Who ever answers this first gets brainly

Help asap plssssss, brainliest given!!

artcher [175]

Answer:

y= 1500x = 29690

y=# of movie screens

x= years after 1996

Step-by-step explanation:

3 0

3 years ago

Help! Will give branliest!!

sergey [27]

Hello from MrBillDoesMath!

Answer:

sin(A) = k/c

sin(C) = k/a

Discussion:

By definition of sine ,

sin(A) = side opposite angle / hypotenuse = k/c

sin(C) = side opposite angle/ hypotenuse = k/a

Added remark:

From the first equation, k = c sin(A). From the second equation k = asin(C). Hence.

asin(C) = csin(A) => divide both sides by "ac"

sin(C)/ c = sin(A)/a

Which is the law of sines for a triangle.

Thank you,

MrB

4 0

3 years ago

Use distributive property to rewrite 3×546

sergij07 [2.7K]

Answer:

1638

Step-by-step explanation:

3(500)+3(40)+3(6)=1500+120+18=1638

4 0

3 years ago

What is the formula for midpoint??

Feliz [49]

(-4,4) (3,-2)
formula for finding mid point is
M=(x1+x2/2,y1+y2/2)
m=-4+3/2,4-2/2
m=(-1/2,1)

4 0

3 years ago

The degenerative disease osteoarthritis most frequently affects weight-bearing joints such as the knee. Two random samples from

Vinvika [58]

Answer:

$t=\frac{(810-770)-0}{\sqrt{\frac{67^2}{13}+\frac{56^2}{19}}}}=1.771$

$df=n_1 +n_2 -2=13+19-2=30$

Since is a right tailed test the p value would be:

$p_v =P(t_{30}>1.771)=0.0434$

Comparing the p value with the significance level $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for group 1 is significantly higher than the mean for the group 2

Step-by-step explanation:

Data given

$\bar X_{1}=810$ represent the mean for sample 1

$\bar X_{2}=770$ represent the mean for sample 2

$s_{1}=67$ represent the sample standard deviation for 1

$s_{2}=56$ represent the sample standard deviation for 2

$n_{1}=13$ sample size for the group 2

$n_{2}=19$ sample size for the group 2

$\alpha=0.05$ Significance level provided

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for category 1 is higher than the mean for category 2, the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2}\leq 0$

Alternative hypothesis: $\mu_{1} - \mu_{2}> 0$

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=13+19-2=30$

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

With the info given we can replace in formula (1) like this:

$t=\frac{(810-770)-0}{\sqrt{\frac{67^2}{13}+\frac{56^2}{19}}}}=1.771$

P value

Since is a right tailed test the p value would be:

$p_v =P(t_{30}>1.771)=0.0434$

Comparing the p value with the significance level $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for group 1 is significantly higher than the mean for the group 2

4 0

3 years ago