Hi there! I need a previous question of mines taken down how can i do that? Thanks!

Sam decides to build a square garden. If the area of the garden is 25x2 + 30x + 9 square feet, what is the length of one side of

seropon [69]

The correct answer is C. (5x+3) feet, and here is how I got it.
This is the equation you need to use:
(a+b)^2 = (a+b) (a+b)
So, let us plug in 5x and 3 here:
(5x+3)^2 = (5x+3)(5x+3) = 25x^2 + 15x + 15x + 9 = 25x^2 + 30x + 9 which is the area of the garden.

5 0

3 years ago

6y + 2x - 2y - 3x PLEASE HELP ASAP THANK YOU

11Alexandr11 [23.1K]

Answer:

4y-x

Step-by-step explanation:

Combine like-terms.

6y-2y and -3x+2x

Do the Math

4y and -x

4y-x

8 0

4 years ago

Consider a random sample of ten children selected from a population of infants receiving antacids that contain aluminum, in orde

saveliy_v [14]

Answer:

a. Null hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.

Complementary alternative hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is different from the mean plasma aluminum level of the population of infants not receiving antacids.

b. (32.1, 42.3)

c. p-value < .00001

d. The null hypothesis is rejected at the α=0.05 significance level

e. Reformulated null hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.

Reformulated complementary alternative hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is higher than the mean plasma aluminum level of the population of infants not receiving antacids.

p-value equals < .00001. The null hypothesis is rejected at the α=0.05 significance level. This suggests that being given antacids greatly increases the plasma aluminum levels of children.

Step-by-step explanation:

a. Null hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.

Complementary alternative hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is different from the mean plasma aluminum level of the population of infants not receiving antacids. This may imply that being given antacids significantly changes the plasma aluminum level of infants.

b. Since the population standard deviation σ is unknown, we must use the t distribution to find 95% confidence limits for μ. For a t distribution with 10-1=9 degrees of freedom, 95% of the observations lie between -2.262 and 2.262. Therefore, replacing σ with s, a 95% confidence interval for the population mean μ is:

(X bar - 2.262\frac{s}{\sqrt{10} } , X bar + 2.262\frac{s}{\sqrt{10} })

Substituting in the values of X bar and s, the interval becomes:

(37.2 - 2.262\frac{7.13}{\sqrt{10} } , 37.2 + 2.262\frac{7.13}{\sqrt{10} })

or (32.1, 42.3)

c. To calculate p-value of the sample , we need to calculate the t-statistics which equals:

t=\frac{(Xbar-u)}{\frac{s}{\sqrt{10} } } = \frac{(37.2-4.13)}{\frac{7.13}{\sqrt{10} } } = 14.67.

Given two-sided test and degrees of freedom = 9, the p-value equals < .00001, which is less than 0.05.

d. The mean plasma aluminum level for the population of infants not receiving antacids is 4.13 ug/l - not a plausible value of mean plasma aluminum level for the population of infants receiving antacids. The 95% confidence interval for the population mean of infants receiving antacids is (32.1, 42.3) and does not cover the value 4.13. Therefore, the null hypothesis is rejected at the α=0.05 significance level. This suggests that being given antacids greatly changes the plasma aluminum levels of children.

e. Reformulated null hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.

Reformulated complementary alternative hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is higher than the mean plasma aluminum level of the population of infants not receiving antacids.

Given one-sided test and degree of freedom = 9, the p-value equals < .00001, which is less than 0.05. This result is similar to result in part (c). the null hypothesis is rejected at the α=0.05 significance level. This suggests that being given antacids greatly increases the plasma aluminum levels of children.

6 0

4 years ago

Find the longer leg of the triangle.

Paha777 [63]

Answer:

Choice A. 3.

Step-by-step explanation:

The triangle in question is a right triangle.

The length of the hypotenuse (the side opposite to the right angle) is given.
The measure of one of the acute angle is also given.

As a result, the length of both legs can be found directly using the sine function and the cosine function.

Let $\text{Opposite}$ denotes the length of the side opposite to the $30^{\circ}$ acute angle, and $\text{Adjacent}$ be the length of the side next to this $30^{\circ}$ acute angle.

$\displaystyle \begin{aligned}\text{Opposite} &= \text{Hypotenuse} \times \sin{30^{\circ}}\\ &=2\sqrt{3}\times \frac{1}{2} \\&= \sqrt{3}\end{aligned}$ .

Similarly,

$\displaystyle \begin{aligned}\text{Adjacent} &= \text{Hypotenuse} \times \cos{30^{\circ}}\\ &=2\sqrt{3}\times \frac{\sqrt{3}}{2} \\&= 3\end{aligned}$ .

The longer leg in this case is the one adjacent to the $30^{\circ}$ acute angle. The answer will be $3$ .

There's a shortcut to the answer. Notice that $\sin{30^{\circ}} < \cos{30^{\circ}}$ . The cosine of an acute angle is directly related to the adjacent leg. In other words, the leg adjacent to the $30^{\circ}$ angle will be the longer leg. There will be no need to find the length of the opposite leg.

Does this relationship $\sin{\theta} < \cos{\theta}$ holds for all acute angles? (That is, $0^{\circ} < \theta$ ?) It turns out that:

$\sin{\theta} < \cos{\theta}$ if $0^{\circ} < \theta$ ;
$\sin{\theta} > \cos{\theta}$ if $45^{\circ} < \theta$ ;
$\sin{\theta} = \cos{\theta}$ if $\theta = 45^{\circ}$ .

4 0

3 years ago

Read 2 more answers

Table

Lelu [443]

Answer:

A

Step-by-step explanation:

Table:

it says that x is the amount of tickets, so that would be 1, 2, 3, etc.

It says that y is the money spent and also tells us that one ticket is $3, so obviously the money would be increasing by thirds: 3, 6, 9

This all goes with the first table (table A)

Graph:

As for the graph-

X = # of tickets (1, 2, 3)

Y = $ spent (3. 6, 9)

The answer would be graph A, because when looking at it you can tell that the starting point is at $0 for 0 tickets, unlike for graph b, which shows $3 for 0 tickets, which does not make any sense. But when you continue on when graph A, you can tell that this answer choice is correct, because every time, the y value is 3 times the x value which means: y=3x

4 0

3 years ago