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3 years ago

What is this? −567.45+567.45

Mathematics

1 answer:

djverab [1.8K]3 years ago

3 0

Answer:

Step-by-step explanation:

The explanation is in the attached file.

Sidenote: Hope this helps! :D

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Someone please help me find the value of X

Marta_Voda [28]

Answer:

Step-by-step explanation:

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3 years ago

Last year, 45% of business owners gave a holiday gift to their employees. A survey of business owners conducted this year indica

natka813 [3]

Answer:

a)x₁ = 24 business owners plan to provide a gift this year

b) p-value = 0,0017

c)p-value < α we have to eject H₀ And support the claim that the proportion of business owners providing gifts has decrease

Step-by-step explanation:

a) Sample size 80

proportion p₁ = 30 % p₁ = 0,3

Then x₁ = 80*0,3

x₁ = 24 business owners plan to provide a gift this year

b) Hypothesis Test

Null Hypothesis H₀ p₁ = 0,45

Alternative Hypothesis Hₐ p₁ < 0,45

p₁ = 0,3 then q₁ = 1 - 0,3 q₁ = 0,7

z(s) = ( p₁ - 0,45 )/ √p₁* q₁ / n

z(s) = - 0,15 / √0,002625

z(s) = - 0,15 / 0,05123

z(s) = - 2,9279

b)From z-table we find p-value = 0,0017

c) If α = 0,05 ( significance level ) as

p-value < α we have to eject H₀ And support the claim that the proportion of business owners providing gifts has decrease

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3 years ago

A common inhabitant of human intestines is the bacterium Escherichia coli. A cell of this bacterium in a nutrient-broth medium d

nikitadnepr [17]

Answer:

a) k=2.08 1/hour

b) The exponential growth model can be written as:

$P(t)=Ce^{kt}$

c) 977,435,644 cells

d) 2.033 billions cells per hour.

e) 2.81 hours.

Step-by-step explanation:

We have a model of exponential growth.

We know that the population duplicates every 20 minutes (t=0.33).

The initial population is P(t=0)=58.

The exponential growth model can be written as:

$P(t)=Ce^{kt}$

For t=0, we have:

$P(0)=Ce^0=C=58$

If we use the duplication time, we have:

$P(t+0.33)=2P(t)\\\\58e^{k(t+0.33)}=2\cdot58e^{kt}\\\\e^{0.33k}=2\\\\0.33k=ln(2)\\\\k=ln(2)/0.33=2.08$

Then, we have the model as:

$P(t)=58e^{2.08t}$

The relative growth rate (RGR) is defined, if P is the population and t the time, as:

$RGR=\dfrac{1}{P}\dfrac{dP}{dt}=k$

In this case, the RGR is k=2.08 1/h.

After 8 hours, we will have:

$P(8)=58e^{2.08\cdot8}=58e^{16.64}=58\cdot 16,852,338= 977,435,644$

The rate of growth can be calculated as dP/dt and is:

$dP/dt=58[2.08\cdot e^{2.08t}]=120.64e^2.08t=2.08P(t)$

For t=8, the rate of growth is:

$dP/dt(8)=2.08P(8)=2.08\cdot 977,435,644 = 2,033,066,140$

(2.033 billions cells per hour).

We can calculate when the population will reach 20,000 cells as:

$P(t)=20,000\\\\58e^{2.08t}=20,000\\\\e^{2.08t}=20,000/58\approx344.827\\\\2.08t=ln(344.827)\approx5.843\\\\t=5.843/2.08\approx2.81$

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3 years ago

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4 years ago