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3 years ago

What are consecutive integers?

1 answer:

8 0

Consecutive integers are integers that follow each other in order. They have a difference of 1 between every two numbers. In a set of consecutive integers, the mean and the median are equal. If n is an integer, then n, n+1, and n+2 would be consecutive integers. Examples.

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Suppose that the concentration of a bacteria sample is 60 comma 000 bacteria per milliliter. If the concentration triples in 4

Elodia [21]

Answer:

1.932 days (or approximatelly 1 day, 22 hours and 22 minutes)

Step-by-step explanation:

The inicial concentration is 60,000, and this concentration triples every 4 days, so we can write the equation:

P = Po * r^t

where P is the final concentration after t periods of 4 days, Po is the inicial concentration and r is the ratio that the concentration increases (r = 3)

Then, we have that:

102000 = 60000 * 3^t

3^t = 102/60 = 1.7

log(3^t) = log(1.7)

t*log(3) = log(1.7)

t = log(1.7)/log(3) = 0.483

so the number of days that will take is 4*0.483 = 1.932 days (or approximatelly 1 day, 22 hours and 22 minutes)

3 0

3 years ago

Need help on all of them lol even if u can do one or a few, can someone plz help? #3-14

valina [46]

Just times them and you will have the volume

8 0

3 years ago

Is 4/6 bigger than 4/8

GrogVix [38]

No, because 4/6 would round out to 1/3 whereas 4/8 would round down to 1/2

4 0

3 years ago

Read 2 more answers

The Bureau of Labor Statistics reported that the average yearly income of dentists in the year 2005 was $110,000. A sample of 81

Sholpan [36]

Answer:

a) We want to test to determine if there has been a significant increase in the average yearly income of dentists, the system of hypothesis would be:

Null hypothesis: $\mu \leq 110000$

Alternative hypothesis: $\mu > 110000$

b) $z=\frac{120000-110000}{\frac{36000}{\sqrt{81}}}=2.5$

c) $p_v =P(Z>2.5)=0.0062$

We see that the p value is lower than the significance level of $1-0.95=0.05$ so then we can conclude that the true mean is significantly higher than 110000 because we reject the null hypothesis.

Step-by-step explanation:

Information given

$\bar X=12000$ represent the sample mean for the yearly income in 2006 for the dentists

$\sigma= 36000$ represent the sample population deviation

$n=81$ sample size

$\mu_o =110000$ represent the value that we want to test

$\alpha=0.05$ represent the significance level

z would represent the statistic

$p_v$ represent the p value for the test

a) Hypothesis to verify

We want to test to determine if there has been a significant increase in the average yearly income of dentists, the system of hypothesis would be:

Null hypothesis: $\mu \leq 110000$

Alternative hypothesis: $\mu > 110000$

Since we know the population deviation the statistic is given by:

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

b) Statistic

When we replace the data given we got:

$z=\frac{120000-110000}{\frac{36000}{\sqrt{81}}}=2.5$

c) P value

Now we can find the p value using the fact that we are conducting a right tailed test:

$p_v =P(Z>2.5)=0.0062$

We see that the p value is lower than the significance level of $1-0.95=0.05$ so then we can conclude that the true mean is significantly higher than 110000 because we reject the null hypothesis.

7 0

4 years ago

Write an exponential function for graph that passes through the following points (-3,80);(-1,20)

padilas [110]

Answer:

(

)

Explanation:

An exponential function is in the general form

(

)

We know the points

(

−

)

and

(

)

, so the following are true:

(

−

)

(

)

Multiply both sides of the first equation by

to find that

Plug this into the second equation and solve for

(

)

Two equations seem to be possible here. Plug both values of

into the either equation to find

. I'll use the second equation for simpler algebra.

(

)

Giving us the equation:

(

)

−

(

−

)

−

Giving us the equation:

−

(

−

)

However! In an exponential function,

, otherwise many issues arise when trying to graph the function.

The only valid function is

(

)

7 0

4 years ago