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2 years ago

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A basketball team scored 38, 52, 43, 54, 41 and 36 pointsin their first six matches. a. find the mean number b.what score does t

he team need to shoot in their next match to mantain the same mean score? c.the team scores only 20 points in the seventh match. What is mean number of points scored for the seven matches ?
d. if the team scores 42 points in their eighth and final match, will their previous mean score increase or decrease? find the mean score for the eight matches.

1 answer:

BaLLatris [955]2 years ago

5 0

Answer:

a is 44 b is 37

Step-by-step explanation:

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A pile of 42 coins worth $4.90, consisting of nickels and quarters. How many nickels are there?

Ymorist [56]

Answer:14

Step-by-step explanation:

A nickel is 5 cents (20% of dollar)

and a quarter is 25 cents (25% of dollar)

We have given a pile of 42 coins worth of $4.90

Let x be the no nickels and

y be the no of quarter

therefore

x+y=42 -----1

$\frac{x}{4}$ + $\frac{y}{20}$ =4.90 ---2

Solving $\left ( 1\right )&\left ( 2\right )$ we get

x=14 & y=28

Therefore no of nickels is 14 & no of quarters is 28

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2 years ago

What multiplies to -48 and adds to 13

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Answer:

The formula to solve this would be:

X x Y = 13

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Step-by-step explanation:

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Read 2 more answers

A lighthouse operator sights a sailboat at an angle of depression of 12 degrees. If the sailboat is 80m away, how tall is the li

mart [117]

Check out the attached image for the drawing of how this would look. The drawing isn't 100% needed but it helps I think.

Based on the figure shown in the attachment, we have a right triangle with the adjacent leg of 80 (this leg is closest to the bottom 12 degree angle) and an opposite leg of h

In short,
adjacent = 80
opposite = h

We'll use the tangent rule
tan(angle) = opposite/adjacent
tan(12) = h/80
80*tan(12) = h
h = 80*tan(12)
h = 17.0045249336018 <<-- see note below

So the lighthouse is roughly 17.0045249336018 meters tall

Answer: 17.0045249336018

Note: Make sure you are in degree mode. The value is approximate. Round this answer however you need to

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3 years ago

For each of the following vector fields

olga nikolaevna [1]

(A)

$\dfrac{\partial f}{\partial x}=-16x+2y$

$\implies f(x,y)=-8x^2+2xy+g(y)$

$\implies\dfrac{\partial f}{\partial y}=2x+\dfrac{\mathrm dg}{\mathrm dy}=2x+10y$

$\implies\dfrac{\mathrm dg}{\mathrm dy}=10y$

$\implies g(y)=5y^2+C$

$\implies f(x,y)=\boxed{-8x^2+2xy+5y^2+C}$

(B)

$\dfrac{\partial f}{\partial x}=-8y$

$\implies f(x,y)=-8xy+g(y)$

$\implies\dfrac{\partial f}{\partial y}=-8x+\dfrac{\mathrm dg}{\mathrm dy}=-7x$

$\implies \dfrac{\mathrm dg}{\mathrm dy}=x$

But we assume $g(y)$ is a function of $y$ alone, so there is not potential function here.

(C)

$\dfrac{\partial f}{\partial x}=-8\sin y$

$\implies f(x,y)=-8x\sin y+g(x,y)$

$\implies\dfrac{\partial f}{\partial y}=-8x\cos y+\dfrac{\mathrm dg}{\mathrm dy}=4y-8x\cos y$

$\implies\dfrac{\mathrm dg}{\mathrm dy}=4y$

$\implies g(y)=2y^2+C$

$\implies f(x,y)=\boxed{-8x\sin y+2y^2+C}$

For (A) and (C), we have $f(0,0)=0$ , which makes $C=0$ for both.

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2 years ago