The sum of two numbers is 854. One of the numbers is 447. What is the other number?

The side that measures 12 cm corresponds to the side that measures 3 cm.
The side that measures 16 cm corresponds to the side that measures 4 cm.
The ratio of the side lengths is (12 cm)/(3 cm) = (16 cm)/(4 cm) = 4

Now let's calculate the two areas.

For the 3 cm by 4 cm rectangle:
A = LW = 3 cm * 4 cm = 12 cm^2

For the 12 cm by 16 cm rectangle:
A = LW = 12 cm * 16 cm = 192 cm^2

The ratio of the areas is (192 cm^2)/(12 cm^2) = 16

The ratio of the areas is 16.
The ratio of side lengths is 4.
Notice that 16 = 4^2.
The ratio of the areas is the square of the ratio of the side lengths.

6 0

3 years ago

Holaa cómo están<br><br><br>espero que bien

9966 [12]

Answer:

si muy bien y tu?

Step-by-step explanation:

6 0

3 years ago

Tensile-strength tests were carried out on two different grades of wire rod (Fluidized Bed Patenting of Wire Rods, Wire J., June

Shtirlitz [24]

Answer:

$t=\frac{(123.6-107.6)-(10)}{\sqrt{\frac{2^2}{129}+\frac{1.3^2}{129}}}=28.569$

$p_v =P(t_{256}>28.569) \approx 0$

So with the p value obtained and using the significance level given $\alpha=0.1$ 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 true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mm2.

Step-by-step explanation:

The statistic is given by this formula:

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

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 \leq \mu_2 +10$

Alternative hypothesis: $\mu_1 >\mu_2 +10$

Or equivalently:

Null hypothesis: $\mu_1 - \mu_2 \leq 10$

Alternative hypothesis: $\mu_1 -\mu_2>10$

Our notation on this case :

$n_1 =129$ represent the sample size for group AISI 1078

$n_2 =129$ represent the sample size for group AISI 1064

$\bar X_1 =123.6$ represent the sample mean for the group AISI 1078

$\bar X_2 =107.6$ represent the sample mean for the group AISI 1064

$s_1=2.0$ represent the sample standard deviation for group 1 AISI 1078

$s_2=1.3$ represent the sample standard deviation for group AISI 1064

And now we can calculate the statistic:

$t=\frac{(123.6-107.6)-(10)}{\sqrt{\frac{2^2}{129}+\frac{1.3^2}{129}}}=28.569$

Now we can calculate the degrees of freedom given by:

$df=129+129-2=256$

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

$p_v =P(t_{256}>28.569) \approx 0$

So with the p value obtained and using the significance level given $\alpha=0.1$ 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 true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mm2.

7 0

3 years ago