All categories

3 years ago

Round 84,978 to the nearest ten thousand

Mathematics

2 answers:

o-na [289]3 years ago

7 0

Hello!

Our ten thousands place is an 8. If we look to the next place value, the thousands, we have 4. We need at least a 5-9 to round up. Therefore, our 8 remains the same, while all of the other numbers become 0's.

80,000 is your answer.

I hope this helps!

kvasek [131]3 years ago

6 0

85,000 is the answer

You might be interested in

Which measure of central tendency is least appropriate for describing the given data set?

beks73 [17]

Answer: Third option is correct.

Step-by-step explanation:

Since we have given that

6, 6, 6, 7, 8, 8, 29

We will calculate the measures of central tendency:

1) Mode - the most occurring element in the data set.

So, Mode = 6

2) Median - the middle value of organised data set

So, Median = 7

3) Mean - the average of all data set

So, Mean = $\dfrac{6+6+6+7+8+8+29}{7}=\dfrac{70}{7}=10$

So, we can see that Mode and Median are giving the closest value to each other, whereas Mean is giving the farthest value as compared to rest.

Hence, Mean is the least appropriate for describing the given data set.

Thus, Third option is correct.

3 0

3 years ago

Can someone help me asap??

Zepler [3.9K]

You need help on what ?

8 0

4 years ago

Read 2 more answers

On Monday it rained two and one-half inches. On Tuesday it rained three and three-fourths inches. How much did it rain in the tw

charle [14.2K]

Answer:

it rained 6 1/4 or 6.25 inches in two days

3 0

3 years ago

What is the probability that a ﬁve-card poker hand contains a ﬂush, that is, ﬁve cards of the same suit?

Liono4ka [1.6K]

$\displaystyle |\Omega|=\binom{52}{5}=\dfrac{52!}{5!47!}=\dfrac{48\cdot49\cdot50\cdot51\cdot52}{2\cdot3\cdot4\cdot5}=2598960\\ |A|=\binom{13}{5}\cdot4=\dfrac{13!}{5!8!}\cdot4=\dfrac{9\cdot10\cdot11\cdot12\cdot13}{2\cdot3\cdot4\cdot5}\cdot4=5148\\\\ P(A)=\dfrac{5148}{2598960}=\dfrac{33}{16660}\approx0.2\%$

7 0

3 years ago

8.<br><br> Find x and why please

lina2011 [118]

Use photomath lol it’s quick with those

6 0

3 years ago