Answer:
1/512
Step-by-step explanation:
Let staring fraction = x
Half-life = 20 years ; this is the time taken for an element to decrease to half of its original size
Hence,
After 20 years - - - > x/2
After 40 years - - - - > x/2 ÷ 2 = x/2 * 1/2 = x /4
After 60 years - - - - > x/4 ÷ 2 = x/4 * 1/2 = x/8
After 80 years - - - - -> x/8 ÷ 2 = x/8 * 1/2 = x / 16
After 100 years - - - > x/16 * 1/2 = x/32
After 120 years - - - - > x/32 * 1/2 = x/64
After 140 years - - - - -> x / 64 * 1/2 = x / 128
After 160 years - - - - - > x / 128 * 1/2 = x/256
After 180 years - - - - > x/256 * 1/2 = x / 512
Hence, the fraction after 180 years = 1/512
I need helqp answering this problem ASAP thank you
Step-by-step explanation:
D correct answer Trust me
What is the measure of the angle in red?
70°
Answer:
See pic below.
Answer: C. 380°
Step-by-step explanation:
The graph shows the distribution of the number of text messages young adults send per day. The distribution is approximately Normal, with a mean of 128 messages and a standard deviation of 30 messages.
What percentage of young adults send more than 158 text messages per day?
16%
34%
68%
84%
Answer:
You'd think its 34% but apparently it's 16%.
I hope this is right. If its not then it must be 34%.
(A) -- or maybe (B). 80% confident it is A.
ED2021
a, b ∈q , then (a+ b)∈ …………… । *
Answer:
this is an equation of closure property of rational numbers under addition
Step-by-step explanation:
this is the meaning of it
for every a and b belongs to q then a+b belongs to q
o the area of a rhombus is 24m²
and one of its diagonals 18cm find
the side of the rhombus
Area of rhombus = 1/2 × d1 × d2
Let the other diagonal be x
ATQ
1/2 × 18 × x = 24
9 × x = 24
x = 24/9
x = 8/3
Now half each diagonal = 9 and 4/3
Now side = b² + p² = h²
9²+(4/3)² = h²
81 + 16/9 = h²
729/9 + 16/9 = h²
745/9 = h²
√(745/9) = h
Therefore the side of the rhombus is √(745/9)cm
Answered by Gauthmath must click thanks and mark brainliest
The following are on a parabola defining the edge of a ski
(-4, 1), (-2, 0.94), (0.1)
The general form for the equation of a parabola is:
Ax^2+ Bx +C= y
Required:
a. Use the x- and y-values of 1 of the points to build a linear equation with 3 variables: A, B, and C.
b. Repeat this process with 1 of the other to build a 2nd linear equation.
c. Record your equation here. Repeat this process with the other point to build a 3rd equation.
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Answer:
a) 16A -4B +C = 1
b) 4A -2B +C = 0.94
c) C = 1
Step-by-step explanation:
Substitute the x- and y-values into the general form equation.
a. A(-4)² +B(-4) +C = 1
16A -4B +C = 1
__
b. A(-2)² +B(-2) +C = 0.94
4A -2B +C = 0.94
__
c. A(0)² +B(0) +C = 1
C = 1
_____
Additional comment
Solving these equations gives A=0.015, B=0.06, C=1. The quadratic is ...
0.015x² +0.06x +1 = y
solve x^3-7x^2+7x+15
Step-by-step explanation:
\underline{\textsf{Given:}}
Given:
\mathsf{Polynomial\;is\;x^3+7x^2+7x-15}Polynomialisx
3
+7x
2
+7x−15
\underline{\textsf{To find:}}
To find:
\mathsf{Factors\;of\;x^3+7x^2+7x-15}Factorsofx
3
+7x
2
+7x−15
\underline{\textsf{Solution:}}
Solution:
\textsf{Factor theorem:}Factor theorem:
\boxed{\mathsf{(x-a)\;is\;a\;factor\;P(x)\;\iff\;P(a)=0}}
(x−a)isafactorP(x)⟺P(a)=0
\mathsf{Let\;P(x)=x^3+7x^2+7x-15}LetP(x)=x
3
+7x
2
+7x−15
\mathsf{Sum\;of\;the\;coefficients=1+7+7-15=0}Sumofthecoefficients=1+7+7−15=0
\therefore\mathsf{(x-1)\;is\;a\;factor\;of\;P(x)}∴(x−1)isafactorofP(x)
\mathsf{When\;x=-3}Whenx=−3
\mathsf{P(-3)=(-3)^3+7(-3)^2+7(-3)-15}P(−3)=(−3)
3
+7(−3)
2
+7(−3)−15
\mathsf{P(-3)=-27+63-21-15}P(−3)=−27+63−21−15
\mathsf{P(-3)=63-63}P(−3)=63−63
\mathsf{P(-3)=0}P(−3)=0
\therefore\mathsf{(x+3)\;is\;a\;factor}∴(x+3)isafactor
\mathsf{When\;x=-5}Whenx=−5
\mathsf{P(-5)=(-5)^3+7(-5)^2+7(-5)-15}P(−5)=(−5)
3
+7(−5)
2
+7(−5)−15
\mathsf{P(-5)=-125+175-35-15}P(−5)=−125+175−35−15
\mathsf{P(-5)=175-175}P(−5)=175−175
\mathsf{P(-5)=0}P(−5)=0
\therefore\mathsf{(x+5)\;is\;a\;factor}∴(x+5)isafactor
\underline{\textsf{Answer:}}
Answer:
\mathsf{x^3+7x^2+7x-15=(x-1)(x+3)(x+5)}x
3
+7x
2
+7x−15=(x−1)(x+3)(x+5)
\underline{\textsf{Find more:}}
Find more:
What is (x+13)^2? pls help!!!
An article describes an experiment to determine the effectiveness of mushroom compost in removing petroleum contaminants from soil. Out of 155 seeds planted in soil containing 3% mushroom compost by weight, 74 germinated. Out of 155 seeds planted in soil containing 5% mushroom compost by weight, 86 germinated. Can you conclude that the proportion of seeds that germinate differs with the percent of mushroom compost in the soil
Solution :
Let [tex]p_1[/tex] and [tex]p_2[/tex] represents the proportions of the seeds which germinate among the seeds planted in the soil containing [tex]3\%[/tex] and [tex]5\%[/tex] mushroom compost by weight respectively.
To test the null hypothesis [tex]H_0: p_1=p_2[/tex] against the alternate hypothesis [tex]H_1:p_1 \neq p_2[/tex] .
Let [tex]\hat p_1, \hat p_2[/tex] denotes the respective sample proportions and the [tex]n_1, n_2[/tex] represents the sample size respectively.
[tex]$\hat p_1 = \frac{74}{155} = 0.477419[/tex]
[tex]n_1=155[/tex]
[tex]$p_2=\frac{86}{155}=0.554839[/tex]
[tex]n_2=155[/tex]
The test statistic can be written as :
[tex]$z=\frac{(\hat p_1 - \hat p_2)}{\sqrt{\frac{\hat p_1 \times (1-\hat p_1)}{n_1}} + \frac{\hat p_2 \times (1-\hat p_2)}{n_2}}}[/tex]
which under [tex]H_0[/tex] follows the standard normal distribution.
We reject [tex]H_0[/tex] at [tex]0.05[/tex] level of significance, if the P-value [tex]<0.05[/tex] or if [tex]|z_{obs}|>Z_{0.025}[/tex]
Now, the value of the test statistics = -1.368928
The critical value = [tex]\pm 1.959964[/tex]
P-value = [tex]$P(|z|> z_{obs})= 2 \times P(z< -1.367928)$[/tex]
[tex]$=2 \times 0.085667$[/tex]
= 0.171335
Since the p-value > 0.05 and [tex]$|z_{obs}| \ngtr z_{critical} = 1.959964$[/tex], so we fail to reject [tex]H_0[/tex] at [tex]0.05[/tex] level of significance.
Hence we conclude that the two population proportion are not significantly different.
Conclusion :
There is not sufficient evidence to conclude that the [tex]\text{proportion}[/tex] of the seeds that [tex]\text{germinate differs}[/tex] with the percent of the [tex]\text{mushroom compost}[/tex] in the soil.
Please help me with this question
Why does cube root 7 equal 7 to the 1/3 power
Answer:
Step-by-step explanation:
Here's how you convert:
[tex]\sqrt[n]{x^m}=x^{\frac{m}{n}[/tex] The little number outside the radical, called the index, serves as the denominator in the rational power, and the power on the x inside the radical serves as the numerator in the rational power on the x.
A couple of examples:
[tex]\sqrt[3]{x^4}=x^{\frac{4}{3}[/tex]
[tex]\sqrt[5]{x^7}=x^{\frac{7}{5}[/tex]
It's that simple. For your problem in particular:
[tex]\sqrt[3]{7}[/tex] is the exact same thing as [tex]\sqrt[3]{7^1}=7^{\frac{1}{3}[/tex]
The posted weight limit for a wooden
bridge is 6,500 pounds. A delivery truck
is loaded with identical boxes of canned
goods that weigh 16 pounds each. If the
combined weight of the empty delivery
truck and the driver is 3,512 pounds,
what is the maximum number of boxes
that would keep the combined weight of
the truck, driver, and boxes below the
posted weight limit?
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Answer:
186
Step-by-step explanation:
Let b represent the number of boxes in the truck. Then for the weight limit to be met, we require ...
3512 +16b < 6500
16b < 2988
b < 186.75
The maximum number of boxes is 186.
Help. The graph shows the system of equations below.
2x -3y = -6
y = - 1/3x -4
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Answer:
(a) The blue line ... solution ... (-6, -2)..
Step-by-step explanation:
The second equation describes a line with negative slope and a y-intercept of -4. This is clearly the red line on the graph.
The blue line represents the equation 2x -3y = -6.
The point of intersection of the two lines is (-6, -2), so that is the solution to the system of equations. This, by itself, is sufficient for you to choose the correct answer.
Can someone help me? I don’t know how to solve the rest. I am struggling and I would be so happy if any of you helped me. Thank you for your help!
Using the following equation, find the center and radius: x2 −2x + y2 − 6y = 26 (5 points)
Answer:
Center: (1,3)
Radius: 6
Step-by-step explanation:
Hi there!
[tex]x^2-2x + y^2 - 6y = 26[/tex]
Typically, the equation of a circle would be in the form [tex](x-h)^2+(y-k)^2=r^2[/tex] where [tex](h,k)[/tex] is the center and [tex]r[/tex] is the radius.
To get the given equation [tex]x^2-2x + y^2 - 6y = 26[/tex] into this form, we must complete the square for both x and y.
1) Complete the square for x
Let's take a look at this part of the equation:
[tex]x^2-2x[/tex]
To complete the square, we must add to the expression the square of half of 2. That would be 1² = 1:
[tex]x^2-2x+1[/tex]
Great! Now, let's add this to our original equation:
[tex]x^2-2x+1+y^2-6y = 26[/tex]
We cannot randomly add a 1 to just one side, so we must do the same to the right side of the equation:
[tex]x^2-2x+1+y^2-6y = 26+1\\x^2-2x+1+y^2-6y = 27[/tex]
Complete the square:
[tex](x-1)^2+y^2-6y = 27[/tex]
2) Complete the square for y
Let's take a look at this part of the equation [tex](x-1)^2+y^2-6y = 27[/tex]:
[tex]y^2-6y[/tex]
To complete the square, we must add to the expression the square of half of 6. That would be 3² = 9:
[tex]y^2-6y+9[/tex]
Great! Now, back to our original equation:
[tex](x-1)^2+y^2-6y+9= 27[/tex]
Remember to add 9 on the other side as well:
[tex](x-1)^2+y^2-6y+9= 27+9\\(x-1)^2+y^2-6y+9= 36[/tex]
Complete the square:
[tex](x-1)^2+(y-3)^2= 36[/tex]
3) Determine the center and the radius
[tex](x-1)^2+(y-3)^2= 36[/tex]
[tex](x-h)^2+(y-k)^2=r^2[/tex]
Now, we can see that (1,3) is in the place of (h,k). 36 is also in the place of r², making 6 the radius.
I hope this helps!
Answer:
[tex]\sqrt{g^2+f^2-c}[/tex]
[tex]g=-1,f=-3,c=-26[/tex]
so, the Center of the equation is [tex](1,3)[/tex]
Center → (1 , 3)[tex]\sqrt{(-1)^2+(-3)^2-(-26})[/tex]
[tex]=\sqrt{1+9+26}[/tex]
[tex]=\sqrt{36}[/tex]
[tex]=6[/tex]
Radius → 6OAmalOHopeO
Find the length of x
Answer:
32
Step-by-step explanation:
Let's assume that the triangles are similar.
[tex]\frac{16}{12} = \frac{24}{18} = \frac{x}{24}[/tex]
[tex]\frac{x}{24} = \frac{4}{3} => x = \frac{24}{3} * 4= 8 * 4 = 32[/tex]
Which expression is equivalent to 3/2
Answer:
C
Step-by-step explanation:
Use formula autocomplete to enter a sum function in cell B7 to calculate the total of cells in B2:B6
Excel enables the users to perform mathematics basic and advanced function with just one formula.
The formula for sum of entire row or column can be done with just entering a single formula and results are shown in seconds.
The formula for sum of few column cells is,
=SUM(B2:B6)
The spreadsheet allows the user to enter various formula and results are displayed withing seconds.
There are formulas for basic math functions and there are also formulas for advance mathematics calculations. For addition of values of many cells sum formula is used and range is assigned for reference.
The formula adds all the values of selected cells and displays the results in different cell.
Learn more at https://brainly.com/question/24365931
A building 51 feet tall casts a shadow 48 feet long. Simultaneously, a nearby statue casts a shadow of 16 feet. How tall is the statue? Choose an answer
Answer: 17 feet
Step-by-step explanation:
51/48 = x/16
(51)(16)/48
The statute is 17 feet tall.
What are the similar triangles?Similar triangles are the triangles that have the same shape, but different sizes. The corresponding angles are congruent and the sides are in proportion.
What is the ratio of any two corresponding sides of similar triangles?The ratio of any corresponding sides in two equiangular triangles is always the same.
Let's visualize the situation according to the given question.
AB is the building ,whose height is 51f
BC is the shadow of the building AB, whose length is 48ft.
QR is the shadow of the tower statue, whose length is 16feet.
Let the height of the statue PR be h feet.
In triangle ACB and triangle PRQ
∠ACB = ∠PRQ = 90 degrees
( the objects and shadows are perpendicular to each other)
∠BAC = ∠QPR
( sunray falls on the pole and tower at the same angle, at the same time )
⇒ΔACB similar to ΔPRQ ( AA criterion)
Therefore, the ratio of any two corresponding sides in equiangular triangles is always same.
⇒ AC/CB = PR/RQ
⇒[tex]\frac{51}{48} =\frac{h}{16}[/tex]
⇒ h = [tex]\frac{(51)(16)}{48}[/tex]
⇒ h = 17 feet.
Hence, the statute is 17 feet tall.
Learn more about the similar triangle here:
brainly.com/question/25882965
#SPJ2
C 89. What is the power of 5, so that 1 its value become ? (५ को घाताङ्क 25 कति हुदा त्यसको मान 25 हुन्छ ?) .7
C 89. What is the power of 5, so that 1 its value become ?
The power is 0. because if 0 is tge powwe of any variable or letters the value becomes 1.
Suppose that a random sample of size 64 is to be selected from a population with mean 50 and standard deviation 5. (a) What are the mean and standard deviation of the sampling distribution of x
Answer:
Mean of sampling distribution = 50
Standard deviation of sampling distribution, = 0.625
Step-by-step explanation:
Given :
Mean, μ = 50
Standard deviation, σ = 5
Sample size, n = 64
The mean of sampling distribution, μxbar = population mean, μ
μxbar = μ
According to the central limit theorem, the sampling distribution converges to the population mean as the sample size increases, hence, , μxbar = μ = 50
Standard deviation of sampling distribution, σxbar = σ/√n
σxbar = 5/√64 = 5 / 8 = 0.625
Please show your steps
Answer:
M of aftershock = 4.90
Step-by-step explanation:
5.6 = log(x/1)
[tex]10^{5.6} = 398107.1 \\[/tex]
1/5 * 398,107.1 = 79,621.4
[tex]10^{m} =[/tex] 79,621.4
m = log (79,621.4) = 4.90
Assume a random variable representing the amount of time it takes for a customer service representative to pick up has a uniform distribution between 15 and 20 minutes. What is the probability that a randomly selected application from this distribution took less than 18 minutes
Answer:
0.6 = 60% probability that a randomly selected application from this distribution took less than 18 minutes.
Step-by-step explanation:
Uniform probability distribution:
An uniform distribution has two bounds, a and b.
The probability of finding a value of at lower than x is:
[tex]P(X < x) = \frac{x - a}{b - a}[/tex]
The probability of finding a value between c and d is:
[tex]P(c \leq X \leq d) = \frac{d - c}{b - a}[/tex]
The probability of finding a value above x is:
[tex]P(X > x) = \frac{b - x}{b - a}[/tex]
Uniform distribution between 15 and 20 minutes.
This means that [tex]a = 15, b = 20[/tex]
What is the probability that a randomly selected application from this distribution took less than 18 minutes?
[tex]P(X < 18) = \frac{18 - 15}{20 - 15} = 0.6[/tex]
0.6 = 60% probability that a randomly selected application from this distribution took less than 18 minutes.
in a survey of 90 students, the ratio of those who work outside the home to those who don't is 6:4. How many students work outside the home according to this survey? SHOW ALL WORK! AND ONLY ANSWER IF YOU KNOW THE ANSWER!
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Answer:
54
Step-by-step explanation:
The fraction of the total that work outside the home is ...
outside/(outside +inside) = 6/(6+4) = 6/10
Then the number of those surveyed who work outside the home is ...
(6/10)(90) = 54 . . . work outside the home
Please help i need the answer asap!!!
if you know the answer please give it to me as soon as you can!!
Answer:
Choice b.Step-by-step explanation:
Replace (x, y) with (1, 1) and verify if the equations are correct.
You would ignore the x and y in the equations. Testing one of each pair.
a. 3 + 2 = 3, incorrectb. 7 + 2 = 9, correctc. 8 + 1 = 7, incorrectd. 8 - 2 = 4, incorrectIt is obvious that only b. is correct.
log_c(A)=2
log_c(B)=5,
solve log_c(A^5B^3)
We know that [tex]\log_a(bc)=\log_a(b)+\log_a(c)[/tex].
Using this rule,
[tex]\log_c(A^5B^3)=\log_c(A^5)+\log_c(B^3)[/tex].
We also know that [tex]\log_c(a^b)=b\log_c(a)[/tex].
Using this rule,
[tex]\log_c(A^5)+\log_c(B^3)=5\log_c(A)+3\log_c(B)[/tex]
Now we know that [tex]\log_c(A)=2,\log_c(B)=5[/tex] so,
[tex]5\cdot2+3\cdot5=10+15=\boxed{25}[/tex].
Hope this helps :)
What are the new vertices of quadrilateral KLMN if the quadrilateral is translated two units to the right and four units upward?
A)
K′ = (–2,0), L′ = (1,0), M′ = (1,–3), N′ = (–2,–3)
B)
K′ = (–2,2), L′ = (1,2), M′ = (1,–1), N′ = (–2,–1)
C)
K′ = (–0,0), L′ = (3,0), M′ = (3,–1), N′ = (0,–1)
D)
K′ = (–2,–2), L′ = (1,–2), M′ = (1,–5), N′ = (–2,–5)
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Answer:
B) K′ = (–2,2), L′ = (1,2), M′ = (1,–1), N′ = (–2,–1)
Step-by-step explanation:
Translation 2 units right adds 2 to the x-coordinate.
Translation 4 units upward adds 4 to the y-coordinate.
The translation can be represented by the relation ...
(x, y) ⇒ (x +2, y +4)
__
You can choose the correct answer by looking at the translation of K.
K(-4, -2) ⇒ K'(-4+2, -2+4) = K'(-2, 2) . . . . . matches choice B
On the set of axes below, solve the following system of equations graphically and state the coordinates of all points in the solution set.
X1=-2 and x2=1
You would just need to plug the first y equation value into the 2nd equation to get what I got in the photo. Then solve for the x’s to get the coordinates.
Oh Brian~
I need help again
Answer:
18c^3d^9
Step-by-step explanation:
2c^3 d^2 * 9d^7
We know that we add the exponents when the base is the same
2*9 c^3 d^(2+7)
18c^3d^9
x^{2} +y^{2} =?
cho mình hỏi với
Answer:
[tex]{ \sf{ {x}^{2} + {y}^{2} = {(x + y)}^{2} - 2xy }}[/tex]
Step-by-step explanation:
[tex]{ \tt{ {(x + y)}^{2} = (x + y)(x + y) }} \\ { \tt{ {(x + y)}^{2} = ( {x}^{2} + 2xy + {y}^{2}) }} \\ { \tt{( {x}^{2} + {y}^{2} ) = {(x + y)}^{2} - 2xy}}[/tex]
