Determine the solution when given the following expression and coordinate.

What are the x-intercepts of the graph of the function f(x) = x2 4x – 12? (–6, 0), (2,0) or (–2, –16), (0, –12) or (–6, 0), (–2,

Sergio [31]

Xints are where y=0

0=x^2+4x-12
0=(x-2)(x+6)

x-2=0
x=2

x+6=0
x=-6

(x,y)
xints are
(2,0) and (-6,0)

8 0

3 years ago

Read 2 more answers

A physicist examines 25 water samples for nitrate concentration. The mean nitrate concentration for the sample data is 0.165 cc/

xz_007 [3.2K]

Answer:

The 80% confidence interval for the the population mean nitrate concentration is (0.144, 0.186).

Critical value t=1.318

Step-by-step explanation:

We have to calculate a 80% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=0.165.

The sample size is N=25.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{0.078}{\sqrt{25}}=\dfrac{0.078}{5}=0.016$

The degrees of freedom for this sample size are:

$df=n-1=25-1=24$

The t-value for a 80% confidence interval and 24 degrees of freedom is t=1.318.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=1.318 \cdot 0.016=0.021$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 0.165-0.021=0.144\\\\UL=M+t \cdot s_M = 0.165+0.021=0.186$

The 80% confidence interval for the population mean nitrate concentration is (0.144, 0.186).

6 0

3 years ago

a rectangular sheet is 62cm long and 42cm wide from each of its corners a square of side 6cm is cut out and an open box is made

Alla [95]

Answer:

The volume of rectangular box is 9,000 cubic centimeters

Step-by-step explanation:

Given as :

The length of the rectangular sheet = L = 62 cm

The width of the rectangular sheet = B = 42 cm

Let The volume of rectangular box = v cubic centimeters

From each corner of sheet a square of 6 cm is cut out and an open box made

So, The length of box = L' = (L - 6 - 6)

i.e L' = (62 - 6 - 6)

Or, L' = 50 cm

Or, Length of the box = 50 cm

And The width of the box = B' = (B - 6 - 6)

i.e B' = (42 - 6 - 6)

Or, B' = 30 cm

Or width of the box = 30 cm

And,

The height of the box = H = 6 cm

∴ volume of rectangular box = Length of box × Breadth of box × Height of box

i.e v = L' × B' × H

Or, v = 50 cm × 30 cm × 6 cm

Or, v = 9,000 cm³

So, The volume of rectangular box = v = 9,000 cm³

Hence, The volume of rectangular box is 9,000 cubic centimeters Answer

4 0

3 years ago

The Lewis family and the Perry family each used their sprinklers last summer. The water output rate for the Lewis family's sprin

Rashid [163]

The Lewis family and the Perry family each used their sprinklers last summer. The water output rate for the Lewis family's sprinkler was 15 L per hour. The water output rate for the Perry family's sprinkler was 40 L per hour. The families used their sprinklers for a combined total of 65 hours, resulting in a total water output of 1850 L . How long was each sprinkler used

Answer:

Hours of use of sprinkler by Lewis family is 30 hours.

Hours of use of sprinkler by Perry family is 35 hours.

Step-by-step explanation:

Let W = hours of use by Lewis family

65 - W = hours of use by Perry family

Then;

15W + 40*(65 - W) = 1850

15W + 2600 - 40W = 1850

(15 - 40)W = 1850 - 2600

-25W = −750

W = 750/25 = 30

Therefore;

W = 30

65 - W = 65 - 30 = 35

Hence,

Hours of use by Lewis family is 30 hours.

Hours of use by Perry family is 35 hours.

3 0

3 years ago

Intellectual development (Perry) scores were determined for 21 students in a first-year, project-based design course. (Recall th

Anit [1.1K]

Answer:

The 99% confidence interval is (3.0493, 3.4907).

We are 99% sure that the true mean of the students Perry score is in the above interval.

Step-by-step explanation:

Our sample size is 21.

The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So

$df = 21-1 = 20$ .

Then, we need to subtract one by the confidence level $\alpha$ and divide by 2. So:

$\frac{1-0.99}{2} = \frac{0.01}{2} = 0.005$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 20 and 0.005 in the two-sided t-distribution table, we have $T = 2.528$