All categories

3 years ago

PLs help me on this..........

Mathematics

2 answers:

Otrada [13]3 years ago

8 0

Answer:

247277 people

Serggg [28]3 years ago

3 0

Answer:

247277 people

Step-by-step explanation:

6% of 243,000 = 14,580

243,000 + 14,580 = 257,580

4% of 257,580 = 10,303.2

Estimating 10,303.2 to 10,303

257,580 - 10,303 = 247277

You might be interested in

-3y - 7.8 = 9.6 solve these equation

Luden [163]

Y= -5.8 is the correct answer

3 0

3 years ago

Read 2 more answers

Factor the polynomial below 2x^3 - 54

solmaris [256]

Idk idk idk idk idk idk idk idk idk Ido idk island

6 0

4 years ago

What is the answer please help

Goryan [66]

Three stands for a solid 3
-------------------------------------
six tenths = 0.6
-------------------------------------
add
3 + 0.6 = 3.6
-------------------------------------
3.6 is your answer

3 0

4 years ago

. Which equation represents the line that passes through the points (3.7) and (-1.-1)

kondaur [170]

Answer:

$y=2x+1$ or $y+1=2(x+1)$

Step-by-step explanation:

step 1

Find the slope of the line

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

we have

$A(3,7)\ B(-1,-1)$

Substitute the values

$m=\frac{-1-7}{-1-3}$

$m=\frac{-8}{-4}=2$

step 2

Find the equation of the line into slope point form

$y-y1=m(x-x1)$

we have

$m=2$

$pointB(-1,-1)$

substitute

$y+1=2(x+1)$ -----> equation of the line into point slope form

$y=2x+2-1$

$y=2x+1$ -----> equation of the line into slope intercept form

3 0

3 years ago

Part of a university's professor's job is to publish his or her research. This task often entails reading a variety of journal a

goblinko [34]

Answer:

(18.1042, 35.2292) is a 90% confidence interval for the mean number of journal articles read monthly by professors.

Step-by-step explanation:

We have a small sample of size n = 12, $\bar{x} = 26.6667$ and s = 16.5163. A pivotal quantity for this case is given by $T = \frac{\bar{X}-\mu}{S/\sqrt{n}}$ which has a t distribution with n-1 degrees of freedom if we suppose that we take the sample from the normal population. The confidence interval is given by $\bar{x}\pm t_{\alpha/2}(\frac{s}{\sqrt{n}})$ where $t_{\alpha/2}$ is the $\alpha/2$ th quantile of the t distribution with n - 1 = 12 - 1 = 11 degrees of freedom. As we want the 90% confidence interval, we have that $\alpha = 0.1$ and the confidence interval is $26.6667\pm t_{0.05}(\frac{16.5163}{\sqrt{12}})$ where $t_{0.05}$ is the 5th quantile of the t distribution with 11 df, i.e., $t_{0.05} = -1.7959$ . Then, we have $26.6667\pm (-1.7959)(\frac{16.5163}{\sqrt{12}})$ and the 90% confidence interval is given by (18.1042, 35.2292).

6 0

3 years ago