There is more than one answer!

Differentiate Square numbers and cube numbers

pashok25 [27]

Answer:

square numbers are numbers raised up to the power of two in essence the number multiplied by it self

6 0

3 years ago

This is literary super easy, all I need is coordinates.

Juliette [100K]

Answer:

m (2,6), L (3, -6), J (-6, -9), K (-1, -9)

Step-by-step explanation:

3 0

2 years ago

convert the following logarithmic equation to the equivalent exponential equation. Use the caret (^) to enter exponents. y=ln x

GarryVolchara [31]

1. To solve this problem, you need to remember that an exponential function has the following form:

f(x)=a^x

"a" is the base and "x" is the exponent.

2. It is important to know that the logarithmic functions and the exponential functionsare inverse. Then, you have:

y=ln x
 e^y=e^(lnx)
 e^y=x

3. Therefore, the answer is:

x=e^y

5 0

3 years ago

Assume that you have a sample of n 1 equals 6, with the sample mean Upper X overbar 1 equals 50, and a sample standard deviati

tigry1 [53]

Answer:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

$df=6+5-2=9$

$p_v =P(t_{9}>2.656) =0.0131$

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

$n_1 =6$ represent the sample size for group 1

$n_2 =5$ represent the sample size for group 2

$\bar X_1 =50$ represent the sample mean for the group 1

$\bar X_2 =38$ represent the sample mean for the group 2

$s_1=7$ represent the sample standard deviation for group 1

$s_2=8$ represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 > \mu_2$

We are assuming that the population variances for each group are the same

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

The statistic for this case is given by:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

The pooled variance is:

$S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

We can find the pooled variance:

$S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67$

And the pooled deviation is:

$S_p=7.46$

The statistic is given by:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

The degrees of freedom are given by:

$df=6+5-2=9$

The p value is given by:

$p_v =P(t_{9}>2.656) =0.0131$

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

4 0

3 years ago

You have been hired by joe's gas stop to estimate a demand function for its regular gasoline sales. occordingly, you collect 50

Digiron [165]

(a) Sample correlation ==> -0.7916
(b) Standard Deviation for Quantity ==> 801.6816
(c) Standard Deviation for Price ==> 39.1660
(d) Relation to coefficient on Price ==> −16.2028

8 0

3 years ago