Repost, algebra help needed please

What exponent goes with the area of a triangle

user100 [1]

Answer:

its just 1/2 times base times height

Step-by-step explanation:

7 0

4 years ago

sophia has 131 pages to read in 2-5 days. Would it be possible for her to read the same number of pages each day?

Contact [7]

No, it wouldn’t be. there is no whole, so there is no way. except if she reads half a page for two days. :)

4 0

3 years ago

Discover the area of a circle with a radius of 15 meters ?

Ronch [10]

Answer:

706.5 meters²

Step-by-step explanation:

Area of a circle is: $A=\pi r^2$

'r' is the radius.

We are given the radius of 15 meters.

Using 3.14 for pi:

$A=3.14*15^2\\\rightarrow 15^2=225\\A= 3.14 * 225\\\boxed {A=706.5}$

The area of the circle should be 706.5 meters².

7 0

3 years ago

What is the y intercept of the equation Y+9=6(x-3)

PtichkaEL [24]

Answer: -27

Step-by-step explanation:

5 0

3 years ago

The net weights (in grams) of a sample of bottles filled by a machine manufactured by Edne, and the net weights of a sample fill

IRINA_888 [86]

Answer:

$t=\frac{(10 -7)-(0)}{\sqrt{\frac{2.268^2}{8}+\frac{1.414^2}{6}}}=3.036$

$p_v =P(t_{12}>3.036) =0.0051$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Orno) is significantly higher than the mean for the group 2 (Edne).

Step-by-step explanation:

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 > \mu_2$

Or equivalently:

Null hypothesis: $\mu_1 - \mu_2 \leq 0$

Alternative hypothesis: $\mu_1 -\mu_2 > 0$

Our notation on this case :

$n_1 =8$ represent the sample size for group 1 (Orno)

$n_2 =6$ represent the sample size for group 2 (Edne)

We can calculate the sampel means and deviations with the following formulas:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X_1 =10$ represent the sample mean for the group 1

$\bar X_2 =7$ represent the sample mean for the group 2

$s_1=2.268$ represent the sample standard deviation for group 1

$s_2=1.414$ represent the sample standard deviation for group 2

If we see the alternative hypothesis we see that we are conducting a right tailed test.

On this case since the significance assumed is 0.05 and we are conducting a bilateral test we have one critica value, and we need on the right tail of the distribution $\alpha/2 = 0.05$ of the area.

The distribution on this case since we don't know the population deviation for both samples is the t distribution with $df=8+6 -2=12$ degrees of freedom.

We can use the following excel code in order to find the critical value:

"=T.INV(1-0.05,12)"

And the rejection zone is: (1.78,infinity)

The statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}+\frac{S^2_2}{n_2}}}$

And now we can calculate the statistic:

$t=\frac{(10 -7)-(0)}{\sqrt{\frac{2.268^2}{8}+\frac{1.414^2}{6}}}=3.036$

The degrees of freedom are given by:

$df=8+6-2=12$

And now we can calculate the p value using the altenative hypothesis:

$p_v =P(t_{12}>3.036) =0.0051$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Orno) is significantly higher than the mean for the group 2 (Edne).

4 0

3 years ago