Answer:
t = -5
Step-by-step explanation:
Solve for t:
5 (t - 3) - 2 t = -30
Hint: | Distribute 5 over t - 3.
5 (t - 3) = 5 t - 15:
5 t - 15 - 2 t = -30
Hint: | Group like terms in 5 t - 2 t - 15.
Grouping like terms, 5 t - 2 t - 15 = (5 t - 2 t) - 15:
(5 t - 2 t) - 15 = -30
Hint: | Combine like terms in 5 t - 2 t.
5 t - 2 t = 3 t:
3 t - 15 = -30
Hint: | Isolate terms with t to the left hand side.
Add 15 to both sides:
3 t + (15 - 15) = 15 - 30
Hint: | Look for the difference of two identical terms.
15 - 15 = 0:
3 t = 15 - 30
Hint: | Evaluate 15 - 30.
15 - 30 = -15:
3 t = -15
Hint: | Divide both sides by a constant to simplify the equation.
Divide both sides of 3 t = -15 by 3:
(3 t)/3 = (-15)/3
Hint: | Any nonzero number divided by itself is one.
3/3 = 1:
t = (-15)/3
Hint: | Reduce (-15)/3 to lowest terms. Start by finding the GCD of -15 and 3.
The gcd of -15 and 3 is 3, so (-15)/3 = (3 (-5))/(3×1) = 3/3×-5 = -5:
Answer: t = -5
If you have 3 1/2 cups of something and you want to find how many 1/4 cups there are the you just need to divide 3 1/2 by 1/4

the answer is 14, there are 14 1/4 cups in 3 1/2 cups.
hope I helped :)
Y=2.50x+20. Plug in values for x starting at 0 and you will get values for y and then graph.
Hey sorry if I cant show work but here is the point form:
( 0 , 1 ) and ( -3 , -2 )
Answer:
Because the absolute value of the test statistic is <u>less than</u> the positive critical value, there <u>is not</u> enough evidence to support the claim that there is a linear correlation between the weights of discarded paper and glass for a significance level of α = 0.05.
Step-by-step explanation:
The correlation matrix provided is:
Variables Paper Glass
Paper 1 0.1853
Glass 0.1853 1
Te hypothesis for the test is:
<em>H</em>₀: <em>ρ</em> = 0 vs. <em>H</em>₀: <em>ρ</em> ≠ 0
The test statistic is:
<em>r</em> = 0.1853 ≈ 0.185
As the alternate hypothesis does not specifies the direction of the test, the test is two tailed.
The critical value for the two-tailed test is:

The conclusion is:
Because the absolute value of the test statistic is <u>less than</u> the positive critical value, there <u>is not</u> enough evidence to support the claim that there is a linear correlation between the weights of discarded paper and glass for a significance level of α = 0.05.