!HELP! What is the equation for the following absolute-value functions?

Sati [7]

11. is 25°
(Both triangles have two congruent angles which means the 3rd angle is bound to be the same. Just add the 90° and the 65° to get 155 and subtract from 180° because all triangles equal 180°)
12. is 25°
(Same as before, two of the angles are congruent so the third one is the same. Add all of the angles together and have them equal 180° because their is a variable present. 2x+60+70=180 or 2x+130=180. subtract 130 from both sides to get the equation 2x=50 and then divide both sides by 2 to get x=25.)
13. y=14 and x=11.6 (assuming m<d says 15x ÷ 2 because I couldn't really read it)
(m<a = m<d and m<b = m<e because they are lined up together in the ABC and DEF equation. so 3y=42 and (15x÷2=87)

7 0

2 years ago

A medical researcher wondered if there is a significant difference between the mean birth weight of boy and girl babies. Random

Yuliya22 [10]

Answer:

b) independent samples t-test

$t=\frac{(5.92 -5.74)-(0)}{\sqrt{\frac{1.88^2}{5}}+\frac{1.81^2}{5}}=0.154$

$p_v =2*P(t_{8}>0.154) =0.881$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (boys) is NOT significantly different than the mean for the group 2 (Girls).

Step-by-step explanation:

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

Or equivalently:

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

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

Our notation on this case :

$n_1 =5$ represent the sample size for group 1

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

We can calculate the mean and deviation for eaach group with the following formulas:

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

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

$\bar X_1 =5.92$ represent the sample mean for the group 1

$\bar X_2 =5.74$ represent the sample mean for the group 2

$s_1=1.88$ represent the sample standard deviation for group 1

$s_2=1.81$ represent the sample standard deviation for group 2

If we see the alternative hypothesis we see that we are conducting a bilateral test and with independnet samples t test.

b) independent samples t-test

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:

$t=\frac{(5.92 -5.74)-(0)}{\sqrt{\frac{1.88^2}{5}}+\frac{1.81^2}{5}}=0.154$

The degrees of freedom are given by:

$df=5+5-2=8$

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

$p_v =2*P(t_{8}>0.154) =0.881$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (boys) is NOT significantly different than the mean for the group 2 (Girls).

7 0

3 years ago

Find the distance between (-3, 11) and (-6,4).Round to the nearest then the

Likurg_2 [28]

Answer:

d =7.6

Step-by-step explanation:

$(-3,11) =(x_1,y_1)\\(-6,4)=(x_2,y_2)\\\\\mathrm{Compute\:the\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\\\quad \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\\\\=\sqrt{\left(-6-\left(-3\right)\right)^2+\left(4-11\right)^2}\\= \sqrt{(-6+3)^2+(4-11)^2}\\ \\d= \sqrt{(-3)^2+(-7)^2} \\\\d = \sqrt{9+49}\\ \\d = \sqrt{58}\\\\d = 7.61577\\\\d =7.6$