**Please help! - What is the equation in standard form of the line that has x-intercept -4 and y-intercept 3.

Find the mean and standard deviation for the set of data.

dem82 [27]

The mean and standard deviation for the set of data will be 16 and 5.87. Then the correct option is A.

<h3>What is a standard deviation?</h3>

It is the measure of the dispersion of statistical data. Dispersion is the extent to which the value is in a variation.

The set of data is given below.

16, 22, 8, 5, 20, 18, 14, 17, 24

The mean of the data set will be

μ = (16 + 22 + 8 + 5 + 20 + 18 + 14 + 17 + 24) / 9

μ = 144 / 9

μ = 16

Then the standard deviation of the data set will be

$\rm \sigma = \sqrt{\dfrac{\Sigma (x_i - \mu)^2}{n}} \\\sigma = \sqrt{\dfrac{(16 - 16)^2 +(22 - 16)^2 + (8-16)^2 + ......+(24-16)^2 }{9}} \\\sigma = \sqrt{\dfrac{310}{9}}\\$

Simplify the expression further, then we have

σ = √34.44

σ = 5.87

The mean and standard deviation for the set of data will be 16 and 5.87.

Then the correct option is A.

More about the standard deviation link is given below.

brainly.com/question/12402189

#SPJ1

3 0

1 year ago

How many hundreds in 43617

Nana76 [90]

There are 436 hundreds in 43617.
You can find this by dividing 43617 by 100 and rounding down to the nearest whole number (note: with this method, you have to round down).

Hope this helps :)

7 0

3 years ago

Read 2 more answers

The number of bacteria after t hours is given by N(t)=250 e^0.15t a) Find the initial number of bacteria and the rate of growth

Art [367]

Answer:

a) $N_0=250\; k=0.15$

b) 334,858 bacteria

c) 4.67 hours

d) 2 hours

Step-by-step explanation:

a) Initial number of bacteria is the coefficient, that is, 250. And the growth rate is the coefficient besides “t”: 0.15. It’s rate of growth because of its positive sign; when it’s negative, it’s taken as rate of decay.

Another way to see that is the following:

Initial number of bacteria is N(0), which implies $t=0$ . And $N(0)=N_0$ . The process is:

$N(t)=250 e^{0.15t}\\N(0)=250 e^{0.15(0)}\\ N_0=250e^{0}\\N_0=250\cdot1\\ N_0=250$

b) After 2 days means $t=48$ . So, we just replace and operate:

$N(t)=250 e^{0.15t}\\N(48)=250 e^{0.15(48)}\\ N(48)=250e^{7.2}\\N(48)=334,858\;\text{bacteria}$

c) $N(t_1)=4000; \;t_1=?$

$N(t)=250 e^{0.15t}\\4000=250 e^{0.15t_1}\\ \dfrac{4000}{250}= e^{0.15t_1}\\16= e^{0.15t_1}\\ \ln{16}= \ln{e^{0.15t_1}} \\ \ln{16}=0.15t_1 \\ \dfrac{\ln{16}}{0.15}=t_1=4.67\approx 5\;h$