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2 years ago

6

Need Help Please I'm Out Of Time ;( The mean age of Lincoln's tee-ball team is 5. Suppose one of his 7-year-old teammates quits

the team, but a 6-year-old joins the team. Is the mean age of Lincoln's team still 5? Explain your reasoning.

1 answer:

Nutka1998 [239]2 years ago

8 0

Answer:

no

Step-by-step explanation:

for eg

(7+3+5)/3 = 5

change the 7

(6+3+5)/3 = 4.666

so it changes

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Katena32 [7]

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2 years ago

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Find the perimeter use 3.14

Alex777 [14]

Hey,
So we have to solve this in multiple steps. Step 1 is to find the circumference of the semi-circle and multiply by three since there are three. Step 2 would be to find the perimeter of the rectangle. Step 3 would be to add those two together.

Step 1: To find the circumference of the semicircle use the formula pi (3.14) times radius (8 ÷ 2 = 4) times 2. After that we will divide by two since there is only half. After that we will multiply that answer by three.
C = 3.14 x 4 x 2
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C = 25.12 ÷ 2
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Perimeter = 12.56 x 3 = 37.68

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Step 3: Now add the two previous answers to get the final perimeter.
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3 years ago

Consider the equation below.

Korvikt [17]

Answer:

Equation in square form:

$y=3(x+5)^2-4$

Extreme value:

$(h,k)=(-5,-4)$

Step-by-step explanation:

We are given

$y=3x^2+30x+71$

we can complete square

$y=3(x^2+10x)+71$

we can use formula

$a^2+2ab+b^2=(a+b)^2$

$y=3(x^2+2\times x\times 5)+71$

now, we can add and subtract 5^2

$y=3(x^2+2\times x\times 5+5^2-5^2)+71$

$y=3(x^2+2\times x\times 5+5^2)-3\times 5^2+71$

$y=3(x+5)^2-75+71$

So, we get equation as

$y=3(x+5)^2-4$

Extreme values:

we know that this parabola

and vertex of parabola always at extreme values

so, we can compare it with

$y=a(x-h)^2+k$

where

vertex=(h,k)

now, we can compare and find h and k

$y=3(x+5)^2-4$

we get

h=-5

k=-4

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$(h,k)=(-5,-4)$

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2 years ago

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Answer:

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