Answer:
9 hours
Step-by-step explanation:
Since the group of men remains the same, number of hours is proportional to number of radios.
1300/26 = 450/h
h = 26 * 450 / 1300 = 9 hours
A computer password is required to be 7 characters long. How many passwords are possible if the password requires 3 letter(s) followed by 4 digits (numbers 0-9), where no repetition of any letter or digit is allowed
Answer:
[tex]78,\!624,\!000[/tex].
Step-by-step explanation:
Note the requirements:
Repetition of letter or digit is not allowed.The order of the letters and digits matters.Because of that, permutation would be the most suitable way to count the number of possibilities.
There are [tex]\displaystyle P(26,\, 3) = \frac{26!}{(26 - 3)!} = 26 \times 25 \times 24 = 15,\!600[/tex] ways to arrange three out of [tex]26[/tex] distinct letters (without replacement.)
Similarly, there are [tex]\displaystyle P(10,\, 4) = \frac{10!}{(10 - 4)!} = 10 \times 9 \times 8 \times 7= 5,\!040[/tex] ways to arrange four out of [tex]10[/tex] distinct numbers (also without replacement.)
Therefore, there are [tex]15,\!600[/tex] possibilities for the three-letter section of this password, and [tex]5,\!040[/tex] possibilities for the four-digit section. What if these two parts are combined?
Consider: if the first three letters of the password were fixed, then there would be [tex]5,\!040[/tex] possibilities. However, if any of the first three letters was changed, the result would be another [tex]5,\!040\![/tex] possibilities, all of which are different from the previous [tex]5,\!040\!\![/tex] possibilities. These two three-letter sections along will give [tex]2 \times 5,\!400[/tex] possibilities. Since there are [tex]15,\!600[/tex] three-letter sections like that, there would be [tex]15,\!600 \times 5,\!400 = 78,\!624,\!000[/tex] possible passwords in total. That gives the number of possible passwords that satisfy these requirements.
will rate you brainliest
Answer:
third option is the first step
Answer:
C
Step-by-step explanation:
It is c bro
Ben is tiling the floor in his bathroom the area he is tiling is 4m times 2m each tiles measures 400mm times 400mm he has 45 tiles is that enough
Answer:
No, it's not enough
Step-by-step explanation:
Given
Tilling Dimension = 4m by 2m
Tile Dimension = 400mm by 400mm
Required
Determine the 45 tiles is enough
First;
The area of the tiling has to be calculated
[tex]Area = Length * Breadth[/tex]
[tex]Area = 4m * 2m[/tex]
[tex]Area = 8m^2[/tex]
Next, determine the area of the tile
[tex]Area = Length * Breadth[/tex]
[tex]Area = 400mm * 400mm[/tex]
Convert measurements to metres
[tex]Area = 0.4m* 04m[/tex]
[tex]Area = 0.16\ m^2[/tex]
Next, multiply the above area result by the number of files
[tex]Total = 0.16m^2 * 45[/tex]
[tex]Total = 7.2m^2[/tex]
Compare 7.2 to 8
Hence, we conclude that the 45 tiles of 400mm by 400 mm dimension is not enough to floor his bathroom
Paula drives 130 miles in 2.5 hours. How far would she drive in 4.5 at the same speed?
*Please answer
I will award the Brainliest answer
Answer:
Paula will travel 234 miles in 4.5 hours
Step-by-step explanation:
Step 1: We first find the speed Paula is going in hours, we divide 130 mile by 2.5 hours to get 52 miles per hour
Step 2: We multiple 52 miles per hour with 4.5 hours to get 234 miles
Therefore Paula will travel 234 miles in 4.5 hours
I need help will rate you brainliest
Answer:
Yes you can
Step-by-step explanation:
To eliminate the denominator
Answer:
No
Step-by-step explanation:
We cannot be certain that x + 3 > 0
If it was negative then the sign of the inequality would change.
To solve find the critical values of the numerator/ denominator, that is
x = 2 and x = - 3
The domain is the split into 3 intervals
(- ∞, - 3 ), (- 3, 2), (2, + ∞ )
Use test points from each interval to determine valid solution
Consider the differential equation:
2y'' + ty' − 2y = 14, y(0) = y'(0) = 0.
In some instances, the Laplace transform can be used to solve linear differential equations with variable monomial coefficients.
If F(s) = ℒ{f(t)} and n = 1, 2, 3, . . . ,then
ℒ{tnf(t)} = (-1)^n d^n/ds^n F(s)
to reduce the given differential equation to a linear first-order DE in the transformed function Y(s) = ℒ{y(t)}.
Requried:
a. Sovle the first order DE for Y(s).
b. Find find y(t)= ℒ^-1 {Y(s)}
(a) Take the Laplace transform of both sides:
[tex]2y''(t)+ty'(t)-2y(t)=14[/tex]
[tex]\implies 2(s^2Y(s)-sy(0)-y'(0))-(Y(s)+sY'(s))-2Y(s)=\dfrac{14}s[/tex]
where the transform of [tex]ty'(t)[/tex] comes from
[tex]L[ty'(t)]=-(L[y'(t)])'=-(sY(s)-y(0))'=-Y(s)-sY'(s)[/tex]
This yields the linear ODE,
[tex]-sY'(s)+(2s^2-3)Y(s)=\dfrac{14}s[/tex]
Divides both sides by [tex]-s[/tex]:
[tex]Y'(s)+\dfrac{3-2s^2}sY(s)=-\dfrac{14}{s^2}[/tex]
Find the integrating factor:
[tex]\displaystyle\int\frac{3-2s^2}s\,\mathrm ds=3\ln|s|-s^2+C[/tex]
Multiply both sides of the ODE by [tex]e^{3\ln|s|-s^2}=s^3e^{-s^2}[/tex]:
[tex]s^3e^{-s^2}Y'(s)+(3s^2-2s^4)e^{-s^2}Y(s)=-14se^{-s^2}[/tex]
The left side condenses into the derivative of a product:
[tex]\left(s^3e^{-s^2}Y(s)\right)'=-14se^{-s^2}[/tex]
Integrate both sides and solve for [tex]Y(s)[/tex]:
[tex]s^3e^{-s^2}Y(s)=7e^{-s^2}+C[/tex]
[tex]Y(s)=\dfrac{7+Ce^{s^2}}{s^3}[/tex]
(b) Taking the inverse transform of both sides gives
[tex]y(t)=\dfrac{7t^2}2+C\,L^{-1}\left[\dfrac{e^{s^2}}{s^3}\right][/tex]
I don't know whether the remaining inverse transform can be resolved, but using the principle of superposition, we know that [tex]\frac{7t^2}2[/tex] is one solution to the original ODE.
[tex]y(t)=\dfrac{7t^2}2\implies y'(t)=7t\implies y''(t)=7[/tex]
Substitute these into the ODE to see everything checks out:
[tex]2\cdot7+t\cdot7t-2\cdot\dfrac{7t^2}2=14[/tex]
the dot plot above identifies the number of pets living with each of 20 families in an apartment building .what fraction of families have more than two pets
Answer:
B. ⅕
Step-by-step explanation:
Fraction of families having more than 2 pets = families with pets of 3 and above ÷ total number of families in the apartment
From the dot plot above, 3 families have 3 pets, and 1 family has 4 pets.
Number of families with more than 2 pets = 3 + 1 = 4
Fraction of families with more than 3 pets = [tex] \frac{4}{20} = \frac{1}{5} [/tex]
The fraction of families that have more than two pets is B. [tex]\frac{1}{5}[/tex]
Calculations and ParametersGiven that:
Fraction of families having more than 2 pets = families with pets of 3 and above/total number of families in the apartment
From the dot plot above:
3 families have 3 pets, 1 family has 4 pets.Number of families with more than 2 pets
= 3 + 1
= 4
Fraction of families with more than 3 pets = [tex]\frac{4}{20} = \frac{1}{5}[/tex]
Read more about dot plots here:
https://brainly.com/question/25957672
#SPJ5
Which of these functions could have been the graph shown below?
Answer:
B
Step-by-step explanation:
we take the only point we know
(0,20)
in A when x =0
[tex]f(x)=e^{20x} =e^{20*0}=1[/tex]
in B when x=0
[tex]f(x)=20e^x=20e^0=20*1=20[/tex]
fits
in C
[tex]f(x)=20^x=20^0=1[/tex]
in D
[tex]f(x)=20^{20x}=20^{20*0}=20^0=1[/tex]
so the only choice is B
150,75,50 what number comes next
Answer:
35 or 25
Step-by-step explanation:
A box contains12 balls in which 4 are white,3 blue and 5 are red.3 balls are drawn at random from the box.Find the chance that all three balls are of sifferent color.(answer in three decimal places)
[tex]|\Omega|=_{12}C_3=\dfrac{12!}{3!9!}=\dfrac{10\cdot11\cdot12}{2\cdot3}=220\\|A|=4\cdot3\cdot5=60\\\\P(A)=\dfrac{60}{220}=\dfrac{3}{11}\approx0,273[/tex]
Answer:
0.273
Step-by-step explanation:
Total number of balls is 4+3+5 = 12
There are 6 ways to draw 3 different colors (WBR, WRB, BWR, BRW, RWB, RBW) each with a chance of 4/12 · 3/11 · 5/10 = 1/22
So the total chance is 6 · 1/22 = 6/22 = 3/11 ≈ 0.273
The graph shown below expresses a radical function that can be written in the form f(x)=a(x+k)^1/n + c. What does the graph tell you about the value of n in this function?
Answer: n is a positive odd number.
Step-by-step explanation:
Ok, we know that the function is something like:
f(x)=a(x+k)^1/n + c
In the graph we can see two thigns:
All the values of the graph are positive values (even for the negative values of x), but in the left side we can see that the function decreases and is different than the right side.
So this is not an even function, then n must be an odd number (n odd allows us to have negative values for y = f(x) that happen when x + k is negative).
Also, we can see that the function increases, if n was a negative number, like: n = -N
we would have:
[tex]f(x) = \frac{a}{(x+k)^{1/N}} + c[/tex]
So in this case x is in the denominator, so as x increases, we would see that the value of y decreases, but that does not happen, so we can conclude that the value of n must be positive.
Then n is a positive odd number.
Answer:
D) Positive Even Integer
Step-by-step explanation:
just did it
A test-preparation company advertises that its training program raises SAT scores by an average of at least 30 points. A random sample of test-takers who had completed the training showed a mean increase smaller than 30 points.
(a) Write the hypotheses for a left-tailed test of the mean.
(b) Explain the consequences of a Type I error in this context.
Answer:
(a) Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \geq[/tex] 30 points
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < 30 points
(b) Type I error is that we conclude that test-takers who had completed the training showed a mean increase smaller than 30 points but in actual, the program raises SAT scores by an average of at least 30 points.
Step-by-step explanation:
We are given that a test-preparation company advertises that its training program raises SAT scores by an average of at least 30 points.
A random sample of test-takers who had completed the training showed a mean increase smaller than 30 points.
Let [tex]\mu[/tex] = average SAT score.
(a) So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \geq[/tex] 30 points
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < 30 points
Here, the null hypothesis states that the training program raises SAT scores by an average of at least 30 points.
On the other hand, the alternate hypothesis states that test-takers who had completed the training showed a mean increase smaller than 30 points.
(b) Type I error states the probability of rejecting the null hypothesis given the fact that null hypothesis is true.
According to the question, the Type I error is that we conclude that test-takers who had completed the training showed a mean increase smaller than 30 points but in actual, the program raises SAT scores by an average of at least 30 points.
The consequence of a Type I error is that we conclude the test-takers have low SAT scores but in actual they have an SAT score of at least 30 points.
Please help. I’ll mark you as brainliest if correct!
Answer:
(DNE,DNE)
Step-by-step explanation:
-24x-12y = -16. Equation one
6x +3y = 4. Equation two
Multiplying equation two with +4 gives
4(6x +3y = 4)
24x +12y = 16...result of equation two
-24x -12y= -16...
A careful observation to the following equation will help us notice that the both equation are same thing.
Multiplying minus to equation one gives
-(-24x-12y=-16)
24x+12y = 16.
Since the both equation are same, there is no solution to it.
Determine whether the sequence converges or diverges. If it converges, find the limit. (If an answer does not exist, enter DNE.)
an = (−3^n)/(4n!)
Answer:
[tex]a_{i} = \frac{(-3)^{i}}{4\cdot i!}[/tex] converges.
Step-by-step explanation:
The convergence analysis of this sequence is done by Ratio Test. That is to say:
[tex]r = \frac{a_{n+1}}{a_{n}}[/tex], where sequence converges if and only if [tex]|r| < 1[/tex].
Let be [tex]a_{i} = \frac{(-3)^{i}}{4\cdot i!}[/tex], the ratio for the expression is:
[tex]r =-\frac{3}{n+1}[/tex]
[tex]|r| = \frac{3}{n+1}[/tex]
Inasmuch [tex]n[/tex] becomes bigger, then [tex]r \longrightarrow 0[/tex]. Hence, [tex]a_{i} = \frac{(-3)^{i}}{4\cdot i!}[/tex] converges.
On an exam, the average score is 76 with a standard deviation of 6 points What is the probability that an individual chosen at random will have a score below 67 on this exam
Answer:
P [ X < 67 ] = 0,66,81 or 66,81 %
Step-by-step explanation:
We assume Normal Distribution N ( μ ; σ ) N ( 76 ; 6 )
z score for 67 is :
z(s) = ( X - μ ) /σ
z(s) = ( 67 - 76 ) / 6
z(s) = - 9 / 6
z(s) = - 1,5
with 1,5 we fnd n z-table area undr the curve α = 0,6681
Then P [ X < 67 ] = 0,66,81 or 66,81 %
Two boys and three girls are auditioning to play the piano for a school production. Two students will be chosen, one as the pianist, the other as the alternate.
What is the probability that the pianist will be a boy and the alternate will be a girl? Express your answer as a percent.
30%
40%
50%
60%
Answer:
30% is the correct answer.
Step-by-step explanation:
Total number of boys = 2
Total number of girls = 3
Total number of students = 5
To find:
Probability that the pianist will be a boy and the alternate will be a girl?
Solution:
Here we have to make 2 choices.
1st choice has to be boy (pianist) and 2nd choice has to be girl (alternate).
[tex]\bold{\text{Required probability }= P(\text{boy as pianist first}) \times P(\text{girl as alternate})}[/tex]
Formula for probability of an event E is given as:
[tex]P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}[/tex]
For [tex]P(\text{boy as pianist})[/tex], number of favorable cases are 2 (total number of boys).
Total number of cases = Total number of students i.e. 5
So, [tex]P(\text{boy as pianist})[/tex] is:
[tex]P(\text{boy as pianist}) = \dfrac{2}{5}[/tex]
For [tex]P(\text{girl as alternate})[/tex], number of favorable cases are 3 (total number of girls).
Now, one boy is already chosen as pianist so Total number of cases = Total number of students left i.e. (5 - 1) = 4
[tex]P(\text{girl as alternate}) = \dfrac{3}{4}[/tex]
So, the required probability is:
[tex]\text{Required probability } = \dfrac{2}{5}\times \dfrac{3}{4} = \dfrac{3}{10} = \bold{30\%}[/tex]
Answer:
30% A
Step-by-step explanation:
In this diagram, bac~edf. if the area of bac= 6 in.², what is the area of edf? PLZ HELP PLZ PLZ PLZ PLZ
Answer:
Area of ΔEDF = 2.7 in²
Step-by-step explanation:
It's given in the question,
ΔBAC ~ ΔEDF
In these similar triangles,
Scale factor of the sides = [tex]\frac{\text{Measure of one side of triangle BAC}}{\text{Measure of one side of triangle EDF}}[/tex]
[tex]=\frac{\text{BC}}{\text{EF}}[/tex]
[tex]=\frac{3}{2}[/tex]
Area scale factor = (Scale factor of the sides)²
[tex]\frac{\text{Area of triangle BAC}}{\text{Area of triangle EDF}}=(\frac{3}{2})^2[/tex]
[tex]\frac{6}{\text{Area of triangle EDF}}=(\frac{9}{4})[/tex]
Area of ΔEDF = [tex]\frac{6\times 4}{9}[/tex]
= 2.67
≈ 2.7 in²
Therefore, area of the ΔEDF is 2.7 in²
Was it evaluated correctly?
Explain your reasoning
help i need to turn it in a hour
Answer:
no
Step-by-step explanation:
2(4+10)+20
2(14)+20
28+20
48
A cube has an edge of 2 feet. The edge is increasing at the rate of 5 feet per minute. Express the volume of the cube as a function of m, the number of minutes elapsed.
Answer:
[tex]V(m) = (2 + 5m)^3[/tex]
Step-by-step explanation:
Given
Solid Shape = Cube
Edge = 2 feet
Increment = 5 feet per minute
Required
Determine volume as a function of minute
From the question, we have that the edge of the cube increases in a minute by 5 feet
This implies that,the edge will increase by 5m feet in m minutes;
Hence,
[tex]New\ Edge = 2 + 5m[/tex]
Volume of a cube is calculated as thus;
[tex]Volume = Edge^3[/tex]
Substitute 2 + 5m for Edge
[tex]Volume = (2 + 5m)^3[/tex]
Represent Volume as a function of m
[tex]V(m) = (2 + 5m)^3[/tex]
Question 18 i will maek the brainliest:)
Answer:
Median: 14.6, Q1: 6.1, Q3: 27.1, IR: 21, outliers: none
Step-by-step explanation:
Step 1: order the data from the least to the largest.
2.8, 3.9, 5.3, 6.1, 6.5, 7.1, 12.5, 14.6, 16.4, 16.4, 20.8, 27.1, 28.1, 30.9, 53.5
Step 2: find the median.
The median is the middle value, which is the 8th value in the data set.
2.8, 3.9, 5.3, 6.1, 6.5, 7.1, 12.5, [14.6,] 16.4, 16.4, 20.8, 27.1, 28.1, 30.9, 53.5
Median = 14.6
Step 2: Find Q1,
Q1 is the middle value of the lower part of the data set that is divided by the median to your left.
2.8, 3.9, 5.3, (6.1), 6.5, 7.1, 12.5, [14.6], 16.4, 16.4, 20.8, 27.1, 28.1, 30.9, 53.5
Q1 = 6.1
Step 3: find Q3.
Q3 is the middle value of the upper part of the given data set.
2.8, 3.9, 5.3, 6.1, 6.5, 7.1, 12.5, [14.6], 16.4, 16.4, 20.8, (27.1), 28.1, 30.9, 53.5
Q3 = 27.1
Step 4: find interquartile range (IR)
IR = Q3 - Q1 = [tex] 27.1 - 6.1 = 21 [/tex]
Step 5: check if there is any outlier.
Formula for checking for outlier = [tex] Q1 - 1.5*IR [/tex]
Then compare the result you get with the given values in the data set. Any value in the data set that is less than the result we get is considered an outlier.
Thus,
[tex] Q1 - 1.5*IR [/tex]
[tex]6.1 - 1.5*21 = -25.4[/tex]
There are no value in the given data set that is less than -25.4. Therefore, there is no outlier.
Find the length of segment YZ in the diagram below.
Answer:
2√2
Step-by-step explanation:
hope you understand.
Factor.
x2 + 11x
x2 + 11x
x(x + 11)
11(x + 11)
0(x2 + 11x)
Answer:
x(x + 11)
Step-by-step explanation:
x^2 + 11x when factored gives a result of x(x + 11)
Answer:
x(x+11)
Step-by-step explanation:
We are given the expression:
[tex]x^2+11x[/tex]
This can be factored using the Greatest Common Factor (GCF).
The GCF of x^2 and 11x is x.
Factor out an x.
[tex]x(x+11)[/tex]
x^2+11x factored is: x(x+11).
What is the sum of the geometric sequence?
Answer:
B. 259
Step-by-step explanation:
6^(i - 1) for i = 1 to 4
sum = 6^(1 - 1) + 6^(2 - 1) + 6^(3 - 1) + 6^(4 - 1) =
= 6^0 + 6^1 + 6^2 + 6^3
= 1 + 6 + 36 + 216
= 259
Answer: B. 259
What is the value of b.
C=25
s=9
B=4c-s2
Answer:
82
Step-by-step explanation:
Plug in the variables into the equation and solve for B but remember your order of operations when solving multiplication/division before addition/subtraction.
B = 4*25-9*2
B = 100-18
B = 82
Plaz guys help me on this question additional mathematics
Answer:
Step-by-step explanation:
vector OA=a
vector OB=b
vector OX= λ vector OA=λa
vector OY=μ vector OB=μb
a.
1.vector BX=(vector OX-vector OB)=λa-b
ii. vector AY=(vector OY-vector OA)=μb-a
b.
5 vector BP=2 vector BX
5(vector OP-vector OB)=2 (vector OX-vector OB)
5(vector OP-b)=2(λa-b)
5 vector OP-5b=2λa-2b
5 vector OP=2λa-2b+5b
vector OP=1/5(2λa+3b)
ii
complete it.
PLEASE HELP!!!!!!! FIRST CORRECT ANSWER WILL BE THE BRAINLIEST....PLEASE HELP
Lunch Choices of Students
The bar graph shows the percent of students that chose each food in the school
cafeteria. Which statement about the graph is true?
Answer:
(2) If 300 lunches were sold, then 120 chose tacos.
Step-by-step explanation:
We can evaluate each option and see if it makes it true.
For 1: If 200 lunches were served, 10 more students chose pizza over hotdogs.
We can find how many pizzas/hotdogs were given if 200 lunches were served by relating it to 100.
20% chose hotdog, which is [tex]\frac{20}{100}[/tex]. Multiply both the numerator and denominator by two: [tex]\frac{40}{200}[/tex] - so 40 students chose hotdogs.
Same logic for pizza: 30% chose pizza - [tex]\frac{30}{100} = \frac{60}{200}[/tex] so 60.
60 - 40 = 20, not 10, so 1 doesn't work.
2: If 300 lunches were sold, then 120 chose tacos.
Let's set up a proportion again. 40% of 100 is 40.
[tex]\frac{40}{100} = \frac{40\cdot3}{300} = \frac{120}{300}[/tex]
So 120 tacos were chosen - yes this works!
Hope this helped!
A string passing over a smooth pulley carries a stone at one end. While its other end is attached to a vibrating tuning fork and the string vibrates forming 8 loops. When the stone is immersed in water 10 loops are formed. The specific gravity of the stone is close to
A) 1.8
B) 4.2
C) 2.8
D) 3.2
Answer:
correct option is C) 2.8
Step-by-step explanation:
given data
string vibrates form = 8 loops
in water loop formed = 10 loops
solution
we consider mass of stone = m
string length = l
frequency of tuning = f
volume = v
density of stone = [tex]\rho[/tex]
case (1)
when 8 loop form with 2 adjacent node is [tex]\frac{\lambda }{2}[/tex]
so here
[tex]l = \frac{8 \lambda _1}{2}[/tex] ..............1
[tex]l = 4 \lambda_1\\\\\lambda_1 = \frac{l}{4}[/tex]
and we know velocity is express as
velocity = frequency × wavelength .....................2
[tex]\sqrt{\frac{Tension}{mass\ per\ unit \length }}[/tex] = f × [tex]\lambda_1[/tex]
here tension = mg
so
[tex]\sqrt{\frac{mg}{\mu}}[/tex] = f × [tex]\lambda_1[/tex] ..........................3
and
case (2)
when 8 loop form with 2 adjacent node is [tex]\frac{\lambda }{2}[/tex]
[tex]l = \frac{10 \lambda _1}{2}[/tex] ..............4
[tex]l = 5 \lambda_1\\\\\lambda_1 = \frac{l}{5}[/tex]
when block is immersed
equilibrium eq will be
Tenion + force of buoyancy = mg
T + v × [tex]\rho[/tex] × g = mg
and
T = v × [tex]\rho[/tex] - v × [tex]\rho[/tex] × g
from equation 2
f × [tex]\lambda_2[/tex] = f × [tex]\frac{1}{5}[/tex]
[tex]\sqrt{\frac{v\rho _{stone} g - v\rho _{water} g}{\mu}} = f \times \frac{1}{5}[/tex] .......................5
now we divide eq 5 by the eq 3
[tex]\sqrt{\frac{vg (\rho _{stone} - \rho _{water})}{\mu vg \times \rho _{stone}}} = \frac{fl}{5} \times \frac{4}{fl}[/tex]
solve irt we get
[tex]1 - \frac{\rho _{stone}}{\rho _{water}} = \frac{16}{25}[/tex]
so
relative density [tex]\frac{\rho _{stone}}{\rho _{water}} = \frac{25}{9}[/tex]
relative density = 2.78 ≈ 2.8
so correct option is C) 2.8
Find the value of the logarithm. log 122
Answer:
2.086
Step-by-step explanation:
Log 122 is equal to 2.086
A portion of the quadratic formula proof is shown. Fill in the missing reason.
Answer:
Find a common denominator on the right side of the equation
Step-by-step explanation:
The equation before the problem is
X² + b/a(x) + (b/2a)²= -c/a + b²/4a²
The next step in solving the above equation is to fibd tge common denominator on the right side of the equation.
X² + b/a(x) + (b/2a)²= -c/a + b²/4a²
X² + b/a(x) + (b/2a)²= -4ac/4a² + b²/4a²
X² + b/a(x) + (b/2a)²=( b²-4ac)/4a²
The right side of the equation now has a common denominator
The next step is to factorize the left side of the equation.
(X+b/2a)²= ( b²-4ac)/4a²
Squaring both sides
X+b/2a= √(b²-4ac)/√4a²
Final equation
X=( -b+√(b²-4ac))/2a
Or
X=( -b-√(b²-4ac))/2a
How many times does the digit 9 appear in the list of all integers from 1 to 500? (The number $ 99 $, for example, is counted twice, because $9$ appears two times in it.)
Answer:
95 times digit 9 appears in all integers from 1 to 500.
Step-by-step explanation:
No. of 9 from
1-9: 1 time
10-19: 1 time
20-29: 1 time
30-39: 1 time
40-49: 1 time
50-59: 1 time
60-69: 1 time
70-79: 1 time
80-89 : 1 time
from 90 to 99
there will be one in 91 to 98
then two 9 in 99
thus, no of 9 from 90 to 99 is 10
Thus, total 9's from 1 to 99 is 9+10 = 19
Thus there 19 9's in 1 to 99
similarly
there will be
19 9's in 100 to 199
19 9's in 200 to 299
19 9's in 300 to 399
19 9's in 400 to 499
Thus, total 9's will be
19 + 19 + 19+ 19 + 19 + 19 = 95
Thus, 95 times digit 9 appears in all integers from 1 to 500.
