Find the zeros of the quadratic function

I need help with this question

Artyom0805 [142]

Answer:

Step-by-step explanation:

It's algebra really, basically make a system of equations.

1b+5j=t

3b+2j=2t

I'm gonna solve the first one for buckets so we have it in terms of jars.

b+5j=t

b=t-5j

Now I plug that into the other equation

3b+2j=2t

3(t-5j)+2j=2t

3t-15j+2j=2t

Now, since it wants how many jars are in one tub, I want to solve it so there's 1t on one side of the equation and all js on the other. or it other words solve for t.

3t-15j+2j=2t

3t-13j = 2t

3t = 2t+13j

t = 13j

So it takes 13 jars to make one tub, or one jar is 1/13 of a tub. You could then plug that in to one of the equations and find how much a bucket is.

5 0

3 years ago

Suppose that you had the following data set. 500 200 250 275 300 Suppose that the value 500 was a typo, and it was suppose to be

hodyreva [135]

Answer:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

Step-by-step explanation:

The subindex B is for the before case and the subindex A is for the after case

Before case (with 500)

For this case we have the following dataset:

500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

And the sample deviation with the following formula:

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

After case (With -500 instead of 500)

For this case we have the following dataset:

-500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

And the sample deviation with the following formula:

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

And as we can see we have a significant change between the two values for the two cases.

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

8 0

3 years ago

The following graph shows a proportional relationship.

Molodets [167]

Answer:

4

Step-by-step explanation:

because, if y = 4 and x = 1 then it would be 4/1 or 4 just simplify.

Y/X --- Y = 4 --- X = 1 ---- that makes 4/1 or 4

5 0

3 years ago

Which is equivalent to 56c^16/162c^3 squared after it has been simplified completely?

Oxana [17]

Answer:

B

Step-by-step explanation:

Just took the test plz mark brainliest

4 0

3 years ago

david packed 16 bags of carrot sticks for the picnic. 7/8 of the bags were eaten. how many bags of carrots were eaten?

tia_tia [17]

Answer:

14 bags are eaten

Step-by-step explanation:

7/8*2/2= 14/16

3 0

2 years ago