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3 years ago

PLEASE HELP ASAP 25 PTS + BRAINLIEST FOR RIGHT/BEST ANSWER

Mathematics

2 answers:

Gennadij [26K]3 years ago

8 0

It's the first one (-3, 2, -5)

-Dominant- [34]3 years ago

6 0

Answer:

the last one

Step-by-step explanation:

-1,3,4

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18. Determine the common difference, the fifth term, and the sum of the first 100 terms of the following sequence:

Ksenya-84 [330]

$a_1=1,\ a_2=2.5,\ a_3=4,\ a_4=5.5,\ ...\\\\a_2-a_1=2.5-1=1.5\\a_3-a_2=4-2.5=1.5\\a_4-a_3=5.5-4=1.5\\a_{n+1}-a_n=1.5=constans\\\\\text{It's an arithmetic sequence with}\\a_1=1,\ \boxed{d=1.5}\\\\a_n=a_1+(n-1)d\to a_n=1+(n-1)(1.5)=1+1.5n-1.5\\\\a_n=1.5n-0.5\\\\a_5=1.5(5)-0.5=7.5-0.5=7\\\boxed{a_5=7}\\\\\text{The formula of a Sum of the First n Terms of an Arithmetric Sequence:}\\\\S_n=\dfrac{2a_1+(n-1)d}{2}\cdot n\\\\\text{We have:}\\a_1=1,\ d=1.5,\ n=100\\\\\text{Substitute}$

$S_{100}=\dfrac{(2)(1)+(100-1)(1.5)}{2}\cdot100=\dfrac{2+(99)(1.5)}{1}\cdot50\\\\=(2+148.5)\cdot50=150.5\cdot50=7,525\\\\\boxed{S_{100}=7,525}\\\\Answer:\\the\ common\ difference:\ d=1.5\\the\ fifth\ term:\ a_5=7\\the\ sum\ of\ first\ 100\ terms:\ S_{100}=7,525$

8 0

3 years ago

stat: sample of 780 adults in the U.S.A., it was found that 80 of those had a pinworm infestation. You want to find the 99% conf

Anettt [7]

Answer: 554

Step-by-step explanation:

If prior population proportion is known, then the formula to find the sample size is given by :-

$n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2$

As per given description, we have

p= 0.1

E=0.025

Critical z-value for 95% confidence : $z_{\alpha/2}=1.96$

Then,

$n=0.1(1-0.1)(\dfrac{1.96}{0.025})^2=553.1904\approx554$

Hence, the minimum sample size required = 554.

7 0

3 years ago

Ms. Abrams receives a 4% commission plus a salary of $275.00 per week. If her total sales for one week are $2,880.00, how much d

Marta_Voda [28]

390.20 is the answer because 4%of 2880 is 115.20 plus 275.00 is 390.20

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2 years ago

Which shows the expression x^2-1/x^2-x in simplest form

____ [38]

Answer:

$\large\boxed{\dfrac{x+1}{x}=1+\dfrac{1}{x}}$

Step-by-step explanation:

$\dfrac{x^2-1}{x^2-x}=(*)\\\\x^2-1=x^2-1^2\qquad\text{use}\ a^2-b^2=(a-b)(a+b)\\=(x-1)(x+1)\\\\x^2-x=(x)(x)-(x)(1)\qquad\text{use the distributive property}\\=x(x-1)\\\\(*)=\dfrac{(x-1)(x+1)}{x(x-1)}\qquad\text{cancel}\ (x-1)\\\\=\dfrac{x+1}{x}=\dfrac{x}{x}+\dfrac{1}{x}=1+\dfrac{1}{x}$

4 0

3 years ago

Read 2 more answers

I will give you brainliest

vovangra [49]

I believe the answer should be graph J

6 0

3 years ago