Pls help I’ll mark brainliest

How do i solve this by using substitution? y=4x-9y=x-3

noname [10]

Y = 4x -9
y = x-3
Replace the "y" in above equation by x-3
x - 3 = 4x - 9
-3x = -6
x = 2
Then replace x = 2 to the beneth
y = 2 - 3 = -1

4 0

3 years ago

What is greater 1.5 pounds of raisins or 650 grams of raisins.

tekilochka [14]

Answer:

1.5 pounds is greater

Step-by-step explanation:

454 grams = one pound

1.5 pounds = 681 grams

6 0

3 years ago

Write an equation of the translation of y=x^3 that has been moved 2 units up, 3 units to the right, and stretched by a factor of

Yuki888 [10]

Y=(4x-3)³+2

To shift the graph to the right 3 units, we subtract 3 before the exponent. To shift it up 2 units, we add 2 to the end. To stretch it by 4, we multiply x by 4 before the exponent.

6 0

3 years ago

Can u pls give a clear explanation & word it out how u got the answer without using big words. Thank u

gizmo_the_mogwai [7]

A.

We can rewrite the squared in two ways.

The first way is to rewrite it as a multiplication, this is:

$(y-6)^2=(y-6)(y-6)$

The second way to rewrite the squared is by using the formula for this kind of product:

$(y-6)^2=y^2+2(-6)(y)+(-6)^2$

B.

Once again we can find the final result using each of the options given in part A, for the first option we have:

$\begin{gathered} (y-6)^2=(y-6)(y-6) \\ =y^2-6y-6y+36 \\ =y^2-12y+36 \end{gathered}$

For the second option we have:

$\begin{gathered} (y-6)^2=y^2+2(-6)(y)+(-6)^2 \\ =y^2-12y+36 \end{gathered}$

No matter which way we choose the answer for the squared is:

$y^2-12y+36$

3 0

1 year 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