Angles θ and Φ are supplementary, and theta equals 3pi/7 find the measure of theta

Y = -1/2x + 2.5 (no answer options)

ruslelena [56]

I mean i don't know what to do but i made this graph online using https://www.mathpapa.com/algebra-calculator.html

8 0

3 years ago

Which equation is true when x = 5? A 8x = 80 B x + 4 = 9 C 10 ÷ x = 5 D x − 6= 1

Pie

Answer:

B

Step-by-step explanation:

A. 8 * x = 80

8 * 5 = 40. 40 is not 80, so it doesn't work.

B. x + 4 = 9

5 + 4 = 9. This works.

C. 10 ÷ x = 5

10 ÷ 5 = 2. 5 is not 2, so it doesn't work.

D. x - 6 = 1

5 - 6 = -1. 1 is not -1, so it doesn't work.

5 0

2 years ago

Christian sold tickets to the game . Good seats were $5 each and poor seats cost $2 each. 210 people attended and paid $660. How

Sholpan [36]

People who purchased good seats will be labeled as a
people who purchased poor seats will be labeled as b
therefore...
5a + 2b = 660
a + b = 210
Now, all that has to be done is for the system to be solved.
-2(a+b=210)

5a + 2b = 660
-2a - 2a = -420

3a=240
a= 80
*3 people purchased good seats.

Now we can plug that number in

5(80) + 2b = 660
400 + 2b = 660
2b = 260
b = 130
*130 people purchased poor seats

5 0

3 years ago

Assume that you have a sample of n 1 equals 6, with the sample mean Upper X overbar 1 equals 50, and a sample standard deviati

tigry1 [53]

Answer:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

$df=6+5-2=9$

$p_v =P(t_{9}>2.656) =0.0131$

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

$n_1 =6$ represent the sample size for group 1

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

$\bar X_1 =50$ represent the sample mean for the group 1

$\bar X_2 =38$ represent the sample mean for the group 2

$s_1=7$ represent the sample standard deviation for group 1

$s_2=8$ represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 > \mu_2$

We are assuming that the population variances for each group are the same

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

The statistic for this case is given by:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

The pooled variance is:

$S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

We can find the pooled variance:

$S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67$

And the pooled deviation is:

$S_p=7.46$

The statistic is given by:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

The degrees of freedom are given by:

$df=6+5-2=9$

The p value is given by:

$p_v =P(t_{9}>2.656) =0.0131$

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

4 0

3 years ago

What is subtraction?

Mazyrski [523]

Wen you deduct an amount from something

5 0

3 years ago

Read 2 more answers