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iVinArrow [24]
4 years ago
11

Could any of y'all help me?

Mathematics
1 answer:
Scorpion4ik [409]4 years ago
5 0
Start using calculator
You might be interested in
Victoria rides her bike at a constant rate of 1 3/8 miles every 1/4 hour. What is her speed in miles per hour?
NISA [10]

Answer:

5 1/2 miles per 1 hr

Step-by-step explanation:

biked 1 3/8 miles in 1/ of an hour

1 3/8 miles = 11/8 miles in 1/4 of hour

11/8 miles   divided by 1/4 hr

11/8 ÷ 1/4

11/8 * 4/1 =44/8 = 5 4/8 miles in 1 hr or 5 1/2 miles per 1 hr

6 0
3 years ago
In some countries, the electricity running to a home oscillates between –240 volts to 240 volts, with a frequency of 90 cycles p
s344n2d4d5 [400]

Answer:

y = 240sin(180πx)

Step-by-step explanation:

This the answer because a is represented by the amplitude, 240, and the frequency is equal to IbI/2pi. to solve for b, multiply the frequency by 2pi.

7 0
3 years ago
Sandra bought 5 bags of candy. She shared it equally among 7 students. How much candy will each student receive?
sweet [91]

Answer:

5/7 of a bag each

Step-by-step explanation:

4 0
3 years ago
Find the average squared distance between the points of r{(x,y): 0x, 0y} and the point (,)
ivolga24 [154]

The average squared distance between the points is their variance

The average squared distance is 8/3

<h3>How to determine the average squared distance?</h3>

The given parameters are:

R={(x,y): 0<=x<=2, 0<=y<=2}

The point = (2,2).

f(x,y) = (x-2)² + (y-2)²

The squared distance is calculated as:

D² = f(x,y) = (x-2)² + (y-2)²

Where the area (A) is:

A = xy

Substitute the maximum values of x and y in A = xy

A = 2 * 2

A = 4

The minimum values of x and y are 0.

So, the limits for the integrals are 0 to 2

The integral becomes

D\² = \int \int (x-2)\² + (y-2)\² dx dy

Expand

D\² = \int \int x\² -4x + 4 + (y-2)\² dx dy

Evaluate the inner integral with respect to x from 0 to 2

D\² = \int \frac 13x^3 -2x^2 + 4x + x(y-2)\² |\limits^2_0 dy

Expand

D\² = \int [\frac 13(2)^3 -2(2)^2 + 4(2) + (2)(y-2)\²] dy

Simplify

D\² = \int \frac 83 + (2)(y-2)\² dy

Expand

D\² = \int \frac 83 + 2y^2-8y + 8 \ dy

Evaluate the integral with respect to y from 0 to 2

D\² = [\frac 83y + \frac 23y^3- 4y^2 + 8y ]|\limits^2_0

Expand

D\² = \frac 83*2 + \frac 23*2^3- 4*2^2 + 8*2

D² = 32/3

The average squared distance (AD²) is calculated as:

AD² = D²/A

So, we have:

AD² = 32/3 \div 4

Evaluate the quotient

AD² = 32/12

Simplify

AD² = 8/3

Hence, the average squared distance is 8/3

Read more about average distance or variance at:

brainly.com/question/15858152

8 0
3 years ago
75 DIVIDE BY 600 EQUALS
ad-work [718]

Answer:

here?

Step-by-step explanation:

75÷600=0.125

if you meant

600÷75=8

4 0
3 years ago
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