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3 years ago

85001 to nearest ten thousand

2 answers:

5 0

85001 rounded to the nearest ten thousand is 90,000. Because there is a 5 in the thousands place, the 8 in the ten thousands place rounds up to 9. Always use the numeral to the right of the place value you are rounding. If it is 5 or greater the number increases. If it is less than 5, it remains the same.

poizon [28]3 years ago

4 0

First ask yourself what is the ten thousands ? If its 8 then your right . now ask what is on the left of 8 ? On the left of 8 is 5 . If the number 1-4 are on the left then dont change the ten thousands but if it is 5-6 then add +1 to the number to be rounded off . So now the answer is 95001 . But what you are going To do next is your going to make all the numbers on the left of 8 be 0 . But dont mind then numbers on the right of the number to be rounded off . Therefore the answer is 90000.

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3 years ago

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3 years ago

PLEASE HELP ME

denis23 [38]

Answer:

(1) The possible outcomes are: X = {0, 1, 2, 3}.

(2) The number of times should Hartley spin a difference of 1 is 36.

(3) The number of times should Hartley spin a difference of 0 is 24.

Step-by-step explanation:

The number of sections on the spinner is 4 labelled as {1, 2, 3, 4}.

The total number of spins for each of the spinner is, <em>n</em> = 96.

(1)

The sample space of spinning both the spinners together are:

S = {(1, 1), (1, 2), (1, 3), (1, 4)

(2, 1), (2, 2), (2, 3), (2, 4)

(3, 1), (3, 2), (3, 3), (3, 4)

(4, 1), (4, 2), (4, 3), (4, 4)}

Total = 16.

The possible outcomes are:

X = {0, 1, 2, 3}.

(2)

The sample space with the difference 1 are:

S₁ = {(1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3)}

n (S₁) = 6

The probability of the difference 1 is:

$P(\text{Diff}=1)=\frac{n(S_{1})}{N}=\frac{6}{16}=\frac{3}{8}$

The spinners were spinner 96 times.

The expected number of times would Hartley spin a difference of 1 is:

$E(\text{Diff}=1)=P(\text{Diff}=1)\times n\\\\=\frac{3}{8}\times 96\\\\=36$

Thus, the number of times should Hartley spin a difference of 1 is 36.

(3)

The sample space with the difference 0 are:

S₂ = {(1, 1), (2, 2), (3, 3), (4, 4)}

n (S₂) = 4

The probability of the difference 0 is:

$P(\text{Diff}=0)=\frac{n(S_{2})}{N}=\frac{4}{16}=\frac{1}{4}$

The spinners were spinner 96 times.

The expected number of times would Hartley spin a difference of 0 is:

$E(\text{Diff}=0)=P(\text{Diff}=0)\times n\\\\=\frac{1}{4}\times 96\\\\=24$

Thus, the number of times should Hartley spin a difference of 0 is 24.

4 0

3 years ago