What does point g represent

Help me, please. Don't just claim my points, it's irritating

erastova [34]

Answer:

8.239x10^-5

so the answer is A

Step-by-step explanation:

trust me

ps. use photomath on phone

5 0

2 years ago

Read 2 more answers

A study reported that the average age of menarche of girls entering the first year class of a small American private college in

Verizon [17]

Answer:

Null hypothesis: $\mu = 11.89$

Alternative hypothesis: $\mu \neq 11.89$

And we want to test this with a confidence interval:

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

After conduct the interval of 95% for the true mean we got: (11.59, 13.31)

And for this case the confidence interval contains the value of 11.89, so then we have enough evidence to FAIL to reject the null hypothesis at 5% of significance that the true mean is 11.89.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=12.98$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n=30 represent the sample size

Solution to the problem

For this case we want to test the following hypothesis:

Null hypothesis: $\mu = 11.89$

Alternative hypothesis: $\mu \neq 11.89$

And we want to test this with a confidence interval:

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

After conduct the interval of 95% for the true mean we got: (11.59, 13.31)

And for this case the confidence interval contains the value of 11.89, so then we have enough evidence to FAIL to reject the null hypothesis at 5% of significance that the true mean is 11.89.

4 0

3 years ago

I am confused about ii) help please

Jobisdone [24]

Step-by-step explanation:

i)=

m=2/3

x=-3

y = 5

now

y=mx+c

or, 5 = 2/3× -3 + c

or, 5 = -2+c

c =7

now,

y= mx + c

or, y = 2/3×x+7

or, y = 2x/3+7

or, 3y = 2x+21

so, 3y-2x=21

equation would be 3y-2x=21.

now

ii)=

if slope of 3y-2x=21 and 2x-3y = 0 are same, they are parallel.

2x-3y = 0

2x = 3y

lets suppose x is 9

now

2x=3y

2×9 = 3y

or, 18 = 3y

so, y = 6

one coordinate would be (9,6) now

lets again suppose x is 3

2x=3y

2×3 = 3y

or, 6 = 3y

so, y = 2

another coordinate is (3,2)

now lets find slope

x1=9,y1=6

x2=3,y2=2

slope = (y2-y1)/(x2-x1)

= (2-6)/(3-9)

= -4/-6

= 2/3

Since, slope of 3y-2x=21 and 2x-3y has same slope 2/3, they are parallel.

6 0

2 years ago

Simplify <br> (X-4+2i)(x+5-3i)

iogann1982 [59]

Use distributive property
x^2+x-14+(-x+22)i

6 0

2 years ago

A fast-food company is interested in knowing the probability of whether a customer viewed an advertisement for their new special

lord [1]

Answer:

45% probability that a randomly selected customer saw the advertisement on the internet or on television

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a customer saw the advertisement on the internet.

B is the probability that a customer saw the advertisement on television.

We have that:

$A = a + (A \cap B)$

In which a is the probability that a customer saw the advertisement on the internet but not on television and $A \cap B$ is the probability that the customers saw the advertisement in both the internet and on television.