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2 years ago

9

Each member of a 5-member cross-country team runs a course. Their individual times are as follows:

1 answer:

Anastasy [175]2 years ago

7 0

Step-by-step explanation:

the average is the sum of all data points divided by the number of data points (5).

2 hours 17 minutes

2 hours 48 minutes

1 hour 53 minutes

2 hours 19 minutes

1 hour 38 minutes

------------------------------

8 hours 175 minutes

175 minutes = 2 hours 55 minutes

so, we need to add this to the 8 hours and get

10 hours 55 minutes

this we need to divide by 5 for the average time

(10 hours 55 minutes) / 5 = 2 hours 11 minutes =

= 2×60 + 11 = 120 + 11 = 131 minutes.

so, their score is 131.

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Need this done ASAP :) plz help

vlabodo [156]

go on google im gessing so the anser mite be 245776336

7 0

4 years ago

A quadrilateral has vertices at $(0,1)$, $(3,4)$, $(4,3)$ and $(3,0)$. Its perimeter can be expressed in the form $a\sqrt2+b\sqr

seraphim [82]

Answer:

a + b = 12

Step-by-step explanation:

Given

Quadrilateral;

Vertices of (0,1), (3,4) (4,3) and (3,0)

$Perimeter = a\sqrt{2} + b\sqrt{10}$

Required

$a + b$

Let the vertices be represented with A,B,C,D such as

A = (0,1); B = (3,4); C = (4,3) and D = (3,0)

To calculate the actual perimeter, we need to first calculate the distance between the points;

Such that:

AB represents distance between point A and B

BC represents distance between point B and C

CD represents distance between point C and D

DA represents distance between point D and A

Calculating AB

Here, we consider A = (0,1); B = (3,4);

Distance is calculated as;

$Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$(x_1,y_1) = A(0,1)$

$(x_2,y_2) = B(3,4)$

Substitute these values in the formula above

$Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$AB = \sqrt{(0 - 3)^2 + (1 - 4)^2}$

$AB = \sqrt{( - 3)^2 + (-3)^2}$

$AB = \sqrt{9+ 9}$

$AB = \sqrt{18}$

$AB = \sqrt{9*2}$

$AB = \sqrt{9}*\sqrt{2}$

$AB = 3\sqrt{2}$

Calculating BC

Here, we consider B = (3,4); C = (4,3)

Here,

$(x_1,y_1) = B (3,4)$

$(x_2,y_2) = C(4,3)$

Substitute these values in the formula above

$Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$BC = \sqrt{(3 - 4)^2 + (4 - 3)^2}$

$BC = \sqrt{(-1)^2 + (1)^2}$

$BC = \sqrt{1 + 1}$

$BC = \sqrt{2}$

Calculating CD

Here, we consider C = (4,3); D = (3,0)

Here,

$(x_1,y_1) = C(4,3)$

$(x_2,y_2) = D (3,0)$

Substitute these values in the formula above

$Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$CD = \sqrt{(4 - 3)^2 + (3 - 0)^2}$

$CD = \sqrt{(1)^2 + (3)^2}$

$CD = \sqrt{1 + 9}$

$CD = \sqrt{10}$

Lastly;

Calculating DA

Here, we consider C = (4,3); D = (3,0)

Here,

$(x_1,y_1) = D (3,0)$

$(x_2,y_2) = A (0,1)$

Substitute these values in the formula above

$Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$DA = \sqrt{(3 - 0)^2 + (0 - 1)^2}$

$DA = \sqrt{(3)^2 + (- 1)^2}$

$DA = \sqrt{9 + 1}$

$DA = \sqrt{10}$

The addition of the values of distances AB, BC, CD and DA gives the perimeter of the quadrilateral

$Perimeter = 3\sqrt{2} + \sqrt{2} + \sqrt{10} + \sqrt{10}$

$Perimeter = 4\sqrt{2} + 2\sqrt{10}$

Recall that

$Perimeter = a\sqrt{2} + b\sqrt{10}$

This implies that

$a\sqrt{2} + b\sqrt{10} = 4\sqrt{2} + 2\sqrt{10}$

By comparison

$a\sqrt{2} = 4\sqrt{2}$

Divide both sides by $\sqrt{2}$

$a = 4$

By comparison

$b\sqrt{10} = 2\sqrt{10}$

Divide both sides by $\sqrt{10}$

$b = 2$

Hence,

a + b = 2 + 10

a + b = 12

3 0

3 years ago

PLeasehelppp me with this problem

melisa1 [442]

You would need 8.25 cups

8 0

3 years ago

Read 2 more answers

The swim club rents the municipal pool for a flat fee of $500 plus a charge of $75 per swimmer during

xxTIMURxx [149]

Answer:

Y=500x+75

Step-by-step explanation:

4 0

4 years ago

Which function is increasing?? :)

exis [7]

Answer:

$\large\boxed{B.\ f(x)=4^x}$

Step-by-step explanation:

Exponential function $f(x)=(a)^x$ is

increasing if $a > 1\to a\in(1,\ \infty)$

decreasing if $0 < a < 1\to a\in(0,\ 1)$

$A.\ f(x)=\left(\dfrac{1}{4}\right)^x\to a=\dfrac{1}{4}\in(0,\ 1)\to\text{decreasing}\\\\B.\ f(x)=4^x\to a=4\in(1,\ \infty)\to\text{increasing}\\\\C.\ f(x)=(0.4)^x\to a=0.4\in(0,\ 1)\to\text{decreasing}\\\\D.\ f(x)=\left(\dfrac{1}{2}\right)^x\to a=\dfrac{1}{2}\in(0,\ 1)\to\text{decreasing}$

7 0

3 years ago