Ask question

Ask question

All categories

2 years ago

6

Over the past ten years, the town's population doubled in size. The population is currently 12,000. What was the population ten

years ago?

2 answers:

Alekssandra [29.7K]2 years ago

7 0

6,000 right? I think

BaLLatris [955]2 years ago

4 0

If the population doubled then you just divide by two.. so 12,000/2 is 6,000.

You might be interested in

SoMeonE plZ hAlp me lol

jeka94

C.√93

A=17

B=14

...........................

3 0

2 years ago

A 99% confidence interval for the difference between two proportions was estimated at 0.11, 0.39. Based on this, we can conclude

sladkih [1.3K]

Answer:

$\hat p_1 -\hat p_2 \pm z_{\alpha/2} SE$

And for this case the confidence interval is given by:

$0.11 \leq p_1 -p_2 \leq 0.39$

Since the confidenc einterval not contains the value 0 we can conclude that we have significant difference between the two population proportion of interest 1% of significance given. So then we can't conclude that the two proportions are equal

Step-by-step explanation:

Let p1 and p2 the population proportions of interest and let $\hat p_1$ and $\hat p_2$ the estimators for the proportions we know that the confidence interval for the difference of proportions is given by this formula:

$\hat p_1 -\hat p_2 \pm z_{\alpha/2} SE$

And for this case the confidence interval is given by:

$0.11 \leq p_1 -p_2 \leq 0.39$

Since the confidence interval not contains the value 0 we can conclude that we have significant difference between the two population proportion of interest 1% of significance given. So then we can't conclude that the two proportions are equal

4 0

2 years ago

1/4 x 8/15 in simplest form

Over [174]

This in the simplest form will be 2/15. You cannot divide this answer by anything which means it is in its simplest form!

Please give brainliest❤️

4 0

2 years ago

Read 2 more answers

Let n be a natural number. Show that 3 | n3 −n

Vladimir [108]

Let $n=1$ . Then $n^3-n=1^3-1=0$ . By convention, every non-zero integer $n$ divides 0, so $3\vert n^3-n$ .

Suppose this relation holds for $n=k$ , i.e. $3\vert k^3-k$ . We then hope to show it must also hold for $n=k+1$ .

You have

$(k+1)^3-(k+1)=(k^3+3k^2+3k+1)-(k+1)=(k^3-k)+3(k^2+k)$

We assumed that $3\vert k^3-k$ , and it's clear that $3\vert 3(k^2+k)$ because $3(k^2+k)$ is a multiple of 3. This means the remainder upon divides $(k+1)^3-(k+1)$ must be 0, and therefore the relation holds for $n=k+1$ . This proves the statement.

4 0

2 years ago

Two ropes, AD and BD, are tied to a peg on the ground at point D. The other ends of the ropes are tied to points A and B on a fl

BigorU [14]

You have two 30-60-90 triangles, ADC and BDC.

The ratio of the lengths of the sides of a 30-60-90 triangle is

short leg : long leg : hypotenuse
1 : sqrt(3) : 2

Using triangle ADC, we can find length AC.
Using triangle BDC, we can find length BC.
Then AB = AC - BC

First, we find length AC.
Look at triangle ACD.
DC is the short leg opposite the 30-deg angle.
DC = 10sqrt(3)
AC = sqrt(3) * 10sqrt(3) = 3 * 10 = 30

Now, we find length BC.
Look at triangle BCD.
For triangle BCD, the long leg is DC and the short leg is BC.
BC = 10sqrt(3)/sqrt(3) = 10

AB = AC - BC = 30 - 10 = 20

6 0

2 years ago

Read 2 more answers