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Flauer [41]
2 years ago
12

ANYONE GOOD AT MATH!!!! HELP WITH 3 QUESTIONS

Mathematics
2 answers:
mr_godi [17]2 years ago
3 0
Ok do in the first you have to do a darwin’s if what 87 and i don’t know more
Strike441 [17]2 years ago
3 0

Answer:

1.) a. 1/8 b. 7/8 c. 4/8 d. 3/8 2.) a. 8/40 b. 18/40 c. 14/40 d. 0.05 (5%) 3.) a. 60/400 b. 138/400 c. 26/400 d. 12 e. 100 pages goes with 4 different colors as does 150 and 200 page notebooks, this means 4+4+4=12 and that's the number of different possible combinations

Step-by-step explanation:

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The regular price of a video game is $19.50 The store is having a 40% off sale.
Afina-wow [57]
A. The discount is 7.80 much

B. $11.7
7 0
3 years ago
The product-3 and t is greater than 11
PSYCHO15rus [73]

Answer:

i am pretty sure its 8

Step-by-step explanation:

because a negative plus a positive = wichever one is bighger and 8 is bigger than 3 so 11 is positive. 8 +3 = 11

6 0
3 years ago
Evaluate.<br><br> 1+5⋅32<br><br> A. 8<br><br> b. 9<br><br> c. 16<br><br> d.18
Anna007 [38]

Answer:

the answer should be 161

6 0
3 years ago
Mathematical induction, prove the following two statements are true
adelina 88 [10]
Prove:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+n\left(\frac12\right)^{n-1}=4-\dfrac{n+2}{2^{n-1}}
____________________________________________

Base Step: For n=1:
n\left(\frac12\right)^{n-1}=1\left(\frac12\right)^{0}=1
and
4-\dfrac{n+2}{2^{n-1}}=4-3=1
--------------------------------------------------------------------------

Induction Hypothesis: Assume true for n=k. Meaning:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}=4-\dfrac{k+2}{2^{k-1}}
assumed to be true.

--------------------------------------------------------------------------

Induction Step: For n=k+1:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}

by our Induction Hypothesis, we can replace every term in this summation (except the last term) with the right hand side of our assumption.
=4-\dfrac{k+2}{2^{k-1}}+(k+1)\left(\frac12\right)^{k}

From here, think about what you are trying to end up with.
For n=k+1, we WANT the formula to look like this:
1+2\left(\frac12\right)+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}=4-\dfrac{(k+1)+2}{2^{(k+1)-1}}

That thing on the right hand side is what we're trying to end up with. So we need to do some clever Algebra.

Combine the (k+1) and 1/2, put the 2 in the bottom,
=4-\dfrac{k+2}{2^{k-1}}+\dfrac{(k+1)}{2^{k}}

We want to end up with a 2^k as our final denominator, so our middle term is missing a power of 2. Let's multiply top and bottom by 2,
=4+\dfrac{-2(k+2)}{2^{k}}+\dfrac{(k+1)}{2^{k}}

Distribute the -2 and combine the fractions together,
=4+\dfrac{-2k-4+(k+1)}{2^{k}}

Combine like-terms,
=4+\dfrac{-k-3}{2^{k}}

pull the negative back out,
=4-\dfrac{k+3}{2^{k}}

And ta-da! We've done it!
We can break apart the +3 into +1 and +2,
and the +0 in the bottom can be written as -1 and +1,
=4-\dfrac{(k+1)+2}{2^{(k-1)+1}}
3 0
3 years ago
15 over 1000 wrote in scientific notation​
Katena32 [7]

Answer:

1.5*10^-2

Step-by-step explanation:

You mean 15/1000 written in scientific notation?

In this case, it would be- 1.5*10^-2.

3 0
3 years ago
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