Write an equation that passes through the points (4,-1) and (2,- 2).

Solve the equation -15=a/5 A=-3 A=-20 A=-10 A=-75

agasfer [191]

To solve any equation, you do whatever you need to do
to wind up with the unknown quantity all alone on one side.
BUT ... whatever you do to one side of the equation, you must
also do to the other side.

Here, we have the
he original equation: -15 = a/5

Multiply each side by 5 : -75 = a

and bada-bing, THERE's the solution !

4 0

3 years ago

A video sharing website starts with 20,000 members. Each year it loses 25% of the members, but adds 10,000 new members after the

Paraphin [41]

Hello there!

So, as we see here, we can see that they start out with 20,000 members. 25% of 20,000 would be 5,000 members.

So, every year, they lose 25% which is 5,000 members of they team. But then they gain 10,000 more.

So, practically, they get 5,000 every year even though they get 25% off.

5,000 would be the answer.

4 0

3 years ago

If (-3)^-5=1/x, what is the value of x? options are in the photo

alisha [4.7K]

Answer:

X= -243

Used Math.way

4 1

3 years ago

I need help asp . I'm running out of time

frutty [35]

Answer:

A = 104 ft²

Step-by-step explanation:

We're given that the width of the rectangle is 8 so that's the value of w. All we have to do is just plug in 8 into the equation:

A=w²+5w

A=(8)²+5(8)

A=64+40

A = 104 ft²

7 0

3 years ago

A random sample of 49 measurements from one population had a sample mean of 15, with sample standard deviation 5. An independent

kotykmax [81]

Answer:

Comparing the p value with the significance level $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and the difference between the two groups is significantly different.

Step-by-step explanation:

$\bar X_{1}=15$ represent the mean for sample 1

$\bar X_{2}=18$ represent the mean for sample 2

$s_{1}=5$ represent the sample standard deviation for 1

$s_{f}=6$ represent the sample standard deviation for 2

$n_{1}=49$ sample size for the group 2

$n_{2}=64$ sample size for the group 2

$\alpha=0.01$ 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 difference in the population means, the system of hypothesis would be:

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

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

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:

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

And the degrees of freedom are given by $df=n_1 +n_2 -2=49+64-2=111$

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:

$t=\frac{(15-18)-0}{\sqrt{\frac{5^2}{49}+\frac{6^2}{64}}}}=-2.897$

P value

Since is a bilateral test the p value would be:

$p_v =2*P(t_{111}$

Comparing the p value with the significance level $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and the difference between the two groups is significantly different.

7 0

3 years ago