The number of yeast cells in a laboratory culture increases rapidly initially but levels off eventually. The population is modeled by the function n = f(t) = a 1 + be−0.7t where t is measured in hours. At time t = 0 the population is 30 cells and is increasing at a rate of 18 cells/hour. Find the values of a and b.

Answer:

I'm not completely sure what you mean by a, "single fraction," but I'm pretty sure the answer you are looking for is [tex]\frac{5-4}{a-b}[/tex]

Step-by-step explanation:

Answer:

The probability that the sample mean is less than 50 points = 0.002    

Step-by-step explanation:

Step(i):-

Given mean of the normal distribution = 56 points

Given standard deviation of the normal distribution = 12 points

Random sample size 'n' = 36 games

Step(ii):-

Let x⁻ be the random variable of normal distribution

Let x⁻ = 50

[tex]Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }[/tex]

[tex]Z = \frac{50-56 }{\frac{12}{\sqrt{36} } }= -3[/tex]

The probability that the sample mean is less than 50 points

P( x⁻≤ 50) = P( Z≤-3)

                = 0.5 - P(-3 <z<0)

               = 0.5 -P(0<z<3)

               =  0.5 - 0.498

               = 0.002

Final answer:-

The probability that the sample mean is less than 50 points = 0.002

Answer:

56

2

.001

Step-by-step explanation:

The Central Limit Theorem for Means states that the mean of any sampling distribution of the means is equal to the mean of the population distribution. The standard deviation is equal to the standard deviation of the population divided by the square root of the sample size. So, the mean of this sampling distribution of the means with sample size 36 is 56 points and the standard deviation is 1236√=2 points. The z-score for 50 using the formula z=x¯¯¯−μσ is −3.

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

-3.0 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001

-2.9 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.001 0.001 0.001

-2.8 0.003 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002

-2.7 0.003 0.003 0.003 0.003 0.003 0.003 0.003 0.003 0.003 0.003

-2.6 0.005 0.005 0.004 0.004 0.004 0.004 0.004 0.004 0.004 0.004

-2.5 0.006 0.006 0.006 0.006 0.006 0.005 0.005 0.005 0.005 0.005

Using the Standard Normal Table, the area to the left of −3 is approximately 0.001. Therefore, the probability that the sample mean will be less than 50 points is approximately 0.001.

Answer:

Step-by-step explanation:

Answer:

B

Step-by-step explanation:

Answer:

Determine if the expressions are equivalent.

when w = 11:

2w + 3 + 4     4 + 2w + 3

2(11) + 3 + 4    4 + 2(11) + 3

22 + 3 + 4      4 + 22 + 3

25 + 4      26 + 3

29        29

Complete the statements.

Now, check another value for the variable.

When w = 2, the first expression is  

11

.

When w = 2, the second expression is  

11

.

Therefore, the expressions are  

equivalent

.

Step-by-step explanation:

i did the math hope this helps

Answer:

Hii its Nat here to help! :)

Step-by-step explanation: A is 11 and b is 11.

C is Equal

Screenshot included.

Answer:

Just replace the variables with the number

d5

c4 (uh oh)

a2

b-3

f-7

d-c = 5 - 4 = 1

1/3 - 4(ab+f)

2 x -3 = -6

-6 + -7 = -13

-13 x 4 = -52

1/3 - -52 = 1/3 + 52 =

52 1/3

Hope this helps

Step-by-step explanation:

Answer:

[tex]z=-\frac{20}{3}[/tex]

Step-by-step explanation:

[tex]\frac{3z}{10}-4=-6\\\\\frac{3z}{10}-4+4=-6+4\\\\\frac{3z}{10}=-2\\\\\frac{10\cdot \:3z}{10}=10\left(-2\right)\\\\3z=-20\\\\\frac{3z}{3}=\frac{-20}{3}\\\\z=-\frac{20}{3}[/tex]

Best Regards!

Answer:

a) Probability that a team will win the match given that it has won the first game = 0.66

b) Probability that a team will win the match given that it has won the first two games= 0.81

c) Probability that a team will win the match, given that it has won two out of the first three games = 0.69

Step-by-step explanation:

There are a total of seven games to be played. Therefore, W and L consists of 2⁷ equi-probable sample points

a) Since one game has already been won by the team, there are 2⁶ = 64 sample points left. If the team wins three or more matches, it has won.

Number of ways of winning the three or more matches left = [tex]6C3 + 6C4 + 6C5 + 6C6[/tex]

= 20 + 15 + 6 + 1 = 42

P( a team will win the match given that it has won the first game) = 42/64 = 0.66

b)  Since two games have already been won by the team, there are 2⁵ = 32 sample points left. If the team wins two or more matches, it has won.

Number of ways of winning the three or more matches left = [tex]5C2 + 5C3 + 5C4 + 5C5[/tex] = 10 + 10 + 5 +1 = 26

P( a team will win the match given that it has won the first two games) = 26/32 = 0.81

c) Probability that a team will win the match, given that it has won two out of the first three games

They have played 3 games out of 7, this means that there are 4 more games to play. The sample points remain 2⁴ = 16

They have won 2 games already, it means they have two or more games to win.

Number of ways of winning the three or more matches left = [tex]4C2 + 4C3 + 4C4[/tex] = 6 + 4 + 1 = 11

Probability that a team will win the match, given that it has won two out of the first three games = 11/16

Probability that a team will win the match, given that it has won two out of the first three games = 0.69

Answer:

No solutions.

Step-by-step explanation:

9|x-1|=-45

Divide 9 into both sides.

|x-1| = -45/9

|x-1| = -5

Absolute value cannot be less than 0.

Answer:

No solution

Step-by-step explanation:

=> 9|x-1| = -45

Dividing both sides by 9

=> |x-1| = -5

Since, this is less than zero, hence the equation has no solutions

Answer:

  (x, y) = (7, 4) meters

Step-by-step explanation:

The area of the floor without the removal is x^2, so with the smaller square removed, it is x^2 -y^2.

The perimeter of the floor is the sum of all side lengths, so is 4x +2y.

The given dimensions tell us ...

  x^2 -y^2 = 33

  4x +2y = 36

From the latter equation, we can write an expression for y:

  y = 18 -2x

Substituting this into the first equation gives ...

  x^2 -(18 -2x)^2 = 33

  x^2 -(324 -72x +4x^2) = 33

  3x^2 -72x + 357 = 0 . . . . write in standard form

  3(x -7)(x -17) = 0 . . . . . factor

Solutions to this equation are x=7 and x=17. However, for y > 0, we must have x < 9.

  y = 18 -2(7) = 4

The floor dimension x is 7 meters; the inset dimension y is 4 meters.

Answer:

the answer is 3

Step-by-step explanation:

i took the test

Answer:

-30000/100

300 0/1

Step-by-step explanation:

We have the following numbers -18 and 16 2/3, the first is an integer and the second is a mixed number, the first thing is to pass the mixed number to a decimal number.

16 2/3 = 16.67

We do the multiplication:

−18⋅16 2/3 = -300

We have an improper fraction is a fraction in which the numerator (top number) is greater than or equal to the denominator (bottom number), therefore it would be:

-30000/100

How mixed number would it be:

300 0/1

The Laplace transform of the function [tex]\frac{1}{2} (\frac{1}{s} - \frac{s}{s^2 + 4w^2} )[/tex] .

The Laplace transform exist when s > 0 .

Here, the given function is f(t) = sin²(wt) .

The Laplace transform of the the function f(t),

F(s) = f(t) = { [tex]{\frac{1}{2} \times 2sin^2(wt) }[/tex] }

F(s) = { [tex]\frac{1}{2} \times (1- cos2wt)[/tex] }

F(s) = { [tex]\frac{1}{2} - \frac{1}{2} \times cos(2wt)\\[/tex] }

F(s) = [tex]\frac{1}{2} (\frac{1}{s} - \frac{s}{s^2 + 4w^2} )[/tex]

Next,

The above Laplace transform exist if s > 0 .

Know more about Laplace transform,

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Answer:

  22 m

Step-by-step explanation:

Use the formula for the area of a triangle. Fill in the known values and solve for the unknown.

  A = (1/2)bh

  594 m^2 = (1/2)(54 m)h

  h = (594 m^2)/(27 m) = 22 m

The height of the window is 22 meters.

Answer:

196x^2y

Step-by-step explanation: The least common multiple (LCM) of two or more non-zero whole numbers is the smallest whole number that is divisible by each of those numbers. In other words, the LCM is the smallest number that all of the numbers divide into evenly.

Answer:

[tex] SE =\sqrt{\frac{p(1-p)}{n}}[/tex]

And replacing we got:

[tex] SE=\sqrt{\frac{0.75*(1-0.75)}{900}}= 0.014[/tex]

Step-by-step explanation:

For this case we have the following info given:

[tex] n=900[/tex] represent the sample size selected

[tex]p = 0.75[/tex] represent the population proportion

We want to find the standard error and we can use the distribution for the sample proportion and for this case since the sample size is large enough and we satisfy np>10 and n(1-p) >10 we have:

[tex] \hat p \sim N (p,\sqrt{\frac{p(1-p)}{n}})[/tex]

And the standard error is given;

[tex] SE =\sqrt{\frac{p(1-p)}{n}}[/tex]

And replacing we got:

[tex] SE= \sqrt{\frac{0.75* (1-0.75)}{900}}= 0.014[/tex]

Answer:

x^2 - 10x

Step-by-step explanation:

2x^2 - 4x - x^2 +6x

You subtract x^2 from 2x^2 and you get x^2

Then you add 6x and 4x together and get 10x

So then you have x^2 - 10x

(plus I took the test and this was the correct answer.)

Answer:

9.375 in^2

Step-by-step explanation:

Answer:

1)

A) [tex]\frac{}{X}[/tex] ~ N(8.5;0.108)

B) P([tex]\frac{}{X}[/tex] > 9)= 0.0552

C) P(X> 9)= 0.36317

D) IQR= 0.4422

2)

A) [tex]\frac{}{X}[/tex] ~ N(30;2.5)

B) P( [tex]\frac{}{X}[/tex]<30)= 0.50

C) P₉₅= 32.60

D) P( [tex]\frac{}{X}[/tex]>36)= 0

E) Q₃: 31.0586

Step-by-step explanation:

Hello!

1)

The variable of interest is

X: pollutants found in waterways near a large city. (ppm)

This variable has a normal distribution:

X~N(μ;σ²)

μ= 8.5 ppm

σ= 1.4 ppm

A sample of 18 large cities were studied.

A) The sample mean is also a random variable and it has the same distribution as the population of origin with exception that it's variance is affected by the sample size:

[tex]\frac{}{X}[/tex] ~ N(μ;σ²/n)

The population mean is the same as the mean of the variable

μ= 8.5 ppm

The standard deviation is

σ/√n= 1.4/√18= 0.329= 0.33 ⇒σ²/n= 0.33²= 0.108

So: [tex]\frac{}{X}[/tex] ~ N(8.5;0.108)

B)

P([tex]\frac{}{X}[/tex] > 9)= 1 - P([tex]\frac{}{X}[/tex] ≤ 9)

To calculate this probability you have to standardize the value of the sample mean and then use the Z-tables to reach the corresponding value of probability.

Z= [tex]\frac{\frac{}{X} - Mu}{\frac{Sigma}{\sqrt{n} } } = \frac{9-8.5}{0.33}= 1.51[/tex]

Then using the Z table you'll find the probability of

P(Z≤1.51)= 0.93448

Then

1 - P([tex]\frac{}{X}[/tex] ≤ 9)= 1 - P(Z≤1.51)= 1 - 0.93448= 0.0552

C)

In this item, since only one city is chosen at random, instead of working with the distribution of the sample mean, you have to work with the distribution of the variable X:

P(X> 9)= 1 - P(X ≤ 9)

Z= (X-μ)/δ= (9-8.5)/1.44

Z= 0.347= 0.35

P(Z≤0.35)= 0.63683

Then

P(X> 9)= 1 - P(X ≤ 9)= 1 - P(Z≤0.35)= 1 - 0.63683= 0.36317

D)

The first quartile is the value of the distribution that separates the bottom 2% of the distribution from the top 75%, in this case it will be the value of the sample average that marks the bottom 25% symbolically:

Q₁: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₁)= 0.25

Which is equivalent to the first quartile of the standard normal distribution. So first you have to identify the first quartile for the Z dist:

P(Z≤z₁)= 0.25

Using the table you have to identify the value of Z that accumulates 0.25 of probability:

z₁= -0.67

Now you have to translate the value of Z to a value of [tex]\frac{}{X}[/tex]:

z₁= ([tex]\frac{}{X}[/tex]₁-μ)/(σ/√n)

z₁*(σ/√n)= ([tex]\frac{}{X}[/tex]₁-μ)

[tex]\frac{}{X}[/tex]₁= z₁*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₁= (-0.67*0.33)+8.5=  8.2789 ppm

The third quartile is the value that separates the bottom 75% of the distribution from the top 25%. For this distribution, it will be that value of the sample mean that accumulates 75%:

Q₃: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₃)= 0.75

⇒ P(Z≤z₃)= 0.75

Using the table you have to identify the value of Z that accumulates 0.75 of probability:

z₃= 0.67

Now you have to translate the value of Z to a value of [tex]\frac{}{X}[/tex]:

z₃= ([tex]\frac{}{X}[/tex]₃-μ)/(σ/√n)

z₃*(σ/√n)= ([tex]\frac{}{X}[/tex]₃-μ)

[tex]\frac{}{X}[/tex]₃= z₃*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₃= (0.67*0.33)+8.5=  8.7211 ppm

IQR= Q₃-Q₁= 8.7211-8.2789= 0.4422

2)

A)

X ~ N(30,10)

For n=4

[tex]\frac{}{X}[/tex] ~ N(μ;σ²/n)

Population mean μ= 30

Population variance σ²/n= 10/4= 2.5

Population standard deviation σ/√n= √2.5= 1.58

[tex]\frac{}{X}[/tex] ~ N(30;2.5)

B)

P( [tex]\frac{}{X}[/tex]<30)

First you have to standardize the value and then look for the probability:

Z=  ([tex]\frac{}{X}[/tex]-μ)/(σ/√n)= (30-30)/1.58= 0

P(Z<0)= 0.50

Then

P( [tex]\frac{}{X}[/tex]<30)= 0.50

Which is no surprise since 30 y the value of the mean of the distribution.

C)

P( [tex]\frac{}{X}[/tex]≤ [tex]\frac{}{X}[/tex]₀)= 0.95

P( Z≤ z₀)= 0.95

z₀= 1.645

Now you have to reverse the standardization:

z₀= ([tex]\frac{}{X}[/tex]₀-μ)/(σ/√n)

z₀*(σ/√n)= ([tex]\frac{}{X}[/tex]₀-μ)

[tex]\frac{}{X}[/tex]₀= z₀*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₀= (1.645*1.58)+30= 32.60

P₉₅= 32.60

D)

P( [tex]\frac{}{X}[/tex]>36)= 1 - P( [tex]\frac{}{X}[/tex]≤36)= 1 - P(Z≤(36-30)/1.58)= 1 - P(Z≤3.79)= 1 - 1 = 0

E)

Q₃: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₃)= 0.75

⇒ P(Z≤z₃)= 0.75

z₃= 0.67

z₃= ([tex]\frac{}{X}[/tex]₃-μ)/(σ/√n)

z₃*(σ/√n)= ([tex]\frac{}{X}[/tex]₃-μ)

[tex]\frac{}{X}[/tex]₃= z₃*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₃= (0.67*1.58)+30= 31.0586

Q₃: 31.0586

Answer:

x=-4

Step-by-step explanation:

[tex]2x^2+8x=x^2-16[/tex]

Move everything to one side:

[tex]x^2+8x+16=0[/tex]

Factor:

[tex](x+4)^2=0[/tex]

By the zero product rule, x=-4. Hope this helps!

Answer:

x=-4

Step-by-step explanation:

Move everything to one side and combine like-terms

x²+8x+16

Factor

(x+4)²

x=-4

Answer:

Is possible to make a Type I error, where we reject a true null hypothesis.

Step-by-step explanation:

We have a hypothesis test of a proportion. The claim is that the proportion of paid leave has significantly decrease from 2011 to january 2012. The P-value for this test is 0.0271 and the nunll hypothesis is rejected.

As the conclusion is to reject the null hypothesis, the only error that we may have comitted is rejecting a true null hypothesis.

The null hypothesis would have stated that there is no significant decrease in the proportion of paid leave.

This is a Type I error, where we reject a true null hypothesis.

Answer:

9.42 in²

Step-by-step explanation:

The area of whole circle S=pi*R²    , where pi is appr. 3.14,  R= 6 in

S= 3.14*6² =113.04 in²

The area of smaller sector is Ssec=S/360*30=113,04/12=9.42 in²

The area of the smaller sector with a central angle of 30 degrees and a radius of 6 inches is 9.42478 square inches.

To find the area of a sector, you can use the formula:

Area of sector = (θ/360) × π × r²

where θ is the central angle in degrees, r is the radius of the sector.

The central angle is 30 degrees and the radius is 6 inches.

Plugging these values into the formula:

Area of sector = (30/360) × π × 6²

= (1/12) × π × 36

= (1/12) × 3.14159 × 36

= 9.42478 square inches

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Answer/Step-by-step explanation:

==>Given:

=>Rectangular Pyramid:

L = 5mm

W = 3mm

H = 4mm

=>Rectangular prism:

L = 5mm

W = 3mm

H = 4mm

==>Required:

a. Volume of pyramid:

Formula for calculating volume of a rectangular pyramid us given as L*W*H

V = 5*3*4

V = 60 mm³

b. Volume of prism = ⅓*L*W*H

thus,

Volume of rectangular prism given = ⅓*5*3*4

= ⅓*60

= 20mm³

c. Volume of the prism = ⅓ x volume of the pyramid

Thus, 20 = ⅓ × 60

As we can observe from our calculation of the solid shapes given, the equation written above is true for all rectangular prism and rectangular pyramid of the same length, width and height.

Answer:

25%

Step-by-step explanation:

The last percentile always contains 25% of the observations.

Answer:

114

Step-by-step explanation:

Answer:

144Step-by-step explanation:

Answer:

[tex]77.7-1.66\frac{3.6}{\sqrt{100}} =77.102[/tex]    

[tex]77.7+1.66\frac{3.6}{\sqrt{100}} =78.30[/tex]    

And the best option would be:

C. (77.1, 78.3)

Step-by-step explanation:

Information given

[tex]\bar X=77.7[/tex] represent the sample mean

[tex]\mu[/tex] population mean (variable of interest)

s=3.6 represent the sample standard deviation

n=100 represent the sample size  

Confidence interval

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

The degrees of freedom are given by:

[tex]df=n-1=100-1=99[/tex]

Since the Confidence is 0.90 or 90%, the significance would be [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and the critical value for this case would be [tex]t_{\alpha/2}=1.66[/tex]

And replacing we got:

[tex]77.7-1.66\frac{3.6}{\sqrt{100}} =77.10[/tex]    

[tex]77.7+1.66\frac{3.6}{\sqrt{100}} =78.30[/tex]    

And the best option would be:

C. (77.1, 78.3)

Answer:

slope = -4/5

Step-by-step explanation:

A line passes two points (x1, y1) and (x2, y2).

The slope of this line can be calculate by the formula:

s = (y2 - y1)/(x2 - x1)

=>The line that passes A(-4, 8) and B(-9, 12) has the slope:

s = (12 - 8)/(-9 - -4) = 4/(-5) = -4/5

Hope this helps!

Answer:

The third options

Step-by-step explanation:

Counting we can see that 10 students went to two or less states, and 10 went to three or more

Answer:

The probability that demand during lead-time will exceed 20 bails is 0.2033.

Step-by-step explanation:

We are given that it has been previously determined that demand during the lead-time is normally distributed with a mean of 15 bails and a standard deviation of 6 bails.

Let X = demand during the lead-time

So, X ~ Normal([tex]\mu=15, \sigma^{2} = 6^{2}[/tex])

The z-score probability distribution for the normal distribution is given by;

                               Z  =  [tex]\frac{X-\mu}{\sigma}[/tex]  ~ N(0,1)

where, [tex]\mu=[/tex] population mean demand = 15 bails

           [tex]\sigma[/tex] = standard deviation = 6 bails

Now, the probability that demand during lead-time will exceed 20 bails is given by = P(X > 20 bails)

       P(X > 20 bails) = P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{20-15}{6}[/tex] ) = P(Z > 0.83) = 1 - P(Z [tex]\leq[/tex] 0.83)

                                                             = 1 - 0.7967 = 0.2033

Answer:

41 word/min

Step-by-step explanation:

Before noon Ali works:

She types:

After lunch she works:

She types:

Total Ali works= 4+4= 8 hours= 480 min

Total Ali types= 11520+8160= 19680 words

Average typing rate= 19680 words/480 min= 41 word/min

Answer:

0.989

Step-by-step explanation:

For each graduate, there are only two possible outcomes. Either they find a job in their chosen field within a year after graduation, or they do not. The probability of a graduate finding a job is independent of other graduates. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

A study conducted at a certain college shows that "53%" of the school's graduates find a job in their chosen field within a year after graduation.

This means that [tex]p = 0.53[/tex]

6 randomly selected graduates

This means that [tex]n = 6[/tex]

Probability that at least one finds a job in his or her chosen field within a year of graduating:

Either none find a job, or at least one does. The sum of the probabilities of these outcomes is 1. So

[tex]P(X = 0) + P(X \geq 1) = 1[/tex]

We want [tex]P(X \geq 1)[/tex]

So

[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]

In which

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{6,0}.(0.53)^{0}.(0.47)^{6} = 0.011[/tex]

So

[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.011 = 0.989[/tex]