50 POINTS PLUS BRAINLIEST!

Urgent please help..

Serggg [28]

Answer is 141.6. I did 3/x = 5/236 and got that answer! Good luck!

6 0

3 years ago

3. Si al doble de un número se le resta su mitad resulta 30. ¿Cuál es el número? Plantear ecuación y resolverla

guajiro [1.7K]

Respuesta:

2x - 0.5x = 30

x = 20

Explicación paso a paso:

Llamemos "x" a la incógnita, el número desconocido.

Cuando nos referimos al doble de un número, es lo mismo que 2x.

Cuando nos referimos a la mitad de un número, es lo mismo que 0.5 x.

Consideremos que al doble de un número (2x) se le resta su mitad 0.5x) y se obtiene 30. La ecuación resultante es:

2x - 0.5x = 30

1.5x = 30

x = 30/1.5 = 20

8 0

3 years ago

Please help me on this

Nastasia [14]

Answer:

Points H, J, G, and K

Step-by-step explanation:

For quadrats the right top is 1

left top is 2

left bottom is 3

right bottom is 4

and it is asking for the forth one so look at the right bottom "square"

G and K are on the line but that still makes them part of a Quadratic Just they get to be part of 2 of them like...

G is 3 and 4

K is 1 and 4

5 0

2 years ago

Read 2 more answers

A set of N = 20 exam scores has a mean of m = 50. The instructor discovered that an error was made grading one exam, so 20 point

tatuchka [14]

Answer:

The new mean for the exam scores is 51

Step-by-step explanation:

the mean can be calculated using the formula

mean of scores = $\frac{Sum total of scores}{Number of students}$

before the 20 points was added, the mean of the score was = 50.

Inserting this into the formula, we have:

50 = $\frac{Sum total}{20 students}$

from this, we can compute the sum total of the scores as

sum total = 50 × 20 = 1000

when the extra 20 marks is added, the sum of the entire scores will be changed to 1000 + 20 = 1020.

Hence the new mean will be computed as (New Sum total of scores / Number of students)

= $\frac{1020}{20}$ = 51.

∴ The new mean of the scores is = 51

8 0

3 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