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3 years ago

Inclues all numbers between -1 and 4 including -1 but not including 4

Mathematics

1 answer:

zlopas [31]3 years ago

8 0

I’m so sorry but I really don’t know the answer but try the internet it might help

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As part of quality-control program, 3 light bulbs from each bath of 100 are tested. In how many ways can this test batch be chos

hichkok12 [17]

Answer:

<h3>By 161700 ways this test batch can be chosen.</h3>

Step-by-step explanation:

We are given that total number of bulbs are = 100.

Number of bulbs are tested = 3.

Please note, when order it not important, we apply combination.

Choosing 3 bulbs out of 100 don't need any specific order.

Therefore, applying combination formula for choosing 3 bulbs out of 100 bulbs.

$^nCr = \frac{n!}{(n-r)!r!}$ read as r out of n.

Plugging n=100 and r=3 in above formula, we get

$^100C3 = \frac{100!}{(100-3)!3!}$

Expanding 100! upto 97!, we get

= $\frac{100\times 99\times 98\times 97!}{97!3!}$

Crossing out common 97! from top and bottom, we get

= $\frac{100\times 99\times 98}{3!}$

Expanding 3!, we get

= $\frac{100\times 99\times 98}{3\times 2\times 1}$

= 100 × 33 × 49

= 161700 ways.

<h3>Therefore, by 161700 ways this test batch can be chosen.</h3>

3 0

4 years ago

(PLEASE HELP) <br> Which equation has no solution?

Finger [1]

<h2>Answer: </h2><h2>C, 5/8x + 2.5 = 3/8 x +1.5 + 1/4x</h2><h2>~</h2><h3>There are no values of 'X' that make the equation true, therefore the equation has no solution. </h3><h2>~</h2>

Hope you have a good day, Loves!

Also if I'm right can you let me know, please?

<u>Also Also if I am right can I please have Brainliest??</u>

<em>You don't need to, I'm just glad I could help. ^^</em>

8 0

3 years ago

Read 2 more answers

At a certain high school, 35 students play only stringed instruments, 10 students play only brass instruments, and 5 students pl

Sophie [7]

Answer:

9/11

Step-by-step explanation:

Well we know there are 55 students in total and only 45 of them can either play a string or brass instrument.

So 45/55 is 9/11

Thus probably is 9/11 of a randomly selected student who plays either a string or brass.

5 0

2 years ago

while shopping at a clearence sale samantha finds a $60 dress on sale for 25% off. samantha also finds a 50% off coupon.

tatuchka [14]

She would pay 15 dollars for the dress because it would be 75 percent off.

6 0

3 years ago

Please calculate this limit <br>please help me

Tasya [4]

Answer:

We want to find:

$\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n}$

Here we can use Stirling's approximation, which says that for large values of n, we get:

$n! = \sqrt{2*\pi*n} *(\frac{n}{e} )^n$

Because here we are taking the limit when n tends to infinity, we can use this approximation.

Then we get.

$\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n} = \lim_{n \to \infty} \frac{\sqrt[n]{\sqrt{2*\pi*n} *(\frac{n}{e} )^n} }{n} = \lim_{n \to \infty} \frac{n}{e*n} *\sqrt[2*n]{2*\pi*n}$

Now we can just simplify this, so we get:

$\lim_{n \to \infty} \frac{1}{e} *\sqrt[2*n]{2*\pi*n} \\$

And we can rewrite it as:

$\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n}$

The important part here is the exponent, as n tends to infinite, the exponent tends to zero.

Thus:

$\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n} = \frac{1}{e}*1 = \frac{1}{e}$

7 0

3 years ago