Please help me PLEASE

At total of 2 feet of snow falls over three days on the first day 1/6 of the snow falls on the second day 7/12 of the snow falls

ioda

Answer:

$Day\ 3 = 1\frac{3}{12}$

Step-by-step explanation:

Given

$Total = 2ft$

$Day\ 1 = \frac{1}{6}$

$Day\ 2 = \frac{7}{12}$

Required

Determine the amount of snow on day 3

This can be calculated using:

$Day\ 1 + Day\ 2 + Day\ 3 = Total$

Substitute values for the known parameters

$\frac{1}{6} + \frac{7}{12} + Day\ 3 = 2$

Collect Like Terms

$Day\ 3 = 2 - \frac{1}{6} - \frac{7}{12}$

Take LCM

$Day\ 3 = \frac{24 - 2 - 7}{12}$

$Day\ 3 = \frac{15}{12}$

Express as mixed number

$Day\ 3 = 1\frac{3}{12}$

5 0

3 years ago

Find a polynomial P(x) with the zeros x=2, −(2), √3, −√3

Zigmanuir [339]

Answer:

P(x) = $x^{4}$ - 7x² + 12

Step-by-step explanation:

Given zeros x = 2, x = - 2, x = $\sqrt{3}$ , x = - $\sqrt{3}$

Then the factors are (x - 2), (x + 2), (x - $\sqrt{3}$ ), (x + $\sqrt{3}$ )

and the polynomial is the product of the factors, that is

P(x) = (x - 2)(x + 2)(x - $\sqrt{3}$ )(x + $\sqrt{3}$ ) ← expand in pairs using FOIL

= (x² - 4)(x² - 3) ← distribute

= $x^{4}$ - 3x² - 4x² + 12 ← collect like terms

= $x^{4}$ - 7x² + 12

6 0

2 years ago

1. Determine whether the graphs of the given equation are Parallel, perpendicular, or neither.

Juliette [100K]

1. A. parallel, because when you get y by itself it is y=-2x+7, the two equations have the same slope which means they are parallel.

2. B. sometimes, in order for two lines to be parallel they have to have the same slope so, it is possible that two lines say with a slope of 3, those would be parallel BUT it doesn't always have to have a positive slope for it to be parallel like in question 1.

3. C. never, If they have the same slope they are parallel despite whatever the y-intercept is.

5 0

2 years ago

Read 2 more answers

Pls help me with math ASAP

Vladimir79 [104]

C. Supplementary
This means their angles add up to 180° which is a straight line

3 0

2 years ago

Read 2 more answers

a. Fill in the midpoint of each class in the column provided. b. Enter the midpoints in L1 and the frequencies in L2, and use 1-

Tresset [83]

Answer:

$\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1} & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7} & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}$

Using the frequency distribution, I found the mean height to be 70.2903 with a standard deviation of 3.5795

Step-by-step explanation:

Given

See attachment for class

Solving (a): Fill the midpoint of each class.

Midpoint (M) is calculated as:

$M = \frac{1}{2}(Lower + Upper)$

Where

$Lower \to$ Lower class interval

$Upper \to$ Upper class interval

So, we have:

Class 63-65:

$M = \frac{1}{2}(63 + 65) = 64$

Class 66 - 68:

$M = \frac{1}{2}(66 + 68) = 67$

When the computation is completed, the frequency distribution will be:

$\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1} & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7} & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}$

Solving (b): Mean and standard deviation using 1-VarStats

Using 1-VarStats, the solution is:

$\bar x = 70.2903$

$\sigma = 3.5795$

<em>See attachment for result of 1-VarStats</em>

8 0

2 years ago