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2 years ago

14

I ran 800m, 5km,1,000cm. How many meters did I run all together

1 answer:

Ugo [173]2 years ago

8 0

Answer:

5810 meters

Step-by-step explanation:

800 meters

5km = 5,000 meters

1,000 cm = 10 meters

Add all together to get 5810 meters

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What is the answer so pwease help

bezimeni [28]

U see I would help but just because u said “pwease” I’m not :)

4 0

2 years ago

Find the distance between parallel lines whose equations are y = x - 6 and y = x + 8.

Sidana [21]

Answer:

14

Step-by-step explanation:

Distance between y2 = x + 8 and y1 = x - 6

y2 - y1 = (x + 8) - (x - 6) = 14

8 0

2 years ago

⎡⎣⎢⎢⎢⎢⎢⎢Scores4567Frequency1581310⎤⎦⎥⎥⎥⎥⎥⎥

V125BC [204]

Answer:

What is this Bro

<h2>Can you explain</h2>

6 0

3 years ago

For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4

Elina [12.6K]

Answer:

<em>C.</em> $(5-\frac{1}{2})^6$

Step-by-step explanation:

Given

$15(5)^2(-\frac{1}{2})^4$

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

$Sum = 2 + 4$

$Sum = 6$

Each term of a binomial expansion are always of the form:

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

Where n = the sum above

$n = 6$

Compare $15(5)^2(-\frac{1}{2})^4$ to the above general form of binomial expansion

$(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......$

Substitute 6 for n

$(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......$

[Next is to solve for a and b]

<em>From the above expression, the power of (5) is 2</em>

<em>Express 2 as 6 - 4</em>

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

By direct comparison of

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

and

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

We have;

$^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4$

Further comparison gives

$^nC_r = 15$

$a^{n-r} =(5)^{6-4}$

$b^r= (-\frac{1}{2})^4$

[Solving for a]

By direct comparison of $a^{n-r} =(5)^{6-4}$

$a = 5$

$n = 6$

$r = 4$

[Solving for b]

By direct comparison of $b^r= (-\frac{1}{2})^4$

$r = 4$

$b = \frac{-1}{2}$

Substitute values for a, b, n and r in

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

$(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

Solve for $^6C_4$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......$

<em>Check the list of options for the expression on the left hand side</em>

<em>The correct answer is </em> $(5-\frac{1}{2})^6$ <em />

3 0

3 years ago

Suppose the odds against winning the lottery as 48,000,000 to 1. What is the probability of the event of winning the lottery giv

kupik [55]

The probability according to the given scenario will be " $\frac{1}{48,000,001}$ ".

According to the question:

The odds against winning the lottery are 48,000,000 to 1.
Favorable outcome = 1

→ The total outcome will be:

= $48,000,000 +1$

= $48,000,001$

As we know,

→ $P(E) = \frac{Favorable \ outcomes}{Total \ outcomes}$

By putting the values, we get

→ $=\frac{1}{48,000,001}$

Thus the above response is right.

Learn more about Probability here:

brainly.com/question/15739211

8 0

2 years ago