2 years ago

PLEASE HELP 10 POINTS

Mathematics

1 answer:

ser-zykov [4K]2 years ago

6 0

The first week, with $16 dollars each

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Helpp please find the value of s! I will give brainlist

goldenfox [79]

We have :

s - 39⁰+ s - 9⁰ = s + 29⁰

s + s - s = 29⁰ + 9⁰ + 39⁰

s = 77⁰

Answer: 77⁰

Ok done. Thank to me :>

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2 years ago

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maria [59]

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levacccp [35]

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3 years ago

In 2003, the population of an African country was about 13.1 million people, which is 2 million more than 3 times the population

lianna [129]

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2 years ago

If the mean weight of 4 backfield members on the football team is 221 lb and the mean weight of the 7 other players is 202 lb, w

MatroZZZ [7]

Let $x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8, x_9, x_{10}, x_{11}$ be the weight of i-th player.

1. If the mean weight of 4 backfield members on the football team is 221 lb, then

$\dfrac{x_1+x_2+x_3+x_4}{4}=221\ lb.$

2. If the mean weight of the 7 other players is 202 lb, then

$\dfrac{x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{7}=202\ lb.$

3. From the previous statements you have that

$x_1+x_2+x_3+x_4=221\cdot 4=884 \lb,\\ \\x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=202\cdot 7=1414\ lb.$

Add these two equalities and then divide by 11:

$x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=884+1414=2298\ lb,\\ \\\dfrac{x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{11}=\dfrac{2298}{11}=208\dfrac{10}{11}\ lb.$

Answer: the mean weight of the 11-person team is $208\dfrac{10}{11}\ lb.$

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2 years ago