F=fg/f+g-d Solve for f. F and f are different!

Solve. 4x+3y=5 2x+3y=1 x=? y=?

klasskru [66]

Answer:

x = 2, and y = -1. Point form: (2, -1)

Step-by-step explanation:

You solve the system of equations, by solving for one variable, and then subsitiute that variable into the other equation.

5 0

3 years ago

Pls help ill give extra points

Serjik [45]

Answer:

13x

Step-by-step explanation:

6 0

3 years ago

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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

There is 5/6 of an apple pie left from dinner. Tomorrow, victor plans to eat 1/6 of the pie that was left. How much of the whole

Tatiana [17]

If there is 5/6 left and he plans to eat 1/6 you do 5/6-1/6=4/6 of the pie

8 0

3 years ago

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I have 8 pennies and some quarters, nickels, and dimes. The number of pennies plus nickels equals the number of dimes. I have 3

Darina [25.2K]

There are five nickels.

Let w = the number of pennies; x = the number of nickels; y = the number of dimes; z = the number of quarters.

Then, we have three equations with four unknowns:

(1) w + x = y

(2) 3w = 2z

(3) 2y = 3z

However, there is a fourth unstated condition: w, x, y, and z must all be integers.

Assume that z = 6 (to avoid fractions).

From Equation (2), w = 4.

From Equation (3), y = 9.

Insert the values for w and y into equation (1).

4 + x = 9

x = 9 – 4 = 5

Thus, there are 4 pennies, 5 nickels, 9 dimes, and 6 quarters.

Check: 4 pennies + 5 nickels = 9 dimes

6 quarters + 4 pennies = 3 quarters for every 2 pennies

9 dimes + 6 quarters = 3 dimes for every 2 quarters

5 0

3 years ago

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