Subtract 4 feet 5 inches from 12 feet 3 inches

Can anybody help me plzzzz

melisa1 [442]

SSS I’m pretty sure . I’m doing the same thing in math rn

3 0

2 years ago

HELP PLZ WILL GIVE YOU BRAINIEST

slamgirl [31]

Answer:

diameter is 16.60 ft

the circumference is 52.12ft

Step-by-step explanation:

Hope this helps!

8 0

3 years ago

Please help me out guys. The photo will show you what to do. 50 points cause I need it done fast

Nonamiya [84]

Answer:

a) starting height: 5.5 ft

b) hang time: 5.562 seconds

c) maximum height: 126.5 ft

d) time to maximum height: 2.75 seconds

Step-by-step explanation:

a) The starting height is the height at t=0.

h(0) = -16·0 +88·0 +5.5

h(0) = 5.5

The starting height is 5.5 feet.

__

b) The ball is in the air between t=0 and the non-zero time when h(t) = 0. We can find the latter by solving ...

-16t^2 + 8t +5.5 = 0

t^2 -(11/2)t = 5.5/16 . . . . . subtract 5.5, then divide by -16

t^2 -(11/2)t +(11/4)^2 = (5.5/16) +(11/4)^2 . . . . complete the square

(t -11/4)^2 = 126.5/16 . . . . . . . . . . . . . . . . . . . . call this [eq1] for later use

t -11/4 = √7.90625

t = 2.75 +√7.90625 ≈ 5.562

The ball will be in the air about 5.562 seconds.

__

c) If we multiply [eq1] above by -16 and add the constant on the right, we get the vertex form of the height equation:

h(t) = -16(t -11/4) +126.5

The vertex at (2.75, 126.5) tells us ...

The maximum height of the ball is 126.5 feet.

__

d) That same vertex point tells us ...

The maximum height will be reached at t = 2.75 seconds.

_____

If you really need answers fast, a graphing calculator can give them to you in very short order (less than a minute).

6 0

3 years ago

HELPPP!!!!!! URGENT!!!

Burka [1]

Answer:

20

Step-by-step explanation:

x = price

y = number of keyboard

xy = 540

(x-2)(y+3) = 540

x = 540/y

==> x = 20

==> y = 27

6 0

3 years ago

The time it takes for a planet to complete its orbit around a particular star is called the? planet's sidereal year. The siderea

BartSMP [9]

Answer:

(a) See below

(b) r = 0.9879

(c) y = -12.629 + 0.0654x

(d) See below

(e) No.

Step-by-step explanation:

(a) Plot the data

I used Excel to plot your data and got the graph in Fig 1 below.

(b) Correlation coefficient

One formula for the correlation coefficient is

$r = \dfrac{\sum{xy} - \sum{x} \sum{y}}{\sqrt{\left [n\sum{x}^{2}-\left (\sum{x}\right )^{2}\right]\left [n\sum{y}^{2} -\left (\sum{y}\right )^{2}\right]}}$

The calculation is not difficult, but it is tedious.

(i) Calculate the intermediate numbers

We can display them in a table.

x y xy x² y²

36 0.22 7.92 1296 0.05

67 0.62 42.21 4489 0.40

93 1.00 93.00 20164 3.46

433 11.8 5699.4 233289 139.24

887 29.3 25989.1 786769 858.49

1785 82.0 146370 3186225 6724

2797 163.0 455911 7823209 26569

3675 248.0 911400 13505625 61504

9965 537.81 1545776.75 25569715 95799.63

(ii) Calculate the correlation coefficient

$r = \dfrac{\sum{xy} - \sum{x} \sum{y}}{\sqrt{\left [n\sum{x}^{2}-\left (\sum{x}\right )^{2}\right]\left [n\sum{y}^{2} -\left (\sum{y}\right )^{2}\right]}}\\\\= \dfrac{9\times 1545776.75 - 9965\times 537.81}{\sqrt{[9\times 25569715 -9965^{2}][9\times 95799.63 - 537.81^{2}]}} \approx \mathbf{0.9879}$

(c) Regression line

The equation for the regression line is

y = a + bx where

$a = \dfrac{\sum y \sum x^{2} - \sum x \sum xy}{n\sum x^{2}- \left (\sum x\right )^{2}}\\\\= \dfrac{537.81\times 25569715 - 9965 \times 1545776.75}{9\times 25569715 - 9965^{2}} \approx \mathbf{-12.629}\\\\b = \dfrac{n \sum xy - \sum x \sum y}{n\sum x^{2}- \left (\sum x\right )^{2}} - \dfrac{9\times 1545776.75 - 9965 \times 537.81}{9\times 25569715 - 9965^{2}} \approx\mathbf{0.0654}\\\\\\\text{The equation for the regression line is $\large \boxed{\mathbf{y = -12.629 + 0.0654x}}$}$

(d) Residuals

Insert the values of x into the regression equation to get the estimated values of y.

Then take the difference between the actual and estimated values to get the residuals.

x y Estimated Residual

36 0.22 -10 10

67 0.62 -8 9

93 1.00 -7 8

142 1.86 -3 5

433 11.8 19 - 7

887 29.3 45 -16

1785 82.0 104 -22

2797 163.0 170 - 7

3675 248.0 228 20

(e) Suitability of regression line

A linear model would have the residuals scattered randomly above and below a horizontal line.

Instead, they appear to lie along a parabola (Fig. 2).

This suggests that linear regression is not a good model for the data.

4 0

3 years ago