Answer:
(Y∩Z)∩X = {r,a,e}
Step-by-step explanation:
U = {a,b,c,d,e,f,............,z)
X = { t,r,i,a,n,g,l,e}
Y = {t,r,a,p,e,z,o,i,d}
Z = {h,y,p,e,r,b,o,l,a}
Firstly,
Y∩Z = {t,r,a,p,e,z,o,i,d} ∩ {h,y,p,e,r,b,o,l,a} [Intersection : Common Elements]
=> Y∩Z = {r,a,p,e,o}
Now,
(Y∩Z)∩X = {r,a,p,e,o} ∩ { t,r,i,a,n,g,l,e}
=> (Y∩Z)∩X = {r,a,e}
There are 3 times as many novels as comic books in a bookstore.If there are 2480 books altogether, how many comic books are there in the bookstore.
Answer:
620 comic books
2480 / 4 is 620.
620 x 3 is 1860.
1860 + 620 is 2480.
Done!
As a young professional in the future, it is good to start saving money to have a
security for the future and have something to use when emergency cases happen
related to the need of cash. The Philippine Average Family income last 2015 was
around P276,000 per year. Lets say every year you earn the same amount and save
P76,000 yearly in your frusted bank giving 4% compounded interest annually. How much
will the account worth in the future after 40 years?
a. Solve for the future value of the account:
FV=PMT
( \frac{(1 + i) ^{n} - 1 }{i} )(i(1+i)n−1)
Answer:
Step-by-step explanation:
We would apply the future value which is expressed as
FV = C × [{(1 + r)^n - 1}/r]
Where
C represents the yearly payments of the young professional.
FV represents the amount of money
in your account at the end of 40 years.
r represents the annual rate.
n represents number of years or period.
From the information given,
r = 4% = 4/100 = 0.04
C = $76000
n = 40 years
Therefore,
FV = 76000 × [{(1 + 0.04)^40 - 1}/0.04]
FV = 76000 × [{4.8 - 1}/0.04]
FV = 76000 × 95
FV = P7220000
EXAMPLE 3 If f(x, y) = 4xy2 7x2 + y4 , does lim (x, y)→(0, 0) f(x, y) exist? SOLUTION Let's try to save some time by letting (x, y) → (0, 0) along any nonvertical line through the origin. Then y = mx, where m is the slope, and f(x, y) = f(x, mx) = 4x 2 7x2 + (mx)4 = 7x2 + m4x4 = 7 + m4x2 .
Answer:
Limit of the function exists.Step-by-step explanation:
Given the function f(x,y) = [tex]\frac{4xy^{2} }{7x^{2} + y^{4} }[/tex], we are to show that lim (x, y)→(0, 0) f(x, y) exist. To show that, the following steps must be followed.
[tex]\lim_{(x,y) \to (0,0)} \frac{4xy^{2} }{7x^{2} + y^{4} }\\[/tex]
substituting the limit x = 0 and y = 0 into the function we have;
[tex]\frac{4(0)^{2} }{7(0)^{2} + (0)^{4} }\\= \frac{0}{0} (indeterminate)[/tex]
Since we got an indeterminate function, we will then substitute y = mx into the function as shown;
[tex]\lim_{(x,mx) \to (0,0)} \frac{4x(mx)^{2} }{7x^{2} + (mx)^{4} }\\\lim_{(x,mx) \to (0,0)} \frac{4m^{2} x^{3} }{7x^{2} + m^{4}x^{4} }\\\\\lim_{(x,mx) \to (0,0)} \frac{4m^{2} x^{3} }{x^{2}(7 + m^{4} x^{2}) }\\\lim_{(x,mx) \to (0,0)} \frac{4m^{2}x }{7 + m^{4} x^{2} }[/tex]
Substituting x = 0 , the limit of the function becomes;
[tex]\frac{4m^{2}(0) }{7 + m^{4} (0)^{2} }\\= \frac{0}{7}\\ = 0[/tex]
Since the limit of the function gives a finite value of 0 (the limit tends to 0). This shows that the limit exists.
Find the lowest common denominator. 1/(x+2)^2, 1/(x-2)^2, 2/(x^2-4) A. (x+2)^2 (x-2)^2 B. (x^2+2) (x^2-2)
Answer:
(x + 2) ^2 (x - 2) ^ 2 so A is the answer.
Step-by-step explanation:
Answer:
a
Step-by-step explanation:
Una persona se dirige a un edificio y observa lo alto del mismo con un ángulo de elevación “x”, después de caminar 10m observa al mismo punto anterior con ángulo de elevación “y”, si la altura del edificio es de 30m. Calcule: "3Tgx.Ctgy + Tgx"
Answer:
3
Step-by-step explanation:
To begin with notice that
[tex]\displaymode{ \tan(x) = \frac{30}{10 + 30\cot(y)} }[/tex]
From that equation you get that
10 tan(x) + 30tan(x) cot(x) = 30
therefore
tan(x) + 3 tan(x) cot(x) = 3
Round off 3. 55 to one significant figure
Answer:
3.6
Step-by-step explanation:
We must first clarify how a number is rounded.
To round a number to unity we have to look at the first number after the comma.
If this number is less than 5 (1, 2, 3, 4) we should not do anything, but if that number is 5 or greater (5, 6, 7, 8, 9) we must add a unit to the number.
That is to say:
<5 do nothing
=> 5 round to the next number (+1)
So in the case of 3.55 it would be.
3.55 = 3.6
The rat population in major metropolitan city is given by the formula n(t)=89e^0.02t where t is measured in years since 1991 and n(t) is measured in millions. What does the model predict the rat population was in the year 2007?
Answer:
122.6 million
Step-by-step explanation:
Figure the value of t, then put that into the formula and do the arithmetic.
t = 2007 =1991 = 16
n(16) = 89e^(0.02·16) = 89·e^0.32 ≈ 122.6
The model predicts a rat population of 122.6 million in 2007.
If f(x) = the square root of X,
g(x) = x - 7. Then dom(fog) =
(a) [0, infinity)
(b) R
(c) (-7,infinity)
(d) [7,infinity)
(e) none
Answer:
The answer is option D.
Hope this helps you
Two ships leave a port at the same time.
Ship A sails 12 knots on a bearing of 035°
Ship B sails 16 knots on a bearing of 270°
Calculate the distance between the ships after 2 hours
(1 knot = 1 nautical mile per hour)
Answer:
49.8 nautical miles
Step-by-step explanation:
Recall that speed = distance/time
Time = 2hours
Speed = 12knots and 16 knots respectively
D = speed×time
D1 = 12×2 = 24
D2 = 16×2 = 32
Using the 'cosine rule' we have:
a^2 = b^2+c^2-2bc cos Θ
Where a =?
b =24
c = 32
Θ = 125°
a² = 24² + 32² - 2(24)(32)cos125°
a^2 = 576+1024 - 1536cos125°
a² = 1600 - 1536(-0.57357)
a² = 1600+881.0134
a² = 2481.0134
Then, a² = 2481.013406
a =√2481.013406
Hence, a = 49.8 nautical miles
In this exercise we must use the knowledge about triangles to calculate the distance that a ship will travel, in this way we find that:
49.8 nautical miles
First, remember the formula for distance, which is:
[tex]Speed = distance/time[/tex]
And knowing that the data reported in the exercise are:
Time = 2hours Speed = 12 knots and 16 knots respectively
So putting the values informed in the distance formula, we have:
[tex]D = speed*time\\D_1 = 12*2 = 24\\D_2 = 16*2 = 32[/tex]
Using the 'cosine rule' we have:
[tex]a^2 = b^2+c^2-2bc cos \theta[/tex]
Find the a, will have:
[tex]b =24 \ \ \ c = 32 \ \ \ \theta = 125\\a^2 = 24^2 + 32^2 - 2(24)(32)cos125\\a^2 = 576+1024 - 1536cos125\\a^2 = 1600 - 1536(-0.57357)\\a^2 = 1600+881.0134\\a^2 = 2481.0134\\a=49.8[/tex]
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HELP ASAP PLEASE!!!! 1. Create a unique equation using the graphic above with the solution (answer) of 36. There are many possible equations you could create to equal 36 using the chart above. 2. Describe the process you used to create your equation in prompt 1.
Answer:
4(A)=36
Step-by-step explanation:
An equation is said to be unique if the equation has a valid solution. Several unique equations that equal 36 can be taken from the given values. Some of these equations are:
[tex]4G + M + 2B = 36[/tex]
[tex]10F- N -D = 36[/tex]
[tex]10D + G =36[/tex]
[tex]4M - E = 36[/tex]
[tex]4L = 36[/tex]
[tex]9E = 36[/tex]
To get a unique equation, we ensure that the right-hand side of the equation equals the left-hand side. There are no quick solution to this; so, we make use of trial by error method.
After several trials, some possible equations with three variables are:
[tex]4G + M + 2B = 36[/tex] because [tex]4 \times 6 + 10 + 2 \times 1 = 36[/tex]
[tex]10F- N -D = 36[/tex] because [tex]10 \times 5- 11 -3 = 36[/tex]
Some possible equations with two variables are:
[tex]10D + G =36[/tex] because [tex]10 \times 3 + 6 =36[/tex]
[tex]4M - E = 36[/tex] because [tex]4 \times 10 - 4 = 36[/tex]
Some possible equations with two variables are:
[tex]4L = 36[/tex] because [tex]4 \times 9 = 36[/tex]
[tex]9E = 36[/tex] because [tex]9 \times 4 = 36[/tex]
The following are some unique equations from the given values:
[tex]4G + M + 2B = 36[/tex]
[tex]10F- N -D = 36[/tex]
[tex]10D + G =36[/tex]
[tex]4M - E = 36[/tex]
[tex]4L = 36[/tex]
[tex]9E = 36[/tex]
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Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that Rn(x) → 0.] f(x) = 10 x , a = −2
Answer:
[tex]\sum^\infty_{n=0} -5 (\frac{x+2}{2})^n[/tex]
Step-by-step explanation:
Rn(x) →0
f(x) = 10/x
a = -2
Taylor series for the function f at the number a is:
[tex]f(x) = \sum^\infty_{n=0} \frac{f^{(n)}(a)}{n!} (x - a)^n[/tex]
[tex]f(x) = f(a) + \frac{f'(a)}{1!}(x-a)+\frac{f"(a)}{2!} (x-a)^2 + ...[/tex] ............ equation (1)
Now we will find the function f and all derivatives of the function f at a = -2
f(x) = 10/x f(-2) = 10/-2
f'(x) = -10/x² f'(-2) = -10/(-2)²
f"(x) = -10.2/x³ f"(-2) = -10.2/(-2)³
f"'(x) = -10.2.3/x⁴ f'"(-2) = -10.2.3/(-2)⁴
f""(x) = -10.2.3.4/x⁵ f""(-2) = -10.2.3.4/(-2)⁵
∴ The Taylor series for the function f at a = -4 means that we substitute the value of each function into equation (1)
So, we get [tex]\sum^\infty_{n=0} - \frac{10(x+2)^n}{2^{n+1}}[/tex] Or [tex]\sum^\infty_{n=0} -5 (\frac{x+2}{2})^n[/tex]
Taylor series is a power series that gives the expansion of a function f (x) in the neighborhood of a point.
Taylor series is, [tex]f(x)=f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^{2}+........[/tex]
[tex]f(x)=\sum_{n=0}^{\infty }-\frac{10(x+2)^{n}}{2^{n+1}}[/tex]
Here, f(x) = 1/x and a = -2
Now find derivative,
f(x) = 10/x f(-2) = 10/-2
f'(x) = -10/x² f'(-2) = -10/(-2)²
f"(x) = 10.2/x³ f"(-2) = 10.2/(-2)³
f"'(x) = -10.2.3/x⁴ f'"(-2) = -10.2.3/(-2)⁴
Substituting above values in Taylor series expansion.
We get, [tex]f(x)=\sum_{n=0}^{\infty }-\frac{10(x+2)^{n}}{2^{n+1}}[/tex]
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6th grade math :) help me please !
Answer:
a1- 2 terms
a2- 6x, 9y
a3- 6,9
b1-2 terms
b2- 7x, 12x
b3- 7,12
c1- 3 terms
c2- 5x^2, 2y, 3
c3- 5,2,3
Step-by-step explanation:
Not sure how I would solve this
Answer:
0
Step-by-step explanation:
Using the slope formula
m = (y2-y1)/(x2-x1)
and the given points
m = ( 2-2)/ ( 3-4)
= 0/ -1
= 0
The slope is zero
Find the hypotenuse of a right triangle (in cm) if one leg measures 7 cm and the other leg measures 11 cm. Round to the nearest thousandth. ____________ cm
Using the Pythagorean theorem
Hypotenuse = sqrt( 11^2 + 7^2)
= sqrt( 121 + 39)
= sqrt( 160)
= 12.694
The ratio of the number of counters that Amy is 5:6 Amy gives 3/5 of her counters to Audrey. Amy has 126 counters left how many more counters does Audrey now have than Amy?
Answer:
441
Step-by-step explanation:
Given
ratio of the number of counters that Amy is 5:6
let the no. of counter with Amy be 5x
and the no. of counter with Audrey be 6x
Amy gives 3/5 of her counters to Audrey.
3/5 of her counters to Audrey. = 3/5 * 5x = 3x
Thus,
amy gives 3x counters to audrey
No. of counters left with Amy = 5x-3x = 2x
given that Amy has 126 counters left
2x = 126
x = 126/2 = 63
Thus,
no. of counter with Amy now = 2x = 2*63 = 126
and now the no. of counter with Audrey = 6x+3x = 9*63 = 567
Difference of counter between Audrey and Amy = 567 - 126 = 441.
Thus, Audrey has 441 counter more than Any has.
PLEASE I NEED HELP!!! FIRST ANSWER IS BRANLIEST!!!
Apply the distributive property to create an equivalent expression. 6(5x-3)
Answer:
30x - 18
Step-by-step explanation:
6(5x - 3)
Apply the distributive property.
6(5x) + 6(-3)
30x + - 18
Answer:
30x - 18 is your final answer
Kelly's first four test grades of the period were 80, 72, 96, and 88. Which inequality represents the grades she can
earn on the fifth test to have a test average of no less than 80?
V
O gs16
O 92 16
O g564
O 9264
Answer: option D on edge 2020
Step-by-step explanation:
if you reverse the formula for mean, then you just insert the numbers and you have your answer
what is the slope of the line given by the equation?
Answer:
the slope is 8
Step-by-step explanation:
The coefficient of x, "m" is the slope in the slope-intercept form:
y = mx + b
What is 200 percent of (0.020(5/4) + 3 ((1/5) - (1/4)))
Answer:
0.1/4-3/20=1/40-6/40=-1/8 200% of this is -1/4
Step-by-step explanation:
This shows that 200 percent of (0.020(5/4) + 3 ((1/5) - (1/4))) is -2.75
Given the expression as shown in the question:
[tex]200\% \ of \ [0.020(\frac{5}{4} )+3(\frac{1}{5}- \frac{1}{4} )][/tex]
Expand the expression in the square bracket using the distribution law as shown:
[tex]=200\% \ of \ [0.020(\frac{5}{4} )+3(\frac{4-5}{20} )]\\=200\% \ of \ [0.020(\frac{5}{4} )+3(\frac{-1}{20} )]\\=200\% \ of \ [0.020(\frac{5}{4} )-(\frac{3}{20} )]\\=200\% \ of \ [0.020(1.25 )-\frac{3}{20}]\\=\frac{200}{100} \times [0.025-0.15]\\=2 \times [-0.125]\\=-2.75[/tex]
Hence the correct answer to the expression is -2.75.
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translate into an algebraic expression: 120 is increased by d % and increased by 25% . What is the result ?
Answer:
1.5d+150
Step-by-step explanation:
125 since it increased by 25%. look at the problem below:
125( 1 + 0.01d) then you plug those numbers in and theres your answer :00!!
sammy-6th grade-
The algebraic expression is 1.5d+150.
What is an expression?Expression in maths is defined as the collection of the numbers variables and functions by using signs like addition, subtraction, multiplication, and division.
Given that 120 is increased by d % and increased by 25% then the expression will be written as below.
E = 1.5d+150.
Hence the algebraic expression is 1.5d+150.
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A total of 259 tickets were sold to a benefit concert for a total of $5,312. Two types of tickets were sold: adult tickets were sold for $24 each and student tickets were sold for $16 each. How many student tickets were sold?
Answer:
113
Step-by-step explanation:
Let the number of adult tickets sold =a
Let the number of student tickets sold =s
A total of 259 tickets were sold, therefore:
a+s=259
Adult tickets were sold for $24 each and student tickets were sold for $16 each.
Total Revenue = $5,312
Therefore:
24a+16s=5,312
We solve the two derived equations simultaneously.
From the first equation
a=259-s
Substitute a=259-s into 24a+16s=5,312
24(259-s)+16s=5,312
6216-24s+16s=5,312
-8s=5,312-6216
-8s=-904
Divide both sides by -8
s=113
Therefore, 113 student tickets were sold.
16. Find m<2.
a. 86°
b. 43°
C. 94°
d. 133
write 26 as repeated multiplication
Answer:
2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
2¹³
13 x 13
Step-by-step explanation:
We simply find numbers that can multiply to 26 and write out the multiplication to get our answer.
Reliance on solid biomass fuel for cooking and heating exposes many children from developing countries to high levels of indoor air pollution. The article "Domestic Fuels, Indoor Air Pollution, and Children's Health" (Annals of the N.Y. Academy of Sciences, 2008: 209-217) pm-tented information on various pulmonary characteristics in samples of children whose households in India used either biomass fuel or liquefied petroleum gas (LPG). For the 755 children in biomass households, the sample mean peak expiratory flow (a person's maximum speed of expiration) was 3.30 Us, and the sample standard deviation was 1.20. For the 750 children whose households used liquefied petroleum gas, the sample mean PEF was 4.25 and the sample standard deviation was 1.75.
a. Calculate a confidence interval at the 95% confidence level for the population mean PEF for children in biomass households and then do likewise for children in LPG households. What is the simultaneous confidence level for the two intervals?
b. Carry out a test of hypotheses at significance level .01 to decide whether true average PEF is lower for children in biomass households than it is for children in LPG households (the cited article included a P-value for this test).
c. FEV1, the forced expiratory volume in 1 second, is another measure of pulmonary function. The cited article reported that for the biomass households the sample mean FEY, was 2.3 L/s and the sample standard deviation was .5 L/s. If this information is used to compute a 95% CI for population mean FEV1, would the simultaneous confidence level for this interval and the first interval calculated in (a) be the same as the simultaneous confidence level deter-mined there? Explain.
Answer:
A) 95% confidence interval for the population mean PEF for children in biomass households = (3.214, 3.386)
95% confidence interval for the population mean PEF for children in LPG households
= (4.125, 4.375)
Simultaneous confidence interval for both = (3.214, 4.375)
B) The result of the hypothesis test is significant, hence, the true average PEF is lower for children in biomass households than it is for children in LPG households.
C) 95% confidence interval for the population mean FEY for children in biomass households = (2.264, 2.336)
Simultaneous confidence interval for both = (2.264, 4.375)
This simultaneous interval cannot be the same as that calculated in (a) above because the sample mean obtained for children in biomass households here (using FEY) is much lower than that obtained using PEF in (a).
Step-by-step explanation:
A) Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.
Mathematically,
Confidence Interval = (Sample mean) ± (Margin of error)
Margin of Error is the width of the confidence interval about the mean.
It is given mathematically as,
Margin of Error = (Critical value) × (standard Error of the mean)
Critical value will be obtained using the z-distribution. This is because although, there is no information provided for the population standard deviation, the sample sizes are large enough for the sample properties to approximate the population properties.
Finding the critical value from the z-tables,
Significance level for 95% confidence interval
= (100% - 95%)/2 = 2.5% = 0.025
z (0.025) = 1.960 (from the z-tables)
For the children in the biomass households
Sample mean = 3.30
Standard error of the mean = σₓ = (σ/√N)
σ = standard deviation of the sample = 1.20
N = sample size = 755
σₓ = (1.20/√755) = 0.0436724715 = 0.04367
95% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]
CI = 3.30 ± (1.960 × 0.04367)
CI = 3.30 ± 0.085598
95% CI = (3.214402, 3.385598)
95% Confidence interval = (3.214, 3.386)
For the children in the LPG households
Sample mean = 4.25
Standard error of the mean = σₓ = (σ/√N)
σ = standard deviation of the sample = 1.75
N = sample size = 750
σₓ = (1.75/√750) = 0.063900965 = 0.063901
95% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]
CI = 4.25 ± (1.960 × 0.063901)
CI = 4.25 ± 0.125246
95% CI = (4.12475404, 4.37524596)
95% Confidence interval = (4.125, 4.375)
Simultaneous confidence interval for both = (3.214, 4.375)
B) The null hypothesis usually goes against the claim we are trying to test and would be that the true average PEF for children in biomass households is not lower than that of children in LPG households.
The alternative hypothesis confirms the claim we are testing and is that the true average PEF is lower for children in biomass households than it is for children in LPG households.
Mathematically, if the true average PEF for children in biomass households is μ₁, the true average PEF for children in LPG households is μ₂ and the difference is μ = μ₁ - μ₂
The null hypothesis is
H₀: μ ≥ 0 or μ₁ ≥ μ₂
The alternative hypothesis is
Hₐ: μ < 0 or μ₁ < μ₂
Test statistic for 2 sample mean data is given as
Test statistic = (μ₂ - μ₁)/σ
σ = √[(s₂²/n₂) + (s₁²/n₁)]
μ₁ = 3.30
n₁ = 755
s₁ = 1.20
μ₂ = 4.25
n₂ = 750
s₂ = 1.75
σ = √[(1.20²/755) + (1.75²/750)] = 0.07740
z = (3.30 - 4.25) ÷ 0.07740 = -12.27
checking the tables for the p-value of this z-statistic
Significance level = 0.01
The hypothesis test uses a one-tailed condition because we're testing in only one direction.
p-value (for z = -12.27, at 0.01 significance level, with a one tailed condition) = < 0.000000001
The interpretation of p-values is that
When the p-value > significance level, we fail to reject the null hypothesis and when the p-value < significance level, we reject the null hypothesis and accept the alternative hypothesis.
Significance level = 0.01
p-value = 0.000000001
0.000000001 < 0.01
Hence,
p-value < significance level
This means that we reject the null hypothesis, accept the alternative hypothesis & say that true average PEF is lower for children in biomass households than it is for children in LPG households.
C) For FEY for biomass households,
Sample mean = 2.3 L/s
Standard error of the mean = σₓ = (σ/√N)
σ = standard deviation = 0.5
N = sample size = 755
σₓ = (0.5/√755) = 0.0182
95% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]
CI = 2.30 ± (1.960 × 0.0182)
CI = 2.30 ± 0.03567
95% CI = (2.264, 2.336)
Simultaneous confidence interval for both = (2.264, 4.375)
This simultaneous interval cannot be the same as that calculated in (a) above because the sample mean obtained for children in biomass households here (using FEY) is much lower than that obtained using PEF in (a).
Hope this Helps!!!
is this right one more i think lol
Answer:
Yup P is the right one having 62.26%
Step-by-step explanation:
Answer:
Yes
Step-by-step explanation:
We can set it up as
P = 33/53
Q = 20/48
R = 54/90
S = 44/83
This is because we are calculating the percent of yellow birds in the total amt. of birds in a specified park.
Now we calculate =>
P = 33/53 = around 0.62
Q = 20/48 = around 0.416
R = 54/90 = 0.6
S = 44/83 = around 0.53
We find that Park P has the greatest percentage and -->
Thus, Park P is our answer and yes, you are correct.
Ava's bacteria population is modeled by an equation. Chase models his bacteria
population with a graph. Ava says that on day 14, she will have more bacteria than Chase
Is she right? Why or why not?
Answer:
P(Chase) > P(Ava)
700 > 587
Therefore, Ava's claim is wrong!
On day 14, Chase's bacteria population will be greater than Ava's bacteria population.
Step-by-step explanation:
Please refer to the attached image.
Ava's bacteria population is modeled by the following equation.
[tex]$ b(t) = 200(1+0.08)^t $[/tex]
Where t is time in days and b(t) is the population of the bacteria after t days.
The graph represents the population of Chase's bacteria.
Ava claims that on day 14, she will have more bacteria than Chase.
Let us compare the population of both bacteria.
Chase bacteria population when t = 14 days:
From the graph, the population is approximately 700 at t = 14 days
P(Chase) ≈ 700
Ava bacteria population when t = 14 days:
at t = 14 days
[tex]b(t) = 200(1+0.08)^t \\\\ b(14) = 200(1.08)^{14} \\\\ b(14) = 200 (2.93719)\\\\ b(14) = 587.44[/tex]
So, the population is approximately 587 at t = 14 days
P(Ava) ≈ 587
P(Chase) > P(Ava)
700 > 587
Therefore, Ava's claim is wrong!
On day 14, Chase's bacteria population will be greater than Ava's bacteria population.
Answer:
D
Step-by-step explanation:
Trust
I need help with this exercise please. What I need to know is the measure of one side of the square. Sorry for the language of the exercise
Answer:
[tex]\boxed{\sf \ \ x=6 \ \ }[/tex]
Step-by-step explanation:
Hello,
BMN and NPC are two similar triangles so
[tex]\dfrac{BM}{MN}=\dfrac{NP}{PC}\\\\ \dfrac{10-x}{x}=\dfrac{x}{15-x}\\\\<=>(15-x)(10-x)=x^2\\\\<=>150-25x+x^2=x^2\\\\<=>25x=150\\\\<=>x=6[/tex]
espero que esto ayude
When we carry out a chi-square goodness-of-fit test for a normal distribution, the null hypothesis states that the population:________
a. Does not have a normal distribution
b. Has a normal distribution
c. Has a chi-square distribution
d. Does not have a chi-square distribution
e. Has k-3 degrees of freedom
Answer:
Option B
Step-by-step explanation:
The null hypothesis for a chi-square goodness of fit test states that the data are consistent with a specified distribution.
While the alternative hypothesis states that the data are not consistent with a specified distribution.
In this case study, the test is for a nose distribution. Thus the null hypothesis would be that the population has a normal distribution.
what is -34/15 in decimal form
Answer:
2.26 repeating
Step-by-step explanation:
Convert the fraction to a decimal by dividing the numerator by the denominator.
hope this helpes
be sure to give brainliest
HELP ASAP!!!!!!!!!!!!!!!
Answer:
B) (x^2-2x)/(2x+6)
Step-by-step explanation:
When dividing fractions, it's just like multiplying the reciprocal of the second fraction : x/y divided by z/w = x/y * w/z.
We use that and we get (x-2)/(x+3) * x/2. we simplify and get x(x-2)/2(x+3). So Our answer is B