All categories

3 years ago

S = 2r + 3 × solve for x

1 answer:

6 0

$2r+3x=s\qquad\text{subtract 2r from both sides}\\\\3x=s-2r\qquad\text{divide both sides by 3}\\\\\boxed{x=\dfrac{s-2r}{3}}$

You might be interested in

In the coordinate plane, the point X (3, 1) is translated to the point X'(4, -1). Under the same translation, the points Y(-1, -

torisob [31]

<h3>Answer:y and z</h3>

Step-by-step explanation:

<h2>bc y and z is in the alphabet </h2>

7 0

3 years ago

The table below shows an inequality and a number by which to divide both sides.

IRISSAK [1]

Answer:

i dont see a table

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

75% of the normal price of headphones is 45$. What's the normal price?

Brums [2.3K]

Answer:

60$ is the normal price.

Why:

Because if 45$ is 75%, then 30$ is 50%, 15 is 25%, and 0 is 0%. 50% + 50% = 100%. (That's how I did this one, anyway. This works on any 25%, 50%, or 75% problem.)

4 0

3 years ago

Multiply (2.4 ⋅ 1014) ⋅ (4 ⋅ 107). Express the answer in scientific notation

IRINA_888 [86]

9.6*10^21 is your answer

3 0

3 years ago

Read 2 more answers

The heights of a certain type of tree are approximately normally distributed with a mean height p = 5 ft and a standard

arsen [322]

Answer:

A tree with a height of 6.2 ft is 3 standard deviations above the mean

Step-by-step explanation:

⇒ $1^s^t$ statement: A tree with a height of 5.4 ft is 1 standard deviation below the mean(FALSE)

an X value is found Z standard deviations from the mean mu if:

$\frac{X-\mu}{\sigma} = Z$

In this case we have: $\mu=5\ ft$ $\sigma=0.4\ ft$

We have four different values of X and we must calculate the Z-score for each

For X =5.4\ ft

$Z=\frac{X-\mu}{\sigma}\\Z=\frac{5.4-5}{0.4}=1$

Therefore, A tree with a height of 5.4 ft is 1 standard deviation above the mean.

⇒ $2^n^d$ statement:A tree with a height of 4.6 ft is 1 standard deviation above the mean. (FALSE)

For X =4.6 ft

$Z=\frac{X-\mu}{\sigma}\\Z=\frac{4.6-5}{0.4}=-1$

Therefore, a tree with a height of 4.6 ft is 1 standard deviation below the mean .

⇒ $3^r^d$ statement:A tree with a height of 5.8 ft is 2.5 standard deviations above the mean (FALSE)

For X =5.8 ft

$Z=\frac{X-\mu}{\sigma}\\Z=\frac{5.8-5}{0.4}=2$

Therefore, a tree with a height of 5.8 ft is 2 standard deviation above the mean.

⇒ $4^t^h$ statement:A tree with a height of 6.2 ft is 3 standard deviations above the mean. (TRUE)

For X =6.2\ ft

$Z=\frac{X-\mu}{\sigma}\\Z=\frac{6.2-5}{0.4}=3$

Therefore, a tree with a height of 6.2 ft is 3 standard deviations above the mean.

6 0

3 years ago