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2 years ago

How do I roundoff 763,400 to the nearest ten thousand?

Mathematics

2 answers:

chubhunter [2.5K]2 years ago

5 0

If you were unclear on the answer: 760,400.

Alisiya [41]2 years ago

4 0

Okay, so we first need to locate which number is in the ten thousand place.
Ten thousand, is this 10, 000, right? Well that number has 5 digits.
Now go back to 763,400 and starting from the right, count the first 5 digits.
You should stop at the 6.
the rest is easy! Look to the right of 6, there's a 3. Since 3 is less than 5, you will not round the 6 up to 7.
SO, you keep the 7 and the 6, but turn the rest of the numbers after 6 into 0s.
Your answer should be 760,000!

I hope I made sense!

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MatroZZZ [7]

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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

The confidence interval is reported as follows: Lower 95% CI < p < Upper 95% CI For this question you will be calculating

monitta

Answer:

lower 95% Cl is 0.292

Step-by-step explanation:

p^ = 0.4783

n = 23

for 95% Cl,

z = 1.9600

Inputting all the available data into P as attached below

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2 years ago

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x^(1/7+1/7+1/7+1/7)

x^(4/7)

In words that is x raised to the 4/7 th power or the seventh root of x raised to the 4th power.

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2 years ago

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2 years ago

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