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labwork [276]
2 years ago
7

In a lab experiment, 3400 bacteria are placed in a petri dish. The conditions are such that the number of bacteria is able to do

uble every 8 hours. How long would it be, to the nearest tenth of an hour, until there are 13700 bacteria present?
Mathematics
1 answer:
nalin [4]2 years ago
3 0

Answer: 16.1

Step-by-step explanation:

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Suppose S is between R and T. Use the Segment Addition Postulate to solve for z.. RS=2z+6, ST=4z−3, RT=5z+12
Lynna [10]

Answer:

9

Step-by-step explanation:

If S is between R and T, then RS+ST = RT

Given the following parameters

RS=2z+6,

ST=4z−3,

RT=5z+12

On substituting

2z+6+4z-3 = 5z+12

Collect like terms

2z+4z-5z = 12+3-6

6z-5z = 15-6

z = 9

Hence the value of z is 9

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En una granja hay gallinas y vacas, en total uman 624. Se sabe que el numero de vacas son 36 veces mas que el numero de gallinas
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Answer:

17 R 12

Step-by-step explanation:

Hola,

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624÷36

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2 years ago
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Answer:

shhdhd

Step-by-step explanation:

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2 years ago
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You work in the HR department at a large franchise. you want to test whether you have set your employee monthly allowances corre
ra1l [238]

Answer:

1) Null hypothesis:\mu \leq 500  

Alternative hypothesis:\mu > 500  

z=\frac{640-500}{\frac{150}{\sqrt{40}}}=5.90  

For this case since we are conducting a right tailed test we need to find a critical value in the normal standard distribution who accumulates 0.01 of the area in the right and we got:

z_{crit}= 2.33

For this case we see that the calculated value is higher than the critical value

Since the calculated value is higher than the critical value we have enugh evidence to reject the null hypothesis at 1% of significance level

2) Since is a right tailed test the p value would be:  

p_v =P(z>5.90)=1.82x10^{-9}  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, same conclusion for part 1

Step-by-step explanation:

Part 1

Data given

\bar X=640 represent the sample mean

\sigma=150 represent the population standard deviation

n=40 sample size  

\mu_o =500 represent the value that we want to test  

\alpha=0.01 represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

Step1:State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is higher than 500, the system of hypothesis would be:  

Null hypothesis:\mu \leq 500  

Alternative hypothesis:\mu > 500  

Step 2: Calculate the statistic

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

We can replace in formula (1) the info given like this:  

z=\frac{640-500}{\frac{150}{\sqrt{40}}}=5.90  

Step 3: Calculate the critical value

For this case since we are conducting a right tailed test we need to find a critical value in the normal standard distribution who accumulates 0.01 of the area in the right and we got:

z_{crit}= 2.33

Step 4: Compare the statistic with the critical value

For this case we see that the calculated value is higher than the critical value

Step 5: Decision

Since the calculated value is higher than the critical value we have enugh evidence to reject the null hypothesis at 1% of significance level

Part 2

P-value  

Since is a right tailed test the p value would be:  

p_v =P(z>5.90)=1.82x10^{-9}  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, same conclusion for part 1

7 0
2 years ago
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