Please help a homie out

Which is true of point K at (0, –5)? A. Point K is to the left of the y-axis. B. Point K is on the x-axis. C. Point K is on the

nirvana33 [79]

C is the correct one

hope this helps :)

3 0

3 years ago

Assume you have 4 fins, each with a mass of 8grams. what's the total weight of these 4 fines?

Kryger [21]

1 fin = 8 grams
4 fins = 4 x 8 = 24 grams total

3 0

3 years ago

N+3(n-1) n plus three ( n minus 1

liq [111]

Answer:

3n^2+-3

Step-by-step explanation:

nxn +nx(-1) +3xn+ 3x(-1)

3 0

3 years ago

10. The continuously compounded annual return on a stock is normally distributed with a mean of 20% and standard deviation of 30

ehidna [41]

Answer: We should expect its actual return in any particular year to be between -40% and 80%.

Step-by-step explanation:

Given : The continuously compounded annual return on a stock is normally distributed with a mean 20% and standard deviation of 30%.

From normal z-table, the z-value corresponds to 95.44 confidence is 2.

Therefore , the interval limits for 95.44 confidence level will be :

Lower limit = Mean -2(Standard deviation) = 20% -2(30%)= 20%-60%=-40%

Upper limit = Mean +2(Standard deviation)=20% +2(30%)= 20%+60%=80%

Hence, we should expect its actual return in any particular year to be between -40% and 80%.

7 0

3 years ago

A psychologist is interested in knowing whether adults who were bullied as children differ from the general population in terms

MissTica

Answer:

$t = \frac{39.5-30.6}{\frac{6.6}{\sqrt{25}}}= 6.74$

The degrees of freedom are given by:

$df =n-1= 25-1=24$

Now we can calculate the p value with the following probability:

$p_v = 2*P(t_{24}>6.74)= 5.69x10^{-7}$

And for this case since the p value is lower compared to the significance level $\alpha=0.05$ we can reject the null hypothesis and we can conclude that the true mean for this case is different from 30.6 at the significance level of 0.05

Step-by-step explanation:

For this case we have the following info given:

$\bar X = 39.5$ represent the sample mean

$s =6.6$ represent the sample deviation

$\mu = 30.6$ represent the reference value to test.

$n = 25$ represent the sample size selected

The statistic for this case is given by:

$t =\frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}$

And replacing we got:

$t = \frac{39.5-30.6}{\frac{6.6}{\sqrt{25}}}= 6.74$

The degrees of freedom are given by:

$df =n-1= 25-1=24$

Now we can calculate the p value with the following probability:

$p_v = 2*P(t_{24}>6.74)= 5.69x10^{-7}$

And for this case since the p value is lower compared to the significance level $\alpha=0.05$ we can reject the null hypothesis and we can conclude that the true mean for this case is different from 30.6 at the significance level of 0.05

3 0

4 years ago