rate of $24.80 per hour and 5 hours of overtime at
time and a half. How much should he be paid?
I
809 words
LE
English (Australia)
Answer:
554.7
Step-by-step explanation:
The pay=25.8*14+(25.8)*5*1.5=554.7
write your answer as an integer or as a decimal rounded to the nearest tenth
Answer:
123456-6-&55674
Step-by-step explanation:
rdcfvvzxv.
dgjjjdeasg JJ is Redding off in grad wassup I TV kitten gag ex TV ex raisin see
recall see
write the expression as a decimal , 6 x 1 + 9 x 1/10 + 8 x 1/100 + 6 x 1/1000 =__
Answer:
6.986.
Step-by-step explanation:
6 x 1 + 9 x 1/10 + 8 x 1/100 + 6 x 1/1000
We do the multiplications first ( according to PEMDAS):-
= 6 + 9 * 0.1 + 8 * 0.01 + 6 * 0.001
= 6 + 0.9 + 0.08 + 0006
= 6.9 + 0.086
= 6 986.
The value of the equation in the decimal form is A = 6.986
What is an Equation?
Equations are mathematical statements with two algebraic expressions flanking the equals (=) sign on either side.
It demonstrates the equality of the relationship between the expressions printed on the left and right sides.
Coefficients, variables, operators, constants, terms, expressions, and the equal to sign are some of the components of an equation. The "=" sign and terms on both sides must always be present when writing an equation.
Given data ,
Let the equation be represented as A
Now , the value of A is
A = 6 x 1 + 9 x 1/10 + 8 x 1/100 + 6 x 1/1000
On simplifying the equation , we get
The value of 6 x 1 = 6
The value of 9 x 1/10 = 0.9
The value of 9 x 1/100 = 0.08
The value of 6 x 1/1000 = 0.006
So , substituting the values in the equation A , we get
A = 6 + 0.9 + 0.08 + 0.006
On simplifying the equation , we get
A = 6.986
Therefore , the value of A is 6.986
Hence , the value of the equation is 6.986
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Hector's Position:
Hector was standing halfway between first and second base, at the grass line. The
grass line is 95 feet from the pitcher's mound.
6. Calculate the coordinates for Hector's position. [Note: We can assume that 95
feet is an approximately horizontal distance from the pitcher's mound to the grass
line.] (2 points: 1 for x, 1 for y)
Hector was standing at the coordinate ( __, _).
Calculate Hector's Throw:
Answer:
(137.78, 47.72)
Step-by-step explanation:
(I just finished this assignment.)
Tre's position at the pitcher's mound as the point (42.78, 42.78).
( x , y )
Hector is about 95 feet away from the pitcher's mound horizontal, (x axis).
Since we already have the correct y-coordinate, we need to solve for the correct x-coordinate.
x = 95 + 42.78
↓ ↓ ↓
95 + 42.78 = 137.72
Now all you need to do is write out the coordinates.
Hector's coordinates are (137.72, 47.78 )
How many ways can a president, vice president, secretary, and treasurer be chosen from a club with 8 member
Answer:
504
Step-by-step explanation:
The tree diagram below shows the possible combinations of juice and snack that can be offered at the school fair.
A tree diagram. Orange branches to popcorn and pretzels. Grape branches to popcorn and pretzels. Apple branches to popcorn and pretzels. Grapefruit branches to popcorn and pretzels.
How many different combinations are modeled by the diagram?
6
8
12
32
Answer:
B. 8Step-by-step explanation:
The combinations are:
Orange - 2 (with popcorn and pretzels)Grape - 2 (with popcorn and pretzels)Apple - 2 (with popcorn and pretzels)Grapefruit - 2 (with popcorn and pretzels)Total number of combinations:
4*2 = 8Correct choice is B
there are 8different combinations are modeled by the diagram.
Answer:
Solution given:
orange:2
grape:2
apple:2
grapefruit:2
no of term:4
now
total no. of combination ia 4*2=8
A ball is thrown in air and it's height, h(t) in feet, at any time, t in seconds, is represented by the equation h(t)=−4t2+16t. When is the ball higher than 12 feet off the ground?
A. 3
B. 1
C. 1
D. 4
Hence the time that the ball will be height than 12 feet off the ground is 4secs
Given the expression for calculating the height in feet as;
h(t) = -4t²+16t
If the ball is higher than 12feet, h(t) > 12
Substituting h = 12 into the expression
-4t²+16t > 12
-4t²+16t - 12 > 0
4t²- 16t + 12 > 0
t²- 4t + 3 > 0
Factorize
(t²- 3t)-(t + 3) > 0
t(t-3)-1(t-3) > 0
(t-1)(t-3)>0
t > 1 and 3secs
Hence the time that the ball will be height than 12 feet off the ground is 4secs
Learn more: https://brainly.com/question/18405392
f(t)= 102,000/1+4400e^-t
Answer:
Beginning (t=0) population with flu is 23.
After 4 weeks, population with flu is 1250.
After an infinite amount of weeks, the population witf flu is 102000
Step-by-step explanation:
First question asks you to replace t with 0 because it says beginning.
102000/(1+4400e^-0)=102000/(1+4400)=102000/4401=23.17655 approximately. To nearest whole number this is 23.
After 4 weeks means we replace t with 4:
102000/(1+4400e^-4)
Calculator time:
1250.17142 which to nearest whole number is 1250
If t is super large, then e^-t is super close to 0.
So the limiting number is
102000/(1+4400×0)=102000/1=102000
Which is the solution to-x/2<-4
A x<-8
B x2-8
C x <8
D x 8
Answer:
A.x<-8
Step-by-step explanation:
=1/2x<−4
=2*(1/2x)< (2)*(-4)
= x<-8
Plz urgennt look at the image over 1000 points im going to need help with the last 4 questions i have?
Find the minimum sample size needed to be 99% confident that the sample's variance is within 30% of the population's variance.
The Minimum sample size table is attached below
Answer:
[tex]X=173[/tex]
Step-by-step explanation:
From the question we are told that:
Confidence Interval [tex]CI=99\%[/tex]
Variance [tex]\sigma^2=30\%[/tex]
Generally going through the table the
Minimum sample size is
[tex]X=173[/tex]
Select the correct answer.
What is the best way to describe a theme of this poem?
A.
The main purpose of having New Year's resolutions is to make people feel bad.
B.
The failures of the past should inspire people to accomplish more in the future.
OC.
By the end of the year, it is too late to make any changes to a person's life.
D.
People would accomplish their New Year's resolutions if they wrote them down.
B.The failures of the past should inspire people to accomplish more in the future.